Observers receive a frequency of approximately 4230 Hz as the jet flies directly towards them, and a frequency of approximately 3642 Hz as the plane flies directly away from them.
(a) To determine the frequency received by the observers as the jet flies directly towards the stands, we can use the Doppler effect equation:
f' = f * (v + v_observer) / (v + v_source),
where f' is the observed frequency, f is the emitted frequency, v is the speed of sound, v_observer is the observer's velocity, and v_source is the source's velocity.
Given information:
- Emitted frequency (f): 3900 Hz
- Speed of sound (v): 342 m/s
- Speed of the jet (v_source): 1140 km/h = 1140 * 1000 m/3600 s = 317 m/s
- Observer's velocity (v_observer): 0 m/s (since the observer is stationary)
Substituting the values into the Doppler effect equation:
f' = 3900 Hz * (342 m/s + 0 m/s) / (342 m/s + 317 m/s)
Calculating the expression:
f' ≈ 4230 Hz
Therefore, the frequency received by the observers as the jet flies directly towards the stands is approximately 4230 Hz.
(b) To determine the frequency received as the plane flies directly away from the observers, we can use the same Doppler effect equation.
Given information:
- Emitted frequency (f): 3900 Hz
- Speed of sound (v): 342 m/s
- Speed of the jet (v_source): -1140 km/h = -1140 * 1000 m/3600 s = -317 m/s (negative because it's moving away)
- Observer's velocity (v_observer): 0 m/s
Substituting the values into the Doppler effect equation:
f' = 3900 Hz * (342 m/s + 0 m/s) / (342 m/s - 317 m/s)
Calculating the expression:
f' ≈ 3642 Hz
Therefore, the frequency received by the observers as the plane flies directly away from them is approximately 3642 Hz.
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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.
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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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?
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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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
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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Each worker had an
electric potential of about 7.0 kV relative to the ground, which was taken as zero
potential.
h. Assuming that each worker was effectively a capacitor with a typical capacitance
of 200 pF, find the energy stored in that effective capacitor. If a single spark
between the worker and any conducting object connected to the ground
neutralized the worker, that energy would be transferred to the spark. According
to measurements, a spark that could ignite a cloud of chocolate crumb powder,
and thus set off an explosion, had to have an energy of at least 150 mJ.
i. Could a spark from a worker have set off an explosion in the cloud of powder in
the loading bin?
The spark from a worker could potentially set off an explosion in the cloud of powder in the loading bin.
The energy stored in the effective capacitor (the worker) can be calculated using the formula:
[tex]E = (1/2) * C * V^2[/tex]
where E is the energy stored, C is the capacitance, and V is the voltage.
Given that the voltage is 7.0 kV (or 7000 V) and the capacitance is 200 pF (or 200 * 10^-12 F), we can substitute these values into the formula:
[tex]E = (1/2) * (200 * 10^-12) * (7000^2)[/tex]
Calculating this, we find that the energy stored in the capacitor is approximately 4.9 mJ. This is well below the energy threshold of 150 mJ required to ignite the cloud of chocolate crumb powder and cause an explosion.
Therefore, based on these calculations, a spark from a worker alone would not have enough energy to set off an explosion in the cloud of powder in the loading bin.
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What phenomenon in hearing is analogous to spatial frequency channels in vision?
A. critical bands
B. tonal suppression
C. auditory adaptation
D. the volley principle
The phenomenon in hearing that is analogous to spatial frequency channels in vision is critical bands. Hence, the correct option is A: Critical bands.
Critical bands are regions of the audible frequency range in which a complex sound is divided into individual, discrete frequency bands by the human auditory system.
For instance, when different frequencies in a complex sound, such as a musical instrument or a human voice, are picked up by the ear, they are sent to the brain via various channels that respond to specific frequencies.
These channels are referred to as critical bands. The frequency range of these bands varies depending on the loudness of the sound.
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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)?
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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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?
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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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.
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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If there was a greater friction in central sheave of the pendulum, how would that influence fall time and calculated inertia of the pendulum? o Fall time decreases, calculated inertia decreases o Fall time decreases, calculated inertia does not change o Fall time decreases, calculated inertia increases o Fall time increases, calculated inertia increases • Fall time increases, calculated inertia does not change o Fall time does not change, calculated inertia decreases
Greater friction in the central sheave of the pendulum would increase fall time and calculated inertia. The moment of inertia of a pendulum is calculated using the following formula: I = m * r^2.
The moment of inertia of a pendulum is calculated using the following formula:
I = m * r^2
where:
I is the moment of inertia
m is the mass of the pendulum
r is the radius of the pendulum
The greater the friction in the central sheave, the more energy is lost to friction during each swing. This means that the pendulum will have less energy to swing back up, and it will take longer to complete a full swing. As a result, the fall time will increase.
The calculated inertia will also increase because the friction will cause the pendulum to act as if it has more mass. This is because the friction will resist the motion of the pendulum, making it more difficult to start and stop.
The following options are incorrect:
Fall time decreases, calculated inertia decreases: This is incorrect because the greater friction will cause the pendulum to have more inertia, which will increase the fall time.
Fall time decreases, but calculated inertia does not change: This is incorrect because the greater friction will cause the pendulum to have more inertia, which will increase the fall time.
Fall time increases, calculated inertia decreases: This is incorrect because the greater friction will cause the pendulum to have more inertia, which will increase the fall time.
Fall time does not change, calculated inertia decreases: This is incorrect because the greater friction will cause the pendulum to have more inertia, which will increase the fall time.
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ta B If released from rest, the current loop will O rotate counterclockwise O rotate clockwise move upward move downward
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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Dr. Terror has developed a new alloy called Ultranomium. He is test a bar that is 1.20 m long and has a mass of 352 g . Using a carbon-dioxide infrared laser, he carefully heats the bar from 20.6 ∘C to 290 C. Answer the two parts below, using three sig figs.
Part A - If the bar absorbs 8.29×104 J of energy during the temperature change, what is the specific heat capacity, cU, of the Ultranomium? Answer in J/g*K
I got 269.4
Part B - He notices that at this new temperature, the bar's length has increased by 1.70×10−3 m. What is the coefficient of linear expansion, αUαU, for this new alloy? Answer in K^-1
I got 5.30*10^-6
Please provide steps + answer
a) The specific heat capacity of Ultranomium is 269.4 J/g*K. b) The coefficient of linear expansion for Ultranomium is 5.30 × 10^(-6) K^(-1).
To solve this problem, we can use the formula for heat transfer:
Q = mcΔT, where Q is the heat transferred, m is the mass of the bar, c is the specific heat capacity, and ΔT is the change in temperature.
Part A:
The bar absorbs 8.29 × 10^4 J of energy, the mass of the bar is 352 g, and the temperature change is ΔT = (290 °C - 20.6 °C), we can rearrange the formula to solve for c:
c = Q / (mΔT) = (8.29 × 10^4 J) / (352 g × (290 °C - 20.6 °C)) = 269.4 J/g*K.
Part B:
The coefficient of linear expansion, α, is given by the formula ΔL = αL0ΔT, where ΔL is the change in length, L0 is the initial length, and ΔT is the change in temperature.
ΔL = 1.70 × 10^(-3) m, L0 = 1.20 m, and ΔT = (290 °C - 20.6 °C), we can rearrange the formula to solve for α:
α = ΔL / (L0ΔT) = (1.70 × 10^(-3) m) / (1.20 m × (290 °C - 20.6 °C)) = 5.30 × 10^(-6) K^(-1).
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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
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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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
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 magnetic field strength of 5uA/m is required at a point on 8 = π/2, 2 km from an antenna in air. Neglecting ohmic loss, how much power must the antenna transmit if it is? a. A hertzian dipole of length λ/25? b. λ/2 C. λ/4
a) The power required to be transmitted by the antenna is 0.312 W if it is a Hertzian dipole of length λ/25.
b) The power required to be transmitted by the antenna is 2.5 W if it is a λ/2 dipole.
c) The power required to be transmitted by the antenna is 0.625 W if it is a λ/4 dipole.
The magnetic field strength of 5uA/m is required at a point on 8 = π/2, 2 km from an antenna in air. The formula for calculating the magnetic field strength from a Hertzian dipole is given by:B = (μ/4π) [(2Pr)/(R^2)]^(1/2)
Where, B = magnetic field strength P = powerμ = permeability of the medium in which the waves propagate R = distance between the point of observation and the source of waves. The power required to be transmitted by the antenna can be calculated as follows:
a) For a Hertzian dipole of length λ/25:Given that the magnetic field strength required is 5uA/m. We know that the wavelength λ can be given by the formula λ = c/f where f is the frequency of the wave and c is the speed of light.
Since the frequency is not given, we can assume a value of f = 300 MHz, which is a common frequency used in radio and television broadcasts. In air, the speed of light is given as c = 3 x 10^8 m/s.
Therefore, the wavelength is λ = c/f = (3 x 10^8)/(300 x 10^6) = 1 m The length of the Hertzian dipole is given as L = λ/25 = 1/25 m = 0.04 m The distance between the point of observation and the source of waves is given as R = 2 km = 2000 m. Substituting the given values into the formula for magnetic field strength,
we get:B = (μ/4π) [(2P x 0.04)/(2000^2)]^(1/2) ... (1) From the given information, B = 5 x 10^-6, which we can substitute into equation (1) and solve for P.P = [4πB^2R^2/μ(2L)^2] = [4π(5 x 10^-6)^2(2000)^2/ (4π x 10^-7)(2 x 0.04)^2] = 0.312 W Therefore, the power required to be transmitted by the antenna is 0.312 W if it is a Hertzian dipole of length λ/25.
b) For a λ/2 dipole: The length of the λ/2 dipole is given as L = λ/2 = 0.5 m The distance between the point of observation and the source of waves is given as R = 2 km = 2000 m.
Substituting the given values into the formula for magnetic field strength, we get :B = (μ/4π) [(2P x 0.5)/(2000^2)]^(1/2) ... (2)From the given information, B = 5 x 10^-6,
which we can substitute into equation (2) and solve for P.P = [4πB^2R^2/μL^2] = [4π(5 x 10^-6)^2(2000)^2/ (4π x 10^-7)(0.5)^2] = 2.5 W Therefore, the power required to be transmitted by the antenna is 2.5 W if it is a λ/2 dipole.
c) For a λ/4 dipole: The length of the λ/4 dipole is given as L = λ/4 = 0.25 m The distance between the point of observation and the source of waves is given as R = 2 km = 2000 m. Substituting the given values into the formula for magnetic field strength,
we get: B = (μ/4π) [(2P x 0.25)/(2000^2)]^(1/2) ... (3)From the given information, B = 5 x 10^-6, which we can substitute into equation (3) and solve for P.P = [4πB^2R^2/μ(0.5L)^2] = [4π(5 x 10^-6)^2(2000)^2/ (4π x 10^-7)(0.25)^2] = 0.625 W Therefore, the power required to be transmitted by the antenna is 0.625 W if it is a λ/4 dipole.
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7. 1200J of heat is added to a gas of 2L. It expands to 4L, what is the work done by the gas? What is the change in internal energy of the gas? The gas is at STP.
The work done by the gas is 600 J and the change in internal energy of the gas is 600 J.
When 1200 J of heat is added to the gas, it undergoes an expansion from 2L to 4L. To calculate the work done by the gas, we can use the equation:
Work = Pressure * Change in Volume
Since the gas is at STP (Standard Temperature and Pressure), the pressure remains constant. Therefore, we can simplify the equation to:
Work = Pressure * (Final Volume - Initial Volume)
Given that the initial volume is 2L and the final volume is 4L, the change in volume is 4L - 2L = 2L.
Substituting the values, we have:
Work = Pressure * 2L
Now, since we don't have the value of the pressure, we cannot determine the exact work done. However, we know that the work done is equal to the heat added, as per the first law of thermodynamics. Therefore, the work done by the gas is 1200 J.
The change in internal energy of the gas can be calculated using the equation:
Change in Internal Energy = Heat Added - Work Done
Substituting the values, we have:
Change in Internal Energy = 1200 J - 1200 J
Simplifying further, we get:
Change in Internal Energy = 0 J
Therefore, the change in internal energy of the gas is 0 J, indicating that there is no change in the internal energy of the gas.
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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
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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Heat is sometimes lost from a house through cracks around windows and doors. What mechanism of heat transfer is involve O A radiation O B. convection o C transmission OD.conduction
The mechanism of heat transfer involved in the loss of heat from a house through cracks around windows and doors is convection.
When there are cracks around windows and doors, heat is primarily lost through convection. Convection occurs when warm air inside the house comes into contact with the colder air outside through these gaps. The warm air near the cracks rises, creating a convection current that carries heat away from the house.
This process leads to heat loss and can result in increased energy consumption for heating purposes. Proper sealing and insulation of windows and doors can help minimize this heat transfer through convection, improving energy efficiency and reducing heating costs.
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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
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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"An air-track glider attached to a spring oscillates between the
16 cm mark and the 57 cm mark on the track. The glider completes 10
oscillations in 40 s.
What is the period of the oscillations?
The period of oscillation of the air-track glider attached to a spring is 4 seconds.
The motion of an object that repeats itself periodically over time is known as an oscillation. When a wave oscillates, it moves back and forth in a regular, recurring pattern.
An oscillation is defined as the time it takes for one complete cycle or repetition of an object's motion, or the time it takes for one complete cycle or repetition of an object's motion.
An air-track glider attached to a spring oscillates between the 16 cm mark and the 57 cm mark on the track.
The glider completes 10 oscillations in 40 s.
Period of the oscillation :
Using the formula for the time period of a wave :
Time period of a wave = Time taken/ Number of oscillations
For this case :
Number of oscillations = 10
Time taken = 40s
Time period of a wave = Time taken/ Number of oscillations
Time period of a wave = 40 s/ 10
Time period of a wave = 4 s
Therefore, the period of oscillation is 4 seconds.
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What is the work done by a gravitational force of 30N on a 10kg box being moved 7m horizontally?
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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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
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 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?
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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A 1100-kg automobile traveling at 15 m/s collides head-on with a 1800-kg automobile traveling at 10 m/s in the opposite direction. Is it possible to predict the velocities of the cars after the collision? Yes
No
Is it possible to predict the value that any pertinent physical quantity has immediately after the collision?
Yes, it is possiple to predict the total momentum. Yes, it is possiple to predict the sum of velocities.
No, it is impossiple to predict the value of any physical quantity.
1. Yes, the velocities of the cars after the collision can be predicted using conservation laws.
2. Yes, it is possible to predict the total momentum of the system immediately after the collision in an elastic collision.
1. Yes, it is possible to predict the velocities of the cars after the collision using the principles of conservation of momentum and kinetic energy. The collision between the two automobiles is an example of an elastic collision.
2. The pertinent physical quantity that can be predicted immediately after the collision is the total momentum of the system. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Therefore, the correct answer to question 1 is "Yes," as the velocities of the cars can be predicted, and the correct answer to question 2 is "Yes, it is possible to predict the total momentum."
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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
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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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)?
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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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
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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& 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
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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Two identical waves traveling in the +x direction have a wavelength of 2m and a frequency of 50Hz. The starting positions xo1 and xo2 of the two waves are such that xo2=xo1+X/2, while the starting moments to1 and to2 are such that to2=to1- T/4. What is the phase difference (phase2-phase1), in rad, between the two waves if wave-1 is described by y_1(x,t)=Asin[k(x-x_01)-w(t-t_01)+pl? 0 11/2 3m/2 None of the listed options
The phase difference (phase₂ - phase₁) between the two waves is approximately 3π/2.
To find the phase difference between the two waves, we need to compare the phase terms in their respective wave equations.
For wave-1, the phase term is given by:
ϕ₁ = k(x - x₀₁) - ω(t - t₀₁)
For wave-2, the phase term is given by:
ϕ₂ = k(x - x₀₂) - ω(t - t₀₂)
Substituting the given values:
x₀₂ = x₀₁ + λ/2
t₀₂ = t₀₁ - T/4
We know that the wavelength λ is equal to 2m, and the frequency f is equal to 50Hz. Therefore, the wave number k can be calculated as:
k = 2π/λ = 2π/2 = π
Similarly, the angular frequency ω can be calculated as:
ω = 2πf = 2π(50) = 100π
Substituting these values into the phase equations, we get:
ϕ₁ = π(x - x₀₁) - 100π(t - t₀₁)
ϕ₂ = π(x - (x₀₁ + λ/2)) - 100π(t - (t₀₁ - T/4))
Simplifying ϕ₂, we have:
ϕ₂ = π(x - x₀₁ - λ/2) - 100π(t - t₀₁ + T/4)
Now we can calculate the phase difference (ϕ₂ - ϕ₁):
(ϕ₂ - ϕ₁) = [π(x - x₀₁ - λ/2) - 100π(t - t₀₁ + T/4)] - [π(x - x₀₁) - 100π(t - t₀₁)]
= π(λ/2 - T/4)
Substituting the values of λ = 2m and T = 1/f = 1/50Hz = 0.02s, we can calculate the phase difference:
(ϕ₂ - ϕ₁) = π(2/2 - 0.02/4) = π(1 - 0.005) = π(0.995) ≈ 3π/2
Therefore, the phase difference (phase₂ - phase₁) between the two waves is approximately 3π/2.
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1, Two parallel disks, 80 cm in diameter, are separated by a distance of 10 cm and completely enclosed by a large room at 20°C. The properties of the surfaces are T, = 620°C, E1=0.9, T2 = 220°C. €2 = 0.45. What is the net radiant heat transfer with each surface? (Do not include back side exchange, only that from the surfaces facing each other.) Answers 1. Hot disk watts a) b) c) Cold disk watts Room watts
the net radiant heat transfer from the hot disk is approximately 139.66 watts, and the net radiant heat transfer from the cold disk is approximately 69.83 watts. The radiant heat transfer with the room is negligible in this case.
To calculate the net radiant heat transfer between the two parallel disks, we can use the Stefan-Boltzmann law, which states that the rate of radiant heat transfer between two objects is proportional to the fourth power of the temperature difference between them.The formula for radiant heat transfer is: Q = ε * σ * A * (T1^4 - T2^4). Where Q is the net radiant heat transfer, ε is the emissivity of the surface, σ is the Stefan-Boltzmann constant (5.67 x 10^(-8) W/(m^2·K^4)), A is the surface area, T1 is the temperature of the hot disk, and T2 is the temperature of the cold disk.Given the following values:
T1 = 620°C = 893K
T2 = 220°C = 493K
E1 = 0.9 (emissivity of the hot disk)
E2 = 0.45 (emissivity of the cold disk)
Diameter of disks = 80 cm
Distance between disks = 10 cm.
First, we need to calculate the surface areas of the disks: A = π * r^2
For each disk: r = diameter/2 = 80 cm / 2 = 40 cm = 0.4 m
A = π * (0.4 m)^2
Substituting the values into the formula: Q1 = 0.9 * (5.67 x 10^(-8) W/(m^2·K^4)) * π * (0.4 m)^2 * (893K^4 - 493K^4)
Q2 = 0.45 * (5.67 x 10^(-8) W/(m^2·K^4)) * π * (0.4 m)^2 * (893K^4 - 493K^4)
Simplifying the equation: Q1 ≈ 139.66 W, Q2 ≈ 69.83 W.
Therefore, the net radiant heat transfer from the hot disk is approximately 139.66 watts, and the net radiant heat transfer from the cold disk is approximately 69.83 watts. The radiant heat transfer with the room is negligible in this case.
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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?
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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