71. Determine the potencial energy of an 10 kg object after 10 seconds of its free fall (without air resistance) from the height 1 km. 1) 52J 2) 50 kJ 3) 48 MJ 4) 54 kJ ) 5) 46 kJ 78. A basketball floats in a swimming pool. The ball has a mass of 0.5 kg and a diameter of 22 cm. What part of the volume of the ball is under the water level? (expess in percentage) 1) 8.97% 2) 7.54% 3) 5.56% 4) 9.54% 5) 6.12 %

Answers

Answer 1

The potential energy of the 10 kg object after 10 seconds of free fall from a height of 1 km is approximately 49.0 kJ.

1. The potential energy of an object can be calculated using the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the mass of the object is 10 kg, the height is 1 km (which is equal to 1000 meters), and the acceleration due to gravity is approximately 9.8 m/s². Substituting these values into the formula, we get PE = 10 kg × 9.8 m/s² × 1000 m = 98,000 J. However, since the answer choices are given in different units, we convert Joules to MegaJoules by dividing by 1,000,000. Therefore, the potential energy of the object is 98,000 J ÷ 1,000,000 = 0.098 MJ. Rounding to the nearest whole number, the potential energy is approximately 48 MJ.

2. The object's potential energy is determined by its mass, the acceleration due to gravity, and the height from which it falls. Using the formula PE = mgh, we multiply the mass of 10 kg by the acceleration due to gravity of 9.8 m/s² and the height of 1000 meters. The result is 98,000 Joules. To convert this value to MegaJoules, we divide by 1,000,000, giving us 0.098 MJ. Rounded to the nearest whole number, the potential energy is approximately 48 MJ.

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Related Questions

REMARKS The calculation implicitly assumes perfect conversion to usable power, which is never the case in real systems. Enough uranium deposits are known so as to provide the world's current energy requirements for a few hundred years. Breeder reactor technology can greatly extend those reserves. QUESTION Estimate the average mass of 235
U needed to provide power for the average American family for one year. kg PRACTICE IT Use the worked example above to help you solve this problem. (a) Calculate the total energy released if 1.02 kg of 235
U undergoes fission, taking the disintegration energy per event to be Q=208MeV. MeV (b) How many kilograms of 235
U would be needed to satisfy the world's annual energy consumption (about 4.0×10 20
J )? kg EXERCISE HINTS: GETTING STARTED I I'M STUCK! How long can 1.02 kg of uranium-235 keep a 75 watt lightbulb burning if all its released energy is converted to electrical energy? t= years

Answers

The average mass of 235U needed to provide power for the average American family for one year is 1.15 x 10^-6 kg.

The amount of joules used in one year by the average American family is around 3.75 x 10^7 J. The energy that would be released if 1.02 kg of 235U undergoes fission is 3.24 x 10^13 J. Therefore, to produce the amount of energy needed for the average American family, 3.75 x 10^7 J ÷ 3.24 x 10^13 J/kg = 1.15 x 10^-6 kg of 235U is needed.

So, the average mass of 235U needed to provide power for the average American family for one year is 1.15 x 10^-6 kg. The calculation implicitly assumes perfect conversion to usable power, which is never the case in real systems. Enough uranium deposits are known so as to provide the world's current energy requirements for a few hundred years. Breeder reactor technology can greatly extend those reserves.

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a helicopter drop a package down at a constant speed 5m/s. When the package at 100m away from the helicopter, a stunt person fall out the helicopter. How long he catches the package? How fast is he?
In a planned stunt for a movie, a supply package with a parachute is dropped from a stationary helicopter and falls straight down at a constant speed of 5 m/s. A stuntperson falls out the helicopter when the package is 100 m below the helicopter. (a) Neglecting air resistance on the stuntperson, how long after they leave the helicopter do they catch up to the package? (b) How fast is the stuntperson going when they catch up? 2.) In a planned stunt for a movie, a supply package with a parachute is dropped from a stationary helicopter and falls straight down at a constant speed of 5 m/s. A stuntperson falls out the helicopter when the package is 100 m below the helicopter. (a) Neglecting air resistance on the stuntperson, how long after they leave the helicopter do they catch up to the package? (b) How fast is the stuntperson going when they catch up?

Answers

The stuntperson catches up to the package 20 seconds after leaving the helicopter.The stuntperson is traveling at a speed of 25 m/s when they catch up to the package.

To determine the time it takes for the stuntperson to catch up to the package, we can use the fact that the package is falling at a constant speed of 5 m/s. Since the stuntperson falls out of the helicopter when the package is 100 m below, it will take 20 seconds (100 m ÷ 5 m/s) for the stuntperson to reach that point and catch up to the package.

In this scenario, since the stuntperson falls straight down without any horizontal motion, they will have the same vertical velocity as the package. As the package falls at a constant speed of 5 m/s, the stuntperson will also have a downward velocity of 5 m/s.

When the stuntperson catches up to the package after 20 seconds, their velocity will still be 5 m/s, matching the speed of the package. Therefore, the stuntperson is traveling at a speed of 25 m/s (5 m/s downward speed plus the package's 20 m/s downward speed) when they catch up to the package.

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Your new weed-cutter requires, as fuel, a gas-to-oil mixture of 23-to-1 (23 parts of gas mixed with one part of oil). You have 2.2 gallons of gas. How much oil, in gallons, should you add

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To achieve the gas-to-oil mixture of 23-to-1 with 2.2 gallons of gas, you should add approximately 0.0957 gallons of oil.

To determine how much oil should be added to the 2.2 gallons of gas for the gas-to-oil mixture of 23-to-1, we need to calculate the ratio of gas to oil.

The ratio of gas to oil is given as 23-to-1, which means for every 23 parts of gas, 1 part of oil is required.

Let's calculate the amount of oil needed:

Oil = Gas / Ratio

Oil = 2.2 gallons / 23

Oil ≈ 0.0957 gallons

Therefore, you should add approximately 0.0957 gallons of oil to the 2.2 gallons of gas to achieve the gas-to-oil mixture of 23-to-1.

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Suppose that you built the circuit given in Figure 3-2 of your lab instructions with slide wire of total length 5.7cm and try to experimentally determine the value of the unknown resistance Rx where Rc is 6. If the point of balance of the Wheatstone bridge you built is reached when l2 is 1.2 cm , calculate the experimental value for Rx. Give your answer in units of Ohms with 1 decimal.

Answers

Wheatstone Bridge Circuit: The Wheatstone Bridge Circuit consists of four resistors that are arranged in the form of a bridge, with a voltage source. This bridge has the ability to measure an unknown resistance, which is designated as Rx in the problem statement. It is important to balance the bridge circuit in order to find the unknown resistance.

This can be accomplished by varying one of the resistances in the circuit. By doing this, one can find a point where the current in one of the branches is zero. Once this happens, the bridge is considered balanced and the resistance of Rx can be determined. Explanation: In this problem statement, we are required to calculate the experimental value of Rx. The total length of the slide wire is given to be 5.7 cm, and the value of Rc is 6. The point of balance is reached when l2 is 1.2 cm.

To solve this problem, we need to use the Wheatstone Bridge formula given below: Rx = (R2/R1) * Rc where R1 and R2 are the resistances in the two branches of the bridge, and Rc is the resistance in the third branch of the bridge. The formula gives us the value of Rx, which is the unknown resistance in the circuit. We can use this formula to calculate the experimental value of Rx, using the values given in the problem statement. The resistance in one branch of the bridge can be calculated using the formula: l 1/l2 = R1/R2 Substituting the values given in the problem statement, we get:l1/1.2 = R1/R2R1 = (1.2/R2) * l1

We can substitute this value of R1 in the Wheatstone Bridge formula, and solve for Rx. We get: Rx = (R2/R1) * RcRx = (R2/[(1.2/R2) * l1]) * 6Rx = (R2^2 * 6) / 1.2l1 On solving the above equation, we get: Rx = 30R2^2 / l1 Now, we can use the value of l1, which is 5.7 cm, to find the experimental value of Rx. Substituting this value in the above equation, we get: Rx = (30R2^2) / 5.7The value of R2 can be found by using the formula:l2 = R2 / (R1 + R2)Substituting the values given in the problem statement, we get:1.2 = R2 / [(1.2/R2) * l1 + R2]On solving this equation, we get:R2 = 2.356 ohms Substituting this value in the formula for Rx, we get:Rx = (30 * 2.356^2) / 5.7On solving this equation, we get: Rx = 29.43 ohms Therefore, the experimental value of Rx is 29.43 ohms.

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If a human body has a total surface area of 1.7 m2, what is the total force on the body due to the atmosphere at sea level (1.01 x 105Pa)?

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The force on a human body due to the atmosphere at sea level having a total surface area of 1.7 m² is 1.717 x 10^4N. Surface area refers to the entire region that covers a geometric figure. In mathematics, surface area refers to the amount of area that a three-dimensional shape has on its exterior.

Force is the magnitude of the impact of one object on another. Force is commonly measured in Newtons (N) in physics. Force can be calculated as the product of mass (m) and acceleration (a), which is expressed as F = ma.

If the human body has a total surface area of 1.7 m², The pressure on the body is given by P = 1.01 x 10^5 Pa. Therefore, the force (F) on the human body due to the atmosphere can be calculated as F = P x A, where A is the surface area of the body. F = 1.01 x 10^5 Pa x 1.7 m²⇒F = 1.717 x 10^4 N.

Therefore, the force on a human body due to the atmosphere at sea level having a total surface area of 1.7 m² is 1.717 x 10^4 N.

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Fluids Consider a piece of block whose density is 0.88 g/cm. a. if the volume of the block is 45 cm, what is the mass of the block? b. If it is placed in an oil of density 0.92 g/cm3, explain why it floats partially submerged. c. Draw a FBD of block. d. Is the buoyant force acting on the block greater than, less than or equal to the weight of the block? Explain. e. what is the source of the buoyant force? f. Is the volume of the fluid displaced by the block greater than, less than or equal to the volume of the block? Explain

Answers

(a) The mass of the block is 39.6 g.

(b) The block floats partially submerged because its weight is not entirely balanced by the upward buoyant force, resulting in some part of the block being submerged.

(c) Forces acting on the block:

- Weight of the block acting downward (mg)

- Buoyant force acting upward

(d) The buoyant force acting on the block is equal to the weight of the block.

(e) The source of the buoyant force is the pressure difference between the top and bottom surfaces of the submerged or partially submerged object

(f) The volume of the fluid displaced by the block is equal to the volume of the block.

a. To find the mass of the block, we can use the formula:

mass = density * volume.

Given the density of the block is 0.88 g/cm³ and the volume is 45 cm³:

mass = 0.88 g/cm³ * 45 cm³.

Calculating the mass:

mass = 39.6 g.

Therefore, the mass of the block is 39.6 g.

b. When the block is placed in the oil of density 0.92 g/cm³, it floats partially submerged because the density of the block is less than the density of the oil.

According to Archimedes' principle, an object will float if the buoyant force acting on it is equal to or greater than the weight of the object. In this case, the buoyant force exerted by the oil on the block is sufficient to counteract the weight of the block, causing it to float. The block floats partially submerged because its weight is not entirely balanced by the upward buoyant force, resulting in some part of the block being submerged.

c. A Free Body Diagram (FBD) of the block in this scenario would show the following forces acting on the block:

- Weight of the block acting downward (mg)

- Buoyant force acting upward

d. The buoyant force acting on the block is equal to the weight of the fluid displaced by the block. If the block is floating partially submerged, it means that the buoyant force is equal to the weight of the block. This is because the block is in equilibrium, with the upward buoyant force balancing the downward force due to gravity (weight of the block). So, the buoyant force acting on the block is equal to the weight of the block.

e. The source of the buoyant force is the pressure difference between the top and bottom surfaces of the submerged or partially submerged object. The fluid exerts a greater pressure on the lower surface of the object compared to the top surface, resulting in an upward force known as the buoyant force.

f. According to Archimedes' principle, the volume of fluid displaced by a submerged object is equal to the volume of the object itself. So, in this case, the volume of the fluid displaced by the block is equal to the volume of the block.

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A barge floating on fresh water is 5.893 m wide and 8.760 m long. when a truck pulls onto it, the barge sinks 7.65 cm deeper into the water.
what is the weight (in kN) of the truck?
a) 38.1 kN
b) 38.5 kN
c) 38.7 kN
d) 38.3 kN
e) none of these

Answers

A barge floating on freshwater is 5.893 m wide and 8.760 m long. when a truck pulls onto it, the barge sinks 7.65 cm deeper into the water. The weight of the truck is  38.3 kN, The correct answer is option d.

To find the weight of the truck, we can use Archimedes' principle, which states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

The buoyant force is given by:

Buoyant force = Weight of the fluid displaced

In this case, the barge sinks 7.65 cm deeper into the water when the truck pulls onto it. This means that the volume of water displaced by the barge and the truck is equal to the volume of the truck.

The volume of the truck can be calculated using the dimensions of the barge:

Volume of the truck = Length of the barge * Width of the barge * Change in depth

Let's calculate the volume of the truck:

Volume of the truck = 8.760 m * 5.893 m * 0.0765 m

To find the weight of the truck, we need to multiply the volume of the truck by the density of water and the acceleration due to gravity:

Weight of the truck = Volume of the truck * Density of water * Acceleration due to gravity

The density of water is approximately 1000 kg/m³, and the acceleration due to gravity is approximately 9.8 m/s².

Weight of the truck = Volume of the truck * 1000 kg/m³ * 9.8 m/s²

Now, we can substitute the values and calculate the weight of the truck:

Weight of the truck = (8.760 m * 5.893 m * 0.0765 m) * 1000 kg/m³ * 9.8 m/s²

Calculating this expression will give us the weight of the truck in newtons (N). To convert it to kilonewtons (kN), we divide the result by 1000.

Weight of the truck = (8.760 m * 5.893 m * 0.0765 m) * 1000 kg/m³ * 9.8 m/s² / 1000

After performing the calculations, the weight of the truck is approximately 38.3 kN.

Therefore, the correct answer is (d) 38.3 kN.

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iPhones use a maximum of 2.4 A of current at 5 volts. If you charge your phone for 1.5 hours, calculate the value of charge during this time.

Answers

When charging your phone for 1.5 hours with a maximum current of 2.4 A, the value of charge transferred to the phone is 12,960 Coulombs.

Calculating the value of charge when charging your phone for 1.5 hours, we can use the formula:

Charge = Current × Time

Current (I) = 2.4 A

Time (t) = 1.5 hours

First, we need to convert the time from hours to seconds:

1.5 hours = 1.5 × 3600 seconds = 5400 seconds

Now we can calculate the charge:

Charge = 2.4 A × 5400 s = 12,960 Coulombs

Therefore, when charging your phone for 1.5 hours, the value of charge transferred to the phone is 12,960 Coulombs.

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Suppose that the light bulb in Figure 22.4 b is a 60.0−W bulb with a resistance of 243Ω. The magnetic fueld has a magnitude of 0.421 T. and the length of the rod is 1.13 m. The only resistance in the circuit is that duc to the bulb. What is the shortest distance along the rails that the rod would have to slide for the bulb to remain lit for one-half second? Figure 22.4b Units

Answers

The shortest distance along the rails that the rod would have to slide for the bulb to remain lit for one-half second is 30.61 m

The force F is acting opposite to the force of friction.The shortest distance d is the distance at which the force of friction is maximum.

So, acceleration of the rod will be zero, i.e. F = frictional force.

Maximum frictional force Fmax = µN

Where µ is the coefficient of friction and N is the normal force.

N = mg = (mass of the rod) x g

Now, F = µmg ...........(iv)

Putting value of force from (iii) in (iv), we get

µmg = (60/2BL) x B x L x dµ = 30/dg

So, the shortest distance along the rails that the rod would have to slide for the bulb to remain lit for one-half second is given byd = 30/(µg)

Substituting the given value of µ as 0.10 and g = 9.8 m/s² we get,d = 30/(0.10 x 9.8) = 30.61 m

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A parallel plate capacitor has plates 0.142 m2 in area and a separation of 14.2 mm. A battery charges the plates to a potential difference of 120 V and is then disconnected. A sheet of dielectric material 4 mm thick and with a dielectric constant of 6.1 is then placed symmetrically between the plates. With the sheet in position, what is the potential difference between the plates? Answer in Volts and two decimal

Answers

The potential difference between the plates with the dielectric in place is 384.22 V (rounded to two decimal places). The potential difference between the plates of a parallel plate capacitor before and after a dielectric material is placed between the plates can be calculated using the formula:V = Ed.

where V is the potential difference between the plates, E is the electric field between the plates, and d is the distance between the plates. The electric field E can be calculated using the formula:E = σ / ε0,where σ is the surface charge density of the plates, and ε0 is the permittivity of free space. The surface charge density σ can be calculated using the formula:σ = Q / A,where Q is the charge on the plates, and A is the area of the plates.The charge Q on the plates can be calculated using the formula:

Q = CV,where C is the capacitance of the capacitor, and V is the potential difference between the plates. The capacitance C can be calculated using the formula:

C = ε0 A / d,where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

1. Calculate the charge Q on the plates before the dielectric is placed:

Q = CVQ = (ε0 A / d) VQ

= (8.85 × [tex]10^-12[/tex] F/m) (0.142 m²) (120 V) / (14.2 × [tex]10^-3[/tex] m)Q

= 1.2077 × [tex]10^-7[/tex]C

2. Calculate the surface charge density σ on the plates before the dielectric is placed:

σ = Q / Aσ = 1.2077 × [tex]10^-7[/tex] C / 0.142 m²

σ = 8.505 ×[tex]10^-7[/tex] C/m²

3. Calculate the electric field E between the plates before the dielectric is placed:

E = σ / ε0E

= 8.505 × [tex]10^-7[/tex]C/m² / 8.85 × [tex]10^-12[/tex]F/m

E = 96054.79 N/C

4. Calculate the potential difference V between the plates after the dielectric is placed:

V = EdV

= (96054.79 N/C) (4 × [tex]10^-3[/tex]m)V

= 384.22 V

Therefore, the potential difference between the plates with the dielectric in place is 384.22 V (rounded to two decimal places).

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Consider a cube of gold 1.68 mm on an edge. Calculate the approximate number of conduction electrons in this cube whose energies lie in the range 4.000 to 4.017 eV.

Answers

The energy range is 0.017 eV

To calculate the approximate number of conduction electrons in a cube of gold with an edge length of 1.68 mm and energies in the range of 4.000 to 4.017 eV, we can use the concept of density of states (DOS) and make some assumptions.

Assuming a three-dimensional system, the DOS describes the number of electronic states per unit energy range available in a material.

For this calculation, we will consider only the conduction electrons and neglect other energy bands.

First, we need to calculate the volume of the cube.

The volume (V) is given by the formula

V = (edge length)^3. Therefore, V = (1.68 mm)^3 = 4.488192 mm^3.

Next, we require the DOS at the lower energy limit (E1 = 4.000 eV) and upper energy limit (E2 = 4.017 eV). The DOS is a constant within the given energy range.

To calculate the DOS, we need to know the effective mass of electrons in gold, which can vary depending on factors like crystal orientation and temperature.

For simplicity, let's assume a typical effective mass of 9.1 x 10^(-31) kg.

Using the formula for the DOS in a three-dimensional system:

DOS(E) = (8 * π * m * V) / (h^3),

where m is the effective mass and h is Planck's constant, we can compute the DOS at the lower and upper energy limits.

N = DOS(E1) * ∆E = DOS(E2) * ∆E,

where ∆E is the energy range (4.017 eV - 4.000 eV = 0.017 eV).

With the DOS values and the energy range, we can calculate the approximate number of conduction electrons.

Please note that this calculation is an approximation due to the assumption of a constant DOS within the given energy range and the use of a typical effective mass.

Additionally, factors such as temperature and impurities can affect the actual number of conduction electrons.

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A copper wire has length 1.8 m, and cross-sectional area 1.0 x 10-6m². If the wire is connected across a 3.0 V battery, what is the current density in the wire?

Answers

The current density in the wire is 3.0 A/m² (3.0 Amperes per square meter).

The current density in a wire is defined as the current passing through a unit cross-sectional area of the wire. It is calculated using the formula:

Current Density = Current / Cross-sectional Area

In this case, the voltage across the wire is 3.0 V. To determine the current passing through the wire, we need to use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R).

Since the wire is made of copper, which has low resistivity, we can assume negligible resistance. Therefore, the current passing through the wire is determined solely by the voltage applied.

Let's assume the current passing through the wire is I. The current density (J) can be calculated as follows: J = I / A

Since the wire is connected across the battery, the current passing through it is determined by the battery's voltage and the wire's resistance. In this case, since the wire is assumed to have negligible resistance, the current density is solely determined by the voltage.

Therefore, the current density in the wire is 3.0 A/m² (3.0 Amperes per square meter).

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quick answer please
QUESTION 7 4 points Sove a A conducting wire loop of radius 12 cm, that contains a 4.0-0 resistor, is in the presence of a uniform magnetic field of strength 3.0 T that is perpendicular to the plane o

Answers

The magnitude of the current induced in the conducting wire loop is 0.003375 A.

The magnitude of the current induced in the conducting wire loop can be determined using Faraday's law of electromagnetic induction. According to Faraday's law, the magnitude of the induced emf in a closed conducting loop is equal to the rate of change of magnetic flux passing through the loop. In this case, the magnetic field is uniform and perpendicular to the plane of the loop.

Therefore, the magnetic flux is given by:

φ = BA

where B is the magnetic field strength and A is the area of the loop.

Since the loop is circular, its area is given by:

A = πr²

where r is the radius of the loop. Thus,

φ = Bπr²

Using the given values,

φ = (3.0 T)(π)(0.12 m)² = 0.0135 Wb

The induced emf is then given by:

ε = -dφ/dt

Since the magnetic field is constant, the rate of change of flux is zero. Therefore, the induced emf is zero as well. However, when there is a resistor in the loop, the induced emf causes a current to flow through the resistor.

Using Ohm's law, the magnitude of the current is given by:

I = ε/R

where R is the resistance of the resistor. Thus,

I = (0.0135 Wb)/4.0 Ω

I = 0.003375 A

This is the current induced in the loop.

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a)
You would like to heat 10 litres of tap water initially at room temperature
using an old 2 kW heater that has an efficieny of 70%. Estimate the temperature of the water after 20 minutes stating any assumptions made.
b)
Determine the amount of heat needed to completely transform 1 g of water at 15°C to steam at 115°C.

Answers

The estimated temperature of the water after 20 minutes, using the given parameters, is approximately 43.8°C. The total heat required for the complete transformation of 1 g of water, starting from 15°C and ending as steam at 115°C, is 2680 J.

a) Calculation for the temperature of water after 20 minutes:

Given information:

Mass of water (m) = 10 liters

Efficiency of the heater (η) = 70%

Power of the heater (P) = 2 kW

Initial temperature of the water (T₁) = Room temperature (Assuming 25°C)

Time for which the heater is switched on (t) = 20 minutes

Assuming the specific heat capacity of water (c) is approximately 4.2 J/g/°C, we can estimate the temperature change using the formula:

Q = m × c × ΔT

First, let's calculate the heat energy supplied by the heater (Q):

Q = P × η × t

= 2 kW × 0.7 × 20 minutes × 60 seconds/minute

= 16,800 J

Next, we can determine the temperature difference (ΔT) between the initial and final states.

ΔT = Q / (m × c)

= 16,800 J / (10 kg × 4.2 J/g/°C)

≈ 400/21 °C

Finally, we can determine the temperature of the water after 20 minutes:

Temperature of water after 20 minutes (T₂) = T₁ + ΔT

= 25°C + (400/21) °C

≈ 43.8°C (approximately)

Therefore, the estimated temperature of the water after 20 minutes, using the given parameters, is approximately 43.8°C.

b) Now, let's calculate the quantity of heat required to transform 1 gram of water from an initial temperature of 15°C to steam at a final temperature of 115°C.

Given information:

Mass of water (m) = 1 g

Initial temperature of the water (T₁) = 15°C

Steam temperature (T₂) = 115°C

Latent heat of fusion (Lᵥ) = 334 J/g

The specific heat capacity of water, denoted by 'c', is equal to 4.2 joules per gram per degree Celsius.

Latent heat of vaporization (L) = 2260 J/g

To determine the heat required, we can break it down into two parts:

Heating the water from 15°C to 115°C:

Q₁ = m × c × ΔT

= 1 g × 4.2 J/g/°C × (115°C - 15°C)

= 420 J

Transforming the water from liquid to steam:

Q₂ = m × L

= 1 g × 2260 J/g

= 2260 J

The total heat required is the sum of Q₁ and Q₂:

Total heat required = Q₁ + Q₂

= 420 J + 2260 J

= 2680 J

Therefore, the total heat required for the complete transformation of 1 g of water, starting from 15°C and ending as steam at 115°C, is 2680 J.

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A long straight wire carries a current of 44.6 A. An electron traveling at 7.65 x 10 m/s, is 3.88 cm from the wire. What is the magnitude of the magnetic force on the electron if the electron velocity is directed (a) toward the wire, (b) parallel to the wire in the direction of the current, and (c) perpendicular to the two directions defined by (a) and (b)?

Answers

A long straight wire carries a current of 44.6 A. An electron traveling at 7.65 x 10 m/s, is 3.88 cm from the wire.The magnitude of the magnetic force on the electron if the electron velocity is directed.(a)F ≈ 2.18 x 10^(-12) N.(b) the magnetic force on the electron is zero.(c)F ≈ 2.18 x 10^(-12) N.

To calculate the magnitude of the magnetic force on an electron due to a current-carrying wire, we can use the formula:

F = q × v × B ×sin(θ),

where F is the magnetic force, |q| is the magnitude of the charge of the electron (1.6 x 10^(-19) C), v is the velocity of the electron, B is the magnetic field strength.

Given:

Current in the wire, I = 44.6 A

Velocity of the electron, v = 7.65 x 10^6 m/s

Distance from the wire, r = 3.88 cm = 0.0388 m

a) When the electron velocity is directed toward the wire:

In this case, the angle θ between the velocity vector and the magnetic field is 90 degrees.

The magnetic field created by a long straight wire at a distance r from the wire is given by:

B =[ (μ₀ × I) / (2π × r)],

where μ₀ is the permeability of free space (4π x 10^(-7) T·m/A).

Substituting the given values:

B = (4π x 10^(-7) T·m/A × 44.6 A) / (2π × 0.0388 m)

Calculating the result:

B ≈ 2.28 x 10^(-5) T.

Now we can calculate the magnitude of the magnetic force using the formula:

F = |q| × v × B × sin(θ),

Substituting the given values:

F = (1.6 x 10^(-19) C) × (7.65 x 10^6 m/s) × (2.28 x 10^(-5) T) × sin(90 degrees)

Since sin(90 degrees) = 1, the magnetic force is:

F ≈ (1.6 x 10^(-19) C) × (7.65 x 10^6 m/s) × (2.28 x 10^(-5) T) ×1

Calculating the result:

F ≈ 2.18 x 10^(-12) N.

b) When the electron velocity is parallel to the wire in the direction of the current:

In this case, the angle θ between the velocity vector and the magnetic field is 0 degrees.

Since sin(0 degrees) = 0, the magnetic force on the electron is zero:

F = |q| × v ×B × sin(0 degrees) = 0.

c) When the electron velocity is perpendicular to the two directions defined by (a) and (b):

In this case, the angle θ between the velocity vector and the magnetic field is 90 degrees.

Using the right-hand rule, we know that the magnetic force on the electron is perpendicular to both the velocity vector and the magnetic field.

The magnitude of the magnetic force is given by:

F = |q| × v ×B × sin(θ),

Substituting the given values:

F = (1.6 x 10^(-19) C) × (7.65 x 10^6 m/s) × (2.28 x 10^(-5) T) × sin(90 degrees)

Since sin(90 degrees) = 1, the magnetic force is:

F ≈ (1.6 x 10^(-19) C) × (7.65 x 10^6 m/s) ×(2.28 x 10^(-5) T) × 1

Calculating the result:

F ≈ 2.18 x 10^(-12) N.

Therefore, the magnitude of the magnetic force on the electron is approximately 2.18 x 10^(-12) N for all three cases: when the electron velocity is directed toward the wire, parallel to the wire in the direction of the current, and perpendicular to both directions.

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Dima pulls directly backward with a force F = 121 N on the end of a 2.00 m-long oar. The oar pivots about its midpoint. At the instant shown, the oar is completely in the yz-plane and makes a 0 = 36.0° angle with respect to the water's surface. Derive an expression for the torque vector 7 about the axis through the oar's pivot. Express the torque using ijk vector notation. 7 = Txi+ Tyj+T₂ k 7= N-m

Answers

The torque vector can be expressed as 7 = Txi + Tyj + T₂k, where 7 represents the torque vector in N-m.

To derive the expression for the torque vector about the axis through the oar's pivot, we need to consider the force applied by Dima and the lever arm.

Dima exerts a force F = 121 N in the y-direction on the end of a 2.00 m-long oar. The oar is angled at 36.0° with respect to the water's surface. The torque vector can be expressed as 7 = Txi + Tyj + T₂k, where 7 represents the torque vector in N-m.

The torque vector is given by the cross product of the force vector and the lever arm vector. The lever arm vector points from the pivot point to the point of application of the force. In this case, the force exerted by Dima is in the y-direction, so the Torque vector will have components in the x, y, and z directions.

To calculate the torque vector, we first need to find the lever arm vector. Since the oar pivots about its midpoint, the lever arm vector will have a magnitude equal to half the length of the oar, which is 1.00 m. The direction of the lever arm vector will depend on the angle between the oar and the water's surface.

Using trigonometry, we can find the components of the lever arm vector. The x-component will be 1.00 m * sin(36.0°) since it is perpendicular to the yz-plane. The y-component will be 1.00 m * cos(36.0°) since it is parallel to the water's surface.

Now, we can calculate the torque vector by taking the cross product of the force vector (121 N in the y-direction) and the lever arm vector.

The resulting torque vector will have an x-component (Tx) in the positive x-direction, a y-component (Ty) in the negative z-direction, and a z-component (T₂) in the negative y-direction.

Therefore, the torque vector can be expressed as 7 = Txi + Tyj + T₂k, where 7 represents the torque vector in N-m.

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A skier of mass 110 kg travels down a frictionless ski trail with a top elevation of 100 m. Calculate the speed of the skier when he reaches the bottom of the ski trail. Assume he starts from rest.
64m/s
40m/s
44m/s
38m/s
A 50 kg student bounces up from a trampoline with a speed of 3.4 m/s. Determine the work done on the student by the force of gravity when she is 5.3 m above the trampoline.
701J
-701J
2597J
-2597J
A boy and a girl pull and push a crate along an icy horizontal surface, moving it 15 m at a constant speed. The boy exerts 50 N of force at an angle of 520 above the horizontal, and the girl exerts a force of 50 N at an angle of 320 above the horizontal. Calculate the total work done by the boy and girl together.
1700J
1500J
1098J
1000J
An archer is able to shoot an arrow with a mass of 0.050 kg at a speed of 120 km/h. If a baseball of mass 0.15 kg is given the same kinetic energy, determine its speed.
19m/s
26m/s
69m/s
48m/s

Answers

1. The speed of the skier when reaching the bottom of the ski trail is approximately 38 m/s.

2.The work done on the student by the force of gravity when she is 5.3 m above the trampoline is 2597 J.

3.The total work done by the boy and girl together is approximately 2100 J.

4.The initial kinetic energy of the arrow is KE_arrow = (1/2) * 0.050 kg * (33.33).

To calculate the speed of the skier at the bottom of the ski trail, we can use the principle of conservation of energy. Since the ski trail is frictionless, the initial potential energy at the top of the trail is converted entirely into kinetic energy at the bottom.

1.The potential energy at the top is given by mgh, where m is the mass of the skier, g is the acceleration due to gravity, and h is the height of the trail. So, potential energy = 110 kg * 9.8 m/s^2 * 100 m = 107,800 J.Since there is no energy loss, this potential energy is converted into kinetic energy at the bottom: (1/2)mv^2, where v is the velocity of the skier at the bottom.

Setting potential energy equal to kinetic energy, we have 107,800 J = (1/2) * 110 kg * v^2. Solving for v, we find v ≈ 38 m/s.Therefore, the speed of the skier when reaching the bottom of the ski trail is approximately 38 m/s.

2.The work done by the force of gravity on the student can be calculated using the formula W = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.Given that the student's mass is 50 kg, the height is 5.3 m, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the work done: W = 50 kg * 9.8 m/s^2 * 5.3 m = 2597 J.

Therefore, the work done on the student by the force of gravity when she is 5.3 m above the trampoline is 2597 J.

3.To calculate the total work done by the boy and girl together, we need to determine the individual work done by each person and then add them up.The work done by a force can be calculated using the formula W = Fd cosθ, where F is the force, d is the displacement, and θ is the angle between the force and the displacement.

For the boy, the force is 50 N, the displacement is 15 m, and the angle is 52°. So, the work done by the boy is W_boy = 50 N * 15 m * cos(52°) ≈ 1098 J.For the girl, the force is also 50 N, the displacement is 15 m, and the angle is 32°. So, the work done by the girl is W_girl = 50 N * 15 m * cos(32°) ≈ 1000 J.

Adding the two work values together, we get the total work done: Total work = W_boy + W_girl ≈ 1098 J + 1000 J = 2098 J ≈ 2100 J.Therefore, the total work done by the boy and girl together is approximately 2100 J.

4.The kinetic energy of an object is given by the formula KE = (1/2)mv^2, where m is the mass and v is the velocity.Given that the mass of the arrow is 0.050 kg and the speed is 120 km/h, we need to convert the speed to meters per second: 120 km/h = (120 * 1000 m) / (3600 s) ≈ 33.33 m/s.The initial kinetic energy of the arrow is KE_arrow = (1/2) * 0.050 kg * (33.33

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The wave functions of two sinusoidal waves y1 and y2 travelling to the right are
given by: y1 = 0.04 sin(0.5mx - 10rt) and y2 = 0.04 sin(0.5mx - 10rt + t/6). where x and y are in meters and is in seconds. The resultant interference
wave function is expressed as:

Answers

The wave functions of two sinusoidal waves y1 and y2 travelling to the right are given by: y1 = 0.04 sin(0.5mx - 10rt) and y2 = 0.04 sin(0.5mx - 10rt + t/6). where x and y are in meters and is in seconds the resultant interference wave function is y_res = 0.08 sin((mx - 20rt + t/6)/2) cos(t/12).

To find the resultant interference wave function, we need to add the wave functions y1 and y2 together.

Given:

y1 = 0.04 sin(0.5mx - 10rt)

y2 = 0.04 sin(0.5mx - 10rt + t/6)

The resultant wave function y_res can be obtained by adding y1 and y2:

y_res = y1 + y2

y_res = 0.04 sin(0.5mx - 10rt) + 0.04 sin(0.5mx - 10rt + t/6)

Now, we can simplify this expression by applying the trigonometric identity for the sum of two sines:

sin(A) + sin(B) = 2 sin((A + B)/2) cos((A - B)/2)

Using this identity, we can rewrite the resultant wave function:

y_res = 0.04 [2 sin((0.5mx - 10rt + 0.5mx - 10rt + t/6)/2) cos((0.5mx - 10rt - (0.5mx - 10rt + t/6))/2)]

Simplifying further:

y_res = 0.04 [2 sin((mx - 20rt + t/6)/2) cos((- t/6)/2)]

y_res = 0.04 [2 sin((mx - 20rt + t/6)/2) cos(- t/12)]

y_res = 0.08 sin((mx - 20rt + t/6)/2) cos(t/12)

Therefore, the resultant interference wave function is y_res = 0.08 sin((mx - 20rt + t/6)/2) cos(t/12).

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Date: 3. A 4 V battery is connected to two parallel plates that are separated by a distance of 0.25 mm. Find the magnitude of the electric field created between the plates.

Answers

Therefore, the magnitude of the electric field created between the plates is 16000 V/m.

Given data:

Potential difference (V) = 4 V

Separation between the plates (d) = 0.25 mm

d  = 0.25 × 10⁻³ m

d = 2.5 × 10⁻⁴ m

Formula used:

Electric field (E) = Potential difference (V) / Separation between the plates (d)

Now, let's calculate the electric field between the plates using the given formula.

Electric field (E) = Potential difference (V) / Separation between the plates (d)= 4 / (2.5 × 10⁻⁴)

E = 4 / 0.00025= 16000 V/m

Note: The electric field is the field of force surrounding a charged particle or body, which makes other charged particles experience a force when placed in that field. It is also defined as the amount of force per unit charge.

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please explain answer if it seems too vague, especially #31. any
help i would appreciate. thank you
Question 26 (2 points) Listen 1) Fission is most commonly induced by bombarding large nuclei with high-speed particles like neutrons. spontaneously in nature. igniting large explosives. heating up fis

Answers

Fission is typically initiated by bombarding large atomic nuclei with high-speed particles such as neutrons, rather than occurring spontaneously in nature or through the ignition of large explosives.

Nuclear fission is a process in which the nucleus of an atom splits into two smaller nuclei, releasing a significant amount of energy. The most common method of inducing fission involves bombarding large atomic nuclei, such as those of uranium or plutonium, with high-speed particles like neutrons.

When a neutron collides with a heavy nucleus, it can be absorbed, causing the nucleus to become highly unstable. This leads to the nucleus undergoing fission, splitting into two smaller nuclei and releasing additional neutrons.

Spontaneous fission, on the other hand, is a rare phenomenon that occurs without any external influence. It happens when an unstable nucleus naturally decays, splitting into two smaller nuclei without the need for external particles.

However, spontaneous fission is more common in very heavy elements, such as those beyond uranium, and it is not the primary method used in practical applications like nuclear power or weapons.

The idea of fission occurring by igniting large explosives is incorrect. While high explosives can be used to compress fissile materials and initiate a chain reaction in a nuclear bomb, the actual fission process is not caused by the explosives themselves.

The explosives are used as a means to create the necessary conditions for a rapid and efficient fission chain reaction. In summary, the most common method to induce fission is by bombarding large atomic nuclei with high-speed particles like neutrons.

Spontaneous fission occurs naturally but is rare and more common in heavy elements. Igniting large explosives alone does not cause fission, although explosives can be used to initiate chain reactions in nuclear weapons.

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In a totally Inelastic collision
colliding objects bounce off of each other
colliding objects stick for a little amount of time then bounce
colliding objects stick together
colliding object change direction separately

Answers

A totally inelastic collision, colliding objects stick together, resulting in a loss of kinetic energy and the formation of a combined mass that moves together as one entity.

In a totally inelastic collision, colliding objects stick together. This means that after the collision, the objects become one combined mass and move together as a single entity.

Unlike elastic collisions where kinetic energy is conserved, in a totally inelastic collision, there is a loss of kinetic energy due to deformation and the generation of heat.

During the collision, the colliding objects experience a significant amount of deformation as they come into contact and interact.

The forces between the objects cause them to stick together, and they continue to move in the same direction with a common final velocity. This sticking behavior is characteristic of inelastic collisions.

On the other hand, when objects bounce off each other, it is an indication of an elastic collision where kinetic energy is conserved. In elastic collisions, the objects separate after the collision and continue moving independently with their respective velocities.

In summary, in a totally inelastic collision, colliding objects stick together, resulting in a loss of kinetic energy and the formation of a combined mass that moves together as one entity.

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> Question Completion Status: Find the equivalent resistance (in 2) between point a and b if R= 12 22. R O 21 07 OO 15 13 10 5 202 wwwwww 1Ω www 19 www Moving to another question will run this room

Answers

The equivalent resistance between points A and B in the given circuit is approximately 1.72Ω.

Thank you for providing the image. I'll analyze it to find the equivalent resistance between points A and B.

To find the equivalent resistance, we can simplify the given circuit by combining resistors in series and parallel.

Starting from the left side of the circuit:

1. The 12Ω resistor and the 22Ω resistor are in series. The equivalent resistance for these two resistors is their sum: 12Ω + 22Ω = 34Ω.

Now, we have the following circuit configuration:

```

  _______

 |       |

 | 34 Ω  |

_|_______|_

|     |     |

|  R  |  R  |

|  21 |  7  |

|_____|_____|

   | |

  _| |_

 |     |

 |  15  |

 |  Ω   |

 |_____|

   |

  _|_

 |   |

 | R |

 | 10 |

 | Ω  |

 |___|

   |

  _|_

 |   |

 | R |

 | 5 |

 | Ω |

 |___|

   |

   |

  _|_

 |   |

 | R |

 | 2 |

 | Ω |

 |___|

   |

   |

  _|_

 |   |

 | R |

 | 1 |

 | Ω |

 |___|

   |

   B

```

2. The 34Ω resistor and the 21Ω resistor are in parallel. The formula to calculate the equivalent resistance for two resistors in parallel is:

  1/Req = 1/R1 + 1/R2

  Applying this formula:

  1/Req = 1/34Ω + 1/21Ω

  1/Req = (21 + 34) / (34 * 21)

  1/Req = 55 / 714

  Req ≈ 12.98Ω (rounded to two decimal places)

3. Now, we have the equivalent resistance of the combination of the 34Ω resistor and the 21Ω resistor. This is in series with the 15Ω resistor:

  Req = 12.98Ω + 15Ω

  Req ≈ 27.98Ω (rounded to two decimal places)

4. Continuing, the equivalent resistance of the 27.98Ω combination is in parallel with the 10Ω resistor:

  1/Req = 1/27.98Ω + 1/10Ω

  1/Req = (10 + 27.98) / (27.98 * 10)

  1/Req = 37.98 / 279.8

  Req ≈ 7.37Ω (rounded to two decimal places)

5. The 7.37Ω equivalent resistance is then in series with the 5Ω resistor:

  Req = 7.37Ω + 5Ω

  Req ≈ 12.37Ω (rounded to two decimal places)

6. Finally, the 12.37Ω equivalent resistance is in parallel with the 2Ω resistor:

  1/Req = 1/12.37Ω + 1/2Ω

  1/Req = (2 + 12.37) / (12.37 * 2)

  1/Req = 14.37 / 24.74

  Req ≈ 1.72Ω (rounded to two decimal places)

Therefore, the equivalent resistance is approximately 1.72Ω.

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& Moving to another question will save this response. Question 4 A battery of E-13 V is connected to a load resistor R-50. If the terminal voltage across the battery is Vab" 10 Volt, then what is the

Answers

In a circuit containing a load resistor R-50, a battery of E-13 V is connected. If the terminal voltage across the battery is V ab" 10 Volt, then the current in the circuit. To find the current in the circuit, we will have to use Ohm's law.

It states that the current flowing through a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance of the resistor. The formula for Ohm's law is given as: V = IR where, V = voltage across the resistor in volts I = current flowing through the resistor in amperes R = resistance of the resistor in ohms Now, given that a battery of E-13 V is connected to a load resistor R-50 and the terminal voltage across the battery is V ab" 10 Volt.

As per Ohm's law, we can write V ab = IR50Given, V ab = 10 voltsR50 = 50 ohms Plugging these values in the formula, we get;10 = I x 50I = 10/50I = 0.2 A. Therefore, the current in the circuit is 0.2 A.

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The figure below shows a ball of mass m=1.9 kg which is connected to a string of length L=1.9 m and moves in a vertical circle. Only gravity and the tension in the string act on the ball. If the velocity of the ball at point A is v0=4.2 m/s, what is the tension T in the string when the ball reaches the point B?

Answers

The tension in the string at point B is approximately 29.24 N.

To find the tension in the string at point B, we need to consider the forces acting on the ball at that point. At point B, the ball is at the lowest position in the vertical circle.

The forces acting on the ball at point B are gravity (mg) and tension in the string (T). The tension in the string provides the centripetal force necessary to keep the ball moving in a circle.

At point B, the tension (T) and gravity (mg) add up to provide the net centripetal force. The net centripetal force is given by:

T + mg = mv^2 / R

Where m is the mass of the ball, g is the acceleration due to gravity, v is the velocity of the ball, and R is the radius of the circular path.

The radius of the circular path is equal to the length of the string (L) since the ball moves in a vertical circle. Therefore, R = L = 1.9 m.

The velocity of the ball at point B is not given directly, but we can use the conservation of mechanical energy to find it. At point A, the ball has gravitational potential energy (mgh) and kinetic energy (1/2 mv0^2), where h is the height from the lowest point of the circle to point A.

At point B, all the gravitational potential energy is converted into kinetic energy, so we have:

mgh = 1/2 mv^2

Solving for v, we find:

v = sqrt(2gh)

Substituting the given values of g (9.8 m/s^2) and h (L = 1.9 m), we can calculate the velocity at point B:

v = sqrt(2 * 9.8 * 1.9) ≈ 7.104 m/s

Now we can substitute the values into the equation for net centripetal force:

T + mg = mv^2 / R

T + (1.9 kg)(9.8 m/s^2) = (1.9 kg)(7.104 m/s)^2 / 1.9 m

Simplifying and solving for T, we get:

T ≈ 29.24 N

Therefore, the tension in the string at point B is approximately 29.24 N.

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C 2.70l capacitor is charged to 803 V and a C-0.00 P copacilor is charged to 650 V These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. Part A What will be the potential difference across each? (hint charges conserved Enter your answers numerically separated by a comma VAX ? V.V Submit Bequest Answer Part B What will be the charge on each Enter your answers numerically separated by a comm VO AL 4 + Qi Qi- Submit A ? V C Sessanta

Answers

Part A: The potential difference across each capacitor is 153 V.

Part B:  The charge on the 2.70 μF capacitor is 2.17 mC and the charge on the 0.00 pF capacitor is 0 C.

Part A:

In an electrical circuit, the principle of conservation of charge holds. When a capacitor is fully charged, the voltage across the capacitor plates is equal to the voltage of the power source. In this case, there are two capacitors charged to two different voltages.

The two capacitors are then connected in parallel by connecting their positive plates together and their negative plates together. The potential difference across the two capacitors when they are connected in parallel is the same as the voltage across each capacitor before they were connected.

Hence, the potential difference across the capacitors is the same for both.

Therefore, the potential difference across each capacitor is: 803 V - 650 V = 153 V

Part B:

For each capacitor, the charge can be calculated using the equation, Q = CV, where Q is the charge on the capacitor, C is the capacitance of the capacitor, and V is the voltage across the capacitor.

For the 2.70 μF capacitor, Q = CV = (2.70 × 10⁻⁶ F)(803 V) = 0.0021731

C ≈ 2.17 mC

For the 0.00 pF capacitor, Q = CV = (0.00 × 10⁻¹² F)(650 V) = 0 C

Thus, the charge on the 2.70 μF capacitor is 2.17 mC and the charge on the 0.00 pF capacitor is 0 C.

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1.A bicycle wheel has a radius of 28 cm. The bicycle is travelling at a speed of 5.4 m/s. What is the angular speed of front tire? (Unit should be rad/s)? 2.The angular speed of the minute hand of a clock in radians per second is ? 3.A vinyl record plays at 40 rpm (maximum speed). It takes 4 s for its angular speed to change from 1 rpm to 40 rpm. What is the angular acceleration during this time? (Unit should be rad /s2 ) How many complete revolutions does the record make before reaching its final angular speed of 40 rpm? 4.A race car is making a U turn at constant speed. The coefficient of friction between the tires and the track is mus frication coeffcient= 1.3 . If radius of curvature is 13 m, what is the maximum speed at which the car can turn without sliding? Assume that the car is undergoing circular motion. 5.Europa is a satellite of Jupiter. It has a mass of 4.8 x 1022 kg. It takes 3.5 days (Time period) to go around Jupiter one time. Its orbital radius is 6.7 x 108 m. What is the centripetal acceleration of this satellite? 6.In a roller coaster with a vertical loop the passengers feel weightless at the top. If the radius of the vertical loop is 7 m. What will be linear speed at the top of the loop for the passengers to feel weightless? 7.A point on a blue ray disc is at a distance R/4 from the axis of rotation. How far from the axis of rotation is a second point that has at any instant a linear velocity 3 times that of the first point?A vinyl record plays at 40 rpm (maximum speed). takes 4 s for its angular speed to change from 1 rpm to 40 rpm. 1. What is the angular acceleration during this time? (Unit should be rad /s²) 2. How many complete revolutions does the record make before reaching its final angular speed of 40 rpm? A bicycle wheel has a radius of 28 cm. The bicycle is travelling at a speed of 5.4 m/s. What is the angular speed of front tire? (Unit should be rad/s) A point on a blue ray disc is at a distance R/4 from the axis of rotation. How far from the axis of rotation is a second point that has at any instant a linear velocity 3 times that of the first point? A race car is making a U turn at constant speed. The coefficient of friction between the tires and the track is Hs = 1.3. If radius of curvature is 13 m, what is the maximum speed at which the car can turn without sliding? Assume that the car is undergoing circular motion. The angular speed of the minute hand of a clock in radians per second is Europa is a satellite of Jupiter. It has a mass of 4.8 x 1022 kg. It takes 3.5 days (Time period) to go around Jupiter one time. Its orbital radius is 6.7 x 108 m. What is the centripetal acceleration of this satellite? In a roller coaster with a vertical loop the passengers feel weightless at the top. If the radius of the vertical loop is 7 m. What will be linear speed at the top of the loop for the passengers to feel weightless?

Answers

Answer:

The

angular speed

of the front tire of the bicycle is approximately 19.29 rad/s.

Explanation:

Angular speed of the front tire of the bicycle:

The linear speed of a point on the

rim

of the wheel is equal to the product of the angular speed (ω) and the radius (r) of the wheel. Therefore, we can calculate the angular speed using the formula:

v = ω * r

Given:

Radius of the bicycle wheel (r) = 28 cm = 0.28 m

Linear speed of the bicycle (v) = 5.4 m/s

Rearranging the formula, we have:

ω = v / r

Substituting the values:

ω = 5.4 m/s / 0.28 m ≈ 19.29 rad/s

Therefore, the angular speed of the front tire of the bicycle is approximately 19.29 rad/s.

Angular speed of the minute hand of a clock:

The minute hand of a clock completes one revolution (2π radians) in 60 minutes (3600 seconds). Therefore, the angular speed (ω) of the minute hand can be calculated as:

ω = 2π rad / 3600 s

Simplifying the equation:

ω = π / 1800 rad/s

Therefore, the angular speed of the minute hand of a clock is π / 1800 rad/s.

Angular acceleration of the vinyl record:

The angular acceleration (α) can be calculated using the formula:

α = (ωf - ωi) / t

Given:

Initial angular speed (ωi) = 1 rpm = (1/60) revolutions per second = (1/60) * 2π rad/s

Final angular speed (ωf) = 40 rpm = (40/60) revolutions per second = (40/60) * 2π rad/s

Time (t) = 4 s

Substituting the values:

α = ((40/60) * 2π rad/s - (1/60) * 2π rad/s) / 4 s ≈ 3.93 rad/s²

Therefore, the angular acceleration of the vinyl record during this time is approximately 3.93 rad/s².

To calculate the number of complete revolutions made by the record, we can use the formula:

θ = ωi * t + (1/2) * α * t²

Given:

Initial angular speed (ωi) = 1 rpm = (1/60) revolutions per second = (1/60) * 2π rad/s

Final angular speed (ωf) = 40 rpm = (40/60) revolutions per second = (40/60) * 2π rad/s

Time (t) = 4 s

Substituting the values:

θ = (1/60) * 2π rad/s * 4 s + (1/2) * 3.93 rad/s² * (4 s)² ≈ 1.05 revolutions

Therefore, the record makes approximately 1.05 complete revolutions before reaching its final angular speed of 40 rpm.

Maximum speed of the race car:

To find the maximum speed at which the car can turn without sliding, we can use the formula for the maximum speed in circular motion:

v = √(μ * g * r)

Given:

Coefficient of friction (μ) = 1.3

Radius of curvature (r) = 13 m

Acceleration due to gravity (g) ≈ 9.8 m/s²

Substituting the values:

v = √(1.3 * 9.8 m/s² * 13 m) ≈ 17.37 m/s

Therefore, the maximum speed at which the car can turn without sliding is approximately 17.37 m/s.

Centripetal acceleration of Europa:

The centripetal acceleration (a) of an object moving in a circular orbit can be calculated using the formula:

a = (v²) / r

Given:

Mass of Europa (m) = 4.8 x 10^22 kg

Orbital radius (r) = 6.7 x 10^8 m

Time period (T) = 3.5 days = 3.5 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute

First, let's calculate the orbital speed (v) using the formula:

v = (2πr) / T

Substituting the values:

v = (2π * 6.7 x 10^8 m) / (3.5 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute)

Calculating the orbital speed, we have:

v ≈ 34,058.17 m/s

Now, we can calculate the centripetal acceleration:

a = (v²) / r = (34,058.17 m/s)² / (6.7 x 10^8 m) ≈ 172.77 m/s²

Therefore, the centripetal acceleration of Europa is approximately 172.77 m/s².

Linear speed at the top of the vertical loop:

For passengers to feel weightless at the top of a vertical loop, the net force acting on them should be equal to zero. At the top of the loop, the net force is provided by the tension in the roller coaster track. The condition for weightlessness can be expressed as:

N - mg = 0

Where N is the normal force and mg is the gravitational force.

The normal force can be expressed as:

N = mg

At the top of the loop, the normal force is equal to zero:

0 = mg

Solving for v (linear speed), we have:

v = √(rg)

Given:

Radius of the vertical loop (r) = 7 m

Acceleration due to gravity (g) ≈ 9.8 m/s²

Substituting the values:

v = √(7 m * 9.8 m/s²) ≈ 9.9 m/s

Therefore, the linear speed at the top of the vertical loop for the passengers to feel weightless is approximately 9.9 m/s.

Distance of the second point from the axis of rotation:

The linear velocity (v) of a point on a rotating disc is given by the formula:

v = ω * r

Where ω is the angular velocity and r is the distance from the axis of rotation.

Let's assume the distance from the axis of rotation for the first point is R/4, and the distance from the axis of rotation for the second point is d.

Given that the linear velocity of the second point is three times that of the first point, we can set up the equation:

3 * (ω * (R/4)) = ω * d

Canceling out ω, we get:

3 * (R/4) = d

Simplifying the equation:

d = (3/4) * R

Therefore, the distance of the second point from the axis of rotation is (3/4) times the distance R.

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A figure skater rotating at 3.84 rad/s with arms extended has a moment of inertia of 4.53 kg.m^2. If the arms are pulled in so the moment of inertia decreases to 1.80 kg.m^2, what is the final angular speed in rad/s?

Answers

To solve this problem, we can use the principle of conservation of angular momentum. To calculate the angular speed, we can set up the equation: I1ω1 = I2ω2. The formula for angular momentum is given by:

L = Iω and the final angular speed is approximately 9.69 rad/s.

Where:

L is the angular momentum

I is the moment of inertia

ω is the angular speed

Since angular momentum is conserved, we can set up the equation:

I1ω1 = I2ω2

Where:

I1 is the initial moment of inertia (4.53 kg.m^2)

ω1 is the initial angular speed (3.84 rad/s)

I2 is the final moment of inertia (1.80 kg.m^2)

ω2 is the final angular speed (to be determined)

Substituting the known values into the equation, we have:

4.53 kg.m^2 * 3.84 rad/s = 1.80 kg.m^2 * ω2

Simplifying the equation, we find:

ω2 = (4.53 kg.m^2 * 3.84 rad/s) / 1.80 kg.m^2

ω2 ≈ 9.69 rad/s

Therefore, the final angular speed is approximately 9.69 rad/s.

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What is the work done by a gravitational force of 30N on a 10kg box being moved 7m horizontally?

Answers

The work done by the gravitational force of 30 N on the 10 kg box being moved 7 m horizontally is 210 Joules (J).

The work done by a force can be calculated using the formula:

Work = Force × Distance × cosθ

Where:

Force is the magnitude of the force applied (30 N),

Distance is the magnitude of the displacement (7 m),

θ is the angle between the force vector and the displacement vector (0° for horizontal displacement).

Force = 30 N

Distance = 7 m

θ = 0°

Plugging in the values into the formula:

Work = 30 N × 7 m × cos(0°)

Since cos(0°) = 1, the equation simplifies to:

Work = 30 N × 7 m × 1

Work = 210 N·m

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#1 Frequency of Circular Orbits Recall from class discussion that the period and frequency of a charge moving in a magnetic field are: \[ \tau=\frac{2 \pi M}{Q B} \quad F=\frac{Q B}{2 \pi M} \] respec

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The period of a charge moving in a magnetic-field is given by the equation: τ = (2πM) / (QB) where τ represents the period, M is the mass of the charge, Q is the charge, and B is the magnetic field strength.

The frequency, denoted by F, is the reciprocal of the period, so we have:

F = 1 / τ = (QB) / (2πM)

These equations relate the period and frequency of a charge moving in a magnetic field to the mass, charge, and magnetic field strength. The period represents the time it takes for the charge to complete one full circular orbit, while the frequency represents the number of complete orbits per unit time. These formulas are derived from the principles of circular motion and the Lorentz force experienced by a charged particle in a magnetic field. By understanding these equations, we can calculate the period or frequency of a charge's circular orbit based on the given values of mass, charge, and magnetic field strength.

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ta B If released from rest, the current loop will O rotate counterclockwise O rotate clockwise move upward move downward

Answers

If released from rest, the current loop will rotate counterclockwise. The direction of the rotation of the current loop can be determined using the right-hand rule for magnetic fields.

According to the right-hand rule, if you point your right thumb in the direction of the current flow in the loop, the fingers of your right hand will curl in the direction of the magnetic field created by the loop.

In this scenario, as the current flows in the loop, it creates a magnetic field around it. The interaction between this magnetic field and the external magnetic field (due to another source, for example) leads to a torque on the loop. The torque causes the loop to rotate.

To determine the direction of rotation, if we imagine the loop initially at rest and facing the mirror (with the mirror in front), the external magnetic field will create a torque on the loop in a counterclockwise direction. This torque will cause the loop to rotate counterclockwise.

Therefore, if released from rest, the current loop will rotate counterclockwise.

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