When the empty soda bottle sits out in the sun, the air inside the bottle heats up and expands. However, when you put the cap on lightly and place the bottle in the fridge, the air inside the bottle cools down. As a result, the air contracts, leading to a decrease in volume inside the bottle.
When the bottle is exposed to sunlight, the air inside the bottle absorbs heat energy from the sun. This increase in temperature causes the air molecules to gain kinetic energy and move more vigorously, resulting in an expansion of the air volume. Since the cap is lightly placed on the bottle, it allows some air to escape if the pressure inside the bottle becomes too high.
However, when you place the bottle in the fridge, the surrounding temperature decreases. The air inside the bottle loses heat energy to the colder environment, causing the air molecules to slow down and lose kinetic energy. This decrease in temperature leads to a decrease in the volume of the air inside the bottle, as the air molecules become less energetic and occupy less space.
When the empty soda bottle is exposed to sunlight, the air inside expands due to the increase in temperature. However, when the bottle is placed in the fridge, the air inside contracts as it cools down. The cap on the bottle allows for the release of excess pressure during expansion and prevents the bottle from bursting.
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11. In the case of a transverse wave, energy is transmitted A.
in the direction of particle vibration B. at right angles to
particle vibration C. out of phase with particle vibration D. in
all directi
In the case of a transverse wave, energy is transmitted at right angles to particle vibration.
In a transverse wave, such as a wave on a string or an electromagnetic wave, the particles of the medium oscillate up and down or side to side perpendicular to the direction of wave propagation. As these particles move, they transfer energy to neighboring particles, causing them to vibrate as well.
However, the energy itself is transmitted in a direction that is perpendicular to the oscillations of the particles. This means that while the particles move in a certain direction, the energy travels at right angles to their motion, allowing the wave to propagate through the medium.
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5
kg of liquid sulfer at 200°C is cooled down becoming a solid.
200,000 J were transferred from the sulfer to the environment
during this process. what is the final temp of sulfur?
To determine the final temperature of sulfur after it cools down from 200°C to a solid state, we need to consider the amount of energy transferred and the specific heat capacity of sulfur. Let's calculate the final temperature step by step:
Determine the heat transferred:
The amount of energy transferred from the sulfur to the environment is given as 200,000 J.
Calculate the specific heat capacity:
The specific heat capacity of solid sulfur is approximately 0.74 J/g°C.
Convert the mass of sulfur to grams:
Given that we have 5 kg of sulfur, we convert it to grams by multiplying by 1000. So, we have 5,000 grams of sulfur.
Calculate the heat absorbed by sulfur:
The heat absorbed by sulfur can be calculated using the formula: Q = m × c × ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Rearranging the formula, we have ΔT = Q / (m × c).
Substituting the values, we have: ΔT = 200,000 J / (5,000 g × 0.74 J/g°C).
Calculate the final temperature:
Using the value obtained for ΔT, we can calculate the final temperature by subtracting it from the initial temperature of 200°C.
Final temperature = 200°C - ΔT
By calculating the value of ΔT, we find that it is approximately 54.05°C.
Therefore, the final temperature of sulfur after cooling down and becoming a solid is approximately 200°C - 54.05°C = 145.95°C.
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A charge 0.4 nC is placed at (3,-1,2) m and another charge 6.2 nC is placed at (1,1,-3) m. What is the electric field at (-3,-1,2) m
Please show all notes and have answer as a vector.
The electric field at the point (-3, -1, 2) m is (-9.86 x [tex]10^9[/tex] N/C) in the x-direction, (0 N/C) in the y-direction, and (-13.2 x [tex]10^9[/tex] N/C) in the z-direction.
To calculate the electric field at the given point, we can use the principle of superposition. The electric field at a point is the vector sum of the electric fields created by each individual charge at that point.
First, let's calculate the electric field created by the charge of 0.4 nC at (3, -1, 2) m. We can use Coulomb's law:
E1 = (k × q1) / [tex]r_1^2[/tex]
where E1 is the electric field, k is the electrostatic constant (8.99 x [tex]10^9 Nm^2/C^2[/tex]), q1 is the charge (0.4 nC = 0.4 x [tex]10^{-9}[/tex] C), and r1 is the distance from the charge to the point of interest.
Substituting the values, we get:
E1 = (8.99 x [tex]10^9 Nm^2/C^2[/tex] × 0.4 x[tex]10^{-9}[/tex] C) / [tex]\sqrt{(3 - (-3))^2 + (-1 - (-1))^2 + (2 - 2)^2)^2}[/tex]
= 0 N/C (electric field in the y-direction)
Next, let's calculate the electric field created by the charge of 6.2 nC at (1, 1, -3) m:
E2 = (k × q2) / [tex]r_2^2[/tex]
where E2 is the electric field, q2 is the charge (6.2 nC = 6.2 x [tex]10^{-9}[/tex]C), and r2 is the distance from the charge to the point of interest.
Substituting the values, we get:
E2 = (8.99 x[tex]10^9[/tex] [tex]Nm^2/C^2[/tex] × 6.2 x [tex]10^{-9}[/tex] C) /[tex]\sqrt{(1 - (-3))^2 + (1 - (-1))^2 + (-3 - 2)^2)^2}[/tex]
= -13.2 x [tex]10^9[/tex] N/C (electric field in the z-direction)
Since the electric field obeys the principle of superposition, we can add the individual electric fields to get the total electric field at the given point:
E-total = E1 + E2 = (0 N/C, 0 N/C, -13.2 x [tex]10^9[/tex] N/C) + (0 N/C, 0 N/C, -13.2 x [tex]10^9[/tex] N/C) = (-9.86 x [tex]10^9[/tex]N/C, 0 N/C, -13.2 x [tex]10^9[/tex] N/C).
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4. The flat surface of an unoccupied trampoline is 1.0 m above the ground. When stretched down- wards, the upward spring force of the trampoline may be modeled as a linear restoring force. A 50-kg gymnast rests on a trampoline before beginning a routine. [20 points] a) Draw a free-body diagram for the gymnast and state what you know about the magnitude and/or direction of the net force. [3] b) While she is resting on the trampoline, the surface of the trampoline is 5.0 cm lower than before she got on. Find the effective spring constant k of the trampoline. [5] During the routine the gymnast drops from a height of 1.2 metres vertically onto a trampoline. c) How far above the floor is the surface of the trampoline during the lowest part of her bounce? [10] [Hint: ax2 + bx+c=0 (with a, b, c constants) has solutions x = -6£vb2-4ac .] d) If she continues bouncing up and down on the trampoline without any loss of mechanical energy, is her motion simple harmonic? Justify your answer [2] a 2a
The normal force exerted by the trampoline acts upward with a magnitude equal to the weight of the gymnast (mg) to balance the weight. The net force acting on the gymnast is zero since she is at rest. The effective spring constant of the trampoline is 98,000 N/m.
a) Free-body diagram for the gymnast:
The weight of the gymnast acts downward with a magnitude of mg, where m is the mass of the gymnast and g is the acceleration due to gravity.
The normal force exerted by the trampoline acts upward with a magnitude equal to the weight of the gymnast (mg) to balance the weight.
The net force acting on the gymnast is zero since she is at rest.
b) To find the effective spring constant k of the trampoline, we can use Hooke's Law. When the surface of the trampoline is 5.0 cm lower, the displacement is given by Δy = 0.05 m. The weight of the gymnast is balanced by the upward spring force of the trampoline.
Using Hooke's Law:
mg = kΔy
Substituting the given values:
(50 kg)(9.8 m/s²) = k(0.05 m)
Solving for k:
k = (50 kg)(9.8 m/s²) / 0.05 m = 98,000 N/m
Therefore, the effective spring constant of the trampoline is 98,000 N/m.
c) To find the height above the floor during the lowest part of her bounce, we need to consider the conservation of mechanical energy. At the highest point, the gravitational potential energy is maximum, and at the lowest point, it is converted into elastic potential energy of the trampoline.
Using the conservation of mechanical energy:
mgh = 1/2 kx²
Where h is the initial height (1.2 m), k is the spring constant (98,000 N/m), and x is the displacement from the equilibrium position.
At the lowest part of the bounce, the displacement is equal to the initial displacement (0.05 m), but in the opposite direction.
Substituting the values:
(50 kg)(9.8 m/s²)(1.2 m) = 1/2 (98,000 N/m)(-0.05 m)²
Simplifying and solving for h:
h = -[(50 kg)(9.8 m/s²)(1.2 m)] / [1/2 (98,000 N/m)(0.05 m)²] = 0.24 m
Therefore, the surface of the trampoline is 0.24 m above the floor during the lowest part of her bounce.
d) No, her motion is not simple harmonic because she experiences a change in amplitude as she bounces. In simple harmonic motion, the amplitude remains constant, but in this case, the amplitude decreases due to the dissipation of energy through the bounce.
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Find the wavelength of a 10³ Hz EM wave.
Electromagnetic waves, such as light, radio waves, and X-rays, exhibit wave-like behavior and can be characterized by their frequency and wavelength.
Frequency measures the number of wave cycles passing a given point per second, while wavelength represents the distance between two consecutive points on the wave that are in phase.
The wavelength of an electromagnetic (EM) wave can be calculated using the equation:
wavelength = speed of light / frequency.
Given that the frequency of the EM wave is 10^3 Hz, we can substitute this value into the equation to find the wavelength.
The speed of light in a vacuum is a constant value, approximately 3 x 10^8 meters per second.
By dividing the speed of light by the frequency of the wave, we obtain the wavelength.
Therefore, the wavelength of a 10^3 Hz EM wave can be calculated as follows:
wavelength = (3 x 10^8 m/s) / (10^3 Hz) = 3 x 10^5 meters.
Therefore, the wavelength of a 10^3 Hz EM wave is 3 x 10^5 meters.
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When an object is placed 150 cm in front of a lens, the image is formed
75 cm from the lens and on the opposite side of the lens from the object.
What is the power of this lens?
Group of answer choices
+4 D
+3 D
+5 D
–4 D
–2 D
–3 D
–5 D
+2 D
An object is placed 150 cm in front of a lens, and the image is formed 75 cm from the lens and on the opposite side, The power of this lens is +2 D. The correct option is - +2 D.
To find the power of a lens, we can use the lens formula:
1/f = 1/v - 1/u
where f is the focal length of the lens, v is the image distance, and u is the object distance.
Object distance, u = -150 cm (negative sign indicates that the object is on the opposite side of the lens)
Image distance, v = 75 cm
Substituting these values into the lens formula:
1/f = 1/75 - 1/-150
1/f = 2/150 + 1/150
1/f = 3/150
1/f = 1/50
From the lens formula, we can see that the focal length is 50 cm.
The power of a lens is given by the formula:
P = 1/f
Substituting the focal length, we get:
P = 1 m/50 cm
= 100/50
= 2
Therefore, the power of the lens is +2 D. The correct answer is +2 D.
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D Question 4 6.25 pts A space station shaped like a ring rotates in order to generate an acceleration of 12 m/s2. If the station has a radius of 151 m, what is the speed of the outer edge of the ring
The speed of the outer edge of the ring is approximately 42.62 m/s.
To find the speed of the outer edge of the ring, we can use the formula for centripetal acceleration:
a = v^2 / r
Where:
a is the acceleration (given as 12 m/s^2)v is the velocity (speed) of the outer edge of the ring (what we're trying to find)r is the radius of the ring (given as 151 m)Rearranging the formula, we get:
v = √(a * r)
Substituting the given values:
v = √(12 m/s^2 * 151 m)
v ≈ √(1812 m^2/s^2)
v ≈ 42.62 m/s
Therefore, the speed of the outer edge of the ring is approximately 42.62 m/s.
The complete question should be:
A space station shaped like a ring rotates in order to generate an acceleration of 12 m/s2. If the station has a radius of 151 m, what is the speed of the outer edge of the ring in m/s?
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Three charged particles form a triangle: particle 1 with charge Q₁ = 63.0 nC is at xy coordinates (0,3.00 mm), particle 2 with charge Q₂ is at xy coordinates (0,-3.00 mm), particle 3 with charge Q3 = 15.0 nC is at xy coordinates (4.00, 0 mm). In unit-vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q₂ has the following charges?
a) The resulting expression is 44.9737 times the vector (4.00Ȳₓ - 3.00Ȳᵧ), which represents the electrostatic force on particle 3 due to particle 2 when Q₂ is equal to 69.0 nC.
b) The resulting expression is -1.95635 × 10^-4 times the vector (4.00Ȳₓ - 3.00Ȳᵧ), which represents the electrostatic force on particle 3 due to particle 2 when Q₂ is equal to -69.0 nC.
(a) Q₂ = 69.0 nC:
First, we need to calculate the distance between particle 1 and particle 3:
r₁₃ = √[(x₁ - x₃)² + (y₁ - y₃)²]
= √[(0 - 4.00)² + (3.00 - 0)²]
= √[16.00 + 9.00]
= √25.00
= 5.00 mm = 5.00 × 10^-3 m
Next, we calculate the unit vector pointing from particle 1 to particle 3:
Ȳ₃₁ = (x₃ - x₁)Ȳₓ + (y₃ - y₁)Ȳᵧ
= (4.00 - 0)Ȳₓ + (0 - 3.00)Ȳᵧ
= 4.00Ȳₓ - 3.00Ȳᵧ
Now we can calculate the electrostatic force on particle 3 due to particle 1:
F₃₁ = k * |Q₁| * |Q₂| / r₁₃² * Ȳ₃₁
= (8.99 × 10^9 N m²/C²) * (63.0 × 10^-9 C) * (69.0 × 10^-9 C) / (5.00 × 10^-3 m)² * (4.00Ȳₓ - 3.00Ȳᵧ)
= (8.99 × 10^9) * (63.0 × 10^-9) * (-69.0 × 10^-9) / (5.00 × 10^-3)² * (4.00Ȳₓ - 3.00Ȳᵧ)
= (-4.89087 × 10^-5) * (4.00Ȳₓ - 3.00Ȳᵧ)
= -1.95635 × 10^-4 * (4.00Ȳₓ - 3.00Ȳᵧ)
(b) Q₂ = -69.0 nC:
The calculations for distance (r₁₃) and unit vector (Ȳ₃₁) remain the same as in part (a).
Now we can calculate the electrostatic force on particle 3 due to particle 2:
F₃₂ = k * |Q₁| * |Q₂| / r₁₃² * Ȳ₃₁
= (8.99 × 10^9 N m²/C²) * (63.0 × 10^-9 C) * (-69.0 × 10^-9 C) / (5.00 × 10^-3 m)² * (4.00Ȳₓ - 3.00Ȳᵧ)
= (4.49737 × 10^1) * (4.00Ȳₓ - 3.00Ȳᵧ)
= 44.9737 * (4.00Ȳₓ - 3.00Ȳᵧ)
Please note that in both cases, the magnitudes of the charges Q₁ and Q₂ are the same (69.0 × 10^-9 C), but the sign differs.
These calculations give us the electrostatic forces on particle 3 due to the other two particles (Q₁ and Q₂) in unit-vector notation.
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If a ballon is filled to a volume of 3.00 liters at pressue of 2.5 atm what is the volum?
If a ballon is filled to a volume of 3.00 liters at pressue of 2.5 atm then volume of the balloon is 3.00 liters.
According to the information given, the balloon is filled to a volume of 3.00 liters at a pressure of 2.5 atm. Therefore, the volume of the balloon is already specified as 3.00 liters.
Based on the given information, the volume of the balloon is 3.00 liters. No further calculations or analysis are required as the volume is explicitly provided. Therefore, If a ballon is filled to a volume of 3.00 liters at pressue of 2.5 atm then volume of the balloon is 3.00 liters.
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A 3.00-mF and a 5.00-mF capacitor are connected in series across a 30.0-V battery. A 7.00-mF capacitor is then connected in parallel across the 3.00-mF capacitor. Determine the charge stored by the 7.00-mF capacitor.
The charge stored by the 7.00-mF capacitor is Q = 21.0 µC.
Initially, the 3.00-mF and 5.00-mF capacitors are connected in series, resulting in an equivalent capacitance of C_series = 1 / (1/C1 + 1/C2) = 1 / (1/3.00 × 10^(-3) F + 1/5.00 × 10^(-3) F) = 1 / (0.333 × 10^(-3) F + 0.200 × 10^(-3) F) = 1 / (0.533 × 10^(-3) F) = 1.875 × 10^(-3) F.
The potential-difference across the series combination of capacitors is equal to the battery voltage, which is 30.0 V. Using the formula Q = C × V, where Q is the charge stored, C is the capacitance, and V is the voltage, we can calculate the charge stored by the series combination: Q_series = C_series × V = (1.875 × 10^(-3) F) × (30.0 V) = 0.0562 C. Next, the 7.00-mF capacitor is connected in parallel with the 3.00-mF capacitor. The capacitors in parallel share the same potential difference, which is 30.0 V. The total charge stored by the combination of capacitors remains the same, so the charge stored by the 7.00-mF capacitor is equal to the charge stored by the series combination: Q_7.00mF = Q_series = 0.0562 C. Therefore, the charge stored by the 7.00-mF capacitor is 0.0562 C or 21.0 µC.
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Using your understanding of EMF and internal resistance, what differences would you predict between batteries wired in series versus parallel? Come up with an application where series batteries would be preferred and one where parallel batteries would be preferred.
The differences would you predict between batteries wired in series versus parallel is when batteries are wired in series, their EMFs add up while their internal resistances also add up while parallel batteries have the same voltage but lower overall internal resistance.
EMF is an abbreviation for electromotive force. and EMF is a potential difference that exists between two points in a circuit. EMF is produced by a source such as a battery that converts chemical energy into electrical energy. A battery's EMF is the amount of electrical energy produced per unit of charge when electrons flow from the battery's negative terminal to its positive terminal. The internal resistance of a battery is the resistance to current flow that exists inside the battery's cells.
When current flows through a battery, some of the energy is lost as heat because of this resistance. Using these ideas, when batteries are wired in series, their EMFs add up while their internal resistances also add up, this implies that the total voltage of the battery is the sum of all the individual batteries' voltages, while the internal resistance is the sum of all the internal resistances.
Parallel batteries, on the other hand, have the same voltage but lower overall internal resistance since the resistance of each battery is effectively in parallel. The use of series batteries is preferred in applications where high voltages are required, such as in electronic flash units for photography. Parallel batteries are preferred in applications where high currents are required, such as in power tools that require high torque. So therefore the differences would you predict between batteries wired in series versus parallel is when batteries are wired in series, their EMFs add up while their internal resistances also add up while parallel batteries have the same voltage but lower overall internal resistance.
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5.30. The gravitational force between two masses may be written as Mm Mn F = G f = G p² Calculate the divergence of F. p3
To calculate the divergence of a vector field, in this case, the gravitational force field F, we need to take the dot product of the gradient (∇) operator with the vector field. In Cartesian coordinates, the divergence (∇ ·) of a vector field F = Fx i + Fy j + Fz k can be calculated as follows:
∇ · F = (∂Fx/∂x) + (∂Fy/∂y) + (∂Fz/∂z)
F = Gp²
To calculate the divergence, we need to find the partial derivatives of each component of F with respect to its corresponding coordinate. In this case, p = (px, py, pz), and each component is squared:
Fx = G(px)²
Fy = G(py)²
Fz = G(pz)²
∂Fx/∂x = 2Gpx
∂Fy/∂y = 2Gpy
∂Fz/∂z = 2Gpz
∇ · F = (∂Fx/∂x) + (∂Fy/∂y) + (∂Fz/∂z)
= 2Gpx + 2Gpy + 2Gpz
= 2G(px + py + pz)
Therefore, the divergence of the gravitational force field F is 2G times the sum of the components of the vector p, which is (px + py + pz).
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Monochromatic light of wavelength 574 nm illuminates two parallel narrow slits 7.35μm apart. Calculate the angular deviation of the third-order (for m=3 ) bright fringe (a) in radians and (b) in degrees.
The angular deviation of the third-order bright fringe is approximately 0.078 radians and the angular deviation of the third-order bright fringe is approximately 4.47 degrees.
To calculate the angular deviation of the third-order bright fringe,
we can use the formula for the angular position of the bright fringes in a double-slit interference pattern:
(a) In radians:
θ = λ / d
where θ is the angular deviation,
λ is the wavelength of the light,
and d is the distance between the slits.
Given:
λ = 574 nm = 574 × 10^(-9) m
d = 7.35 μm = 7.35 × 10^(-6) m
Substituting these values into the formula, we get:
θ = (574 × 10^(-9) m) / (7.35 × 10^(-6) m)
≈ 0.078 radians
Therefore, the angular deviation of the third-order bright fringe is approximately 0.078 radians.
(b) To convert this value to degrees, we can use the fact that 1 radian is equal to 180/π degrees:
θ_degrees = θ × (180/π)
≈ 0.078 × (180/π)
≈ 4.47 degrees
Therefore, the angular deviation of the third-order bright fringe is approximately 4.47 degrees.
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A beam of alpha particles (a subatomic particle with mass 6.641×10-27 kg and charge 3.20×10-19 C) is accelerated by a potential difference of 2.00 kV and then enters a region 44.0 cm long with mutually perpendicular magnetic and electric fields (a crossed-field region). If the electric field strength is 3.60×106 V/m what magnetic field strength is required so that the alpha particles are undeflected throught the crossed-field region?
To keep alpha particles undeflected in the crossed-field region, a magnetic field strength of 1.20 T is required.
To ensure that alpha particles remain undeflected in the crossed-field region, the electric force experienced by the particles must be balanced by the magnetic force. The electric force is given by Fe = qE, where q is the charge of an alpha particle and E is the electric field strength.
The magnetic force is given by Fm = qvB, where v is the velocity of the alpha particles and B is the magnetic field strength. Since the particles are undeflected, the electric force must equal the magnetic force
Thus, qE = qvB. Solving for B, we get B = (qE)/(qv). Substituting the given values, B = (3.20×10-19 C * 3.60×106 V/m) / (2.00×103 V * 6.641×10-27 kg) = 1.20 T. Therefore, a magnetic field strength of 1.20 T is required for the alpha particles to be undeflected in the crossed-field region.
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In outer space, a constant force is applied to a 33.6 kg probe initially at rest. The probe moves a distance of 102 m in 14 s. Part A What acceleration does this force produce? Express your answer in meters per second squared. IVE ΑΣΦ SMIC ? a= Submit Request Answer What is the magnitude of the force? Express your answer in newtons. 15. ΑΣΦΑ F = Submit ▾ Part B Request Answer PO ? m/s² N
The acceleration produced by a constant force can be calculated using the following formula:f = maWhere:f = force applied on the objectm = mass of the objecta = acceleration produced by the forceRearranging the formula we have:a = f/mWe have m = 33.6 kgf = maLet's find the
acceleration
a first.
To find acceleration, we use the formulaa = (distance traveled)/(time taken)On substituting the values, we get:a = (102 m)/(14 s) = 7.28 m/s²Substituting the value of a = 7.28 m/s² and m = 33.6 kg in f = ma, we have:f = ma = (33.6 kg) × (7.28 m/s²) = 244.608
Acceleration produced by the force is 7.28 m/s² and the magnitude of the force is 244.608 N.Part BNewton's Second Law of Motion states that the acceleration of an object is
directly proportional
to the force applied on it, and inversely proportional to its mass.
Mathematically
, this can be expressed as:f = maIf a constant force is applied to an object, it would accelerate at a constant rate.
The magnitude of the acceleration produced by the force would depend on the magnitude of the force and the mass of the object.If a larger force is applied on an object, it would produce a larger acceleration, and vice versa.Similarly, if the mass of the object is increased, the acceleration produced by the same force would be lower, and vice versa.
In the given question, a constant force is applied on a 33.6 kg probe initially at rest, and it moves a distance of 102 m in 14 s. From the calculations above, we have found that the acceleration produced by the force is 7.28 m/s² and the
magnitude
of the force is 244.608 N.
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(16 points) II. The electric field of an electromagnetic wave traveling in the +x direction through vacuum obeys the equation Ey = (375 N/C) sin[kx - (2.20 x 10'*rad's)t]. (c = 3.0 x 108 m/s) 1. What is the frequency of the wave? 2. What is the wave number (k) and wavelength of this electromagnetic wave? 3. What is the magnetic field of the wave? Express it using sinusoidal function. 4. All electromagnetic (EM) wave is composed of photons. What's the energy of one photon in this given EM wave.
1) The frequency of the given electromagnetic wave is 3.5 x 10^8 Hz.
2) The wave number (k) of this electromagnetic wave is 2.2 x 10^9 rad/s and the wavelength is 2.85 x 10^-2 m.
3) Magnetic field of the wave is the magnetic field and electric field of an electromagnetic wave are related by the equation B = E/c, where c is the speed of light in vacuum. The magnetic field can be expressed as follows :Bz = B sin(kx - wt + φ) = (1.25 x 10^-6) sin(2.2 x 10^9 t - 2.85 x 10^-2 x).
4) The energy of one photon in this given EM wave is 2.32 x 10^-25 J.
1.Frequency of the wave From the given equation Ey = (375 N/C) sin[kx - (2.20 x 10'*rad's)t], we can observe that it has the form y = A sin(wt + φ) where A = 375 N/C, w = 2πf, k = 2.2 x 10^9 rad/s and φ = 0.
Comparing the equations we can find the frequency as follows: w = 2πf∴ f = w/2π = 2.2 x 10^9 /2π = 3.5 x 10^8 Hz The frequency of the given electromagnetic wave is 3.5 x 10^8 Hz.
2. Wave number (k) and wavelength of this electromagnetic wave From the given equation Ey = (375 N/C) sin[kx - (2.20 x 10'*rad's)t], we can observe that it has the form y = A sin(kx - wt + φ) where A = 375 N/C, w = 2πf, k = 2.2 x 10^9 rad/s and φ = 0.
Comparing the equations we can find the wave number as follows :k = 2.2 x 10^9 rad/sλ = 2π/k = 2π/(2.2 x 10^9 rad/s) = 2.85 x 10^-2 m, The wave number (k) of this electromagnetic wave is 2.2 x 10^9 rad/s and the wavelength is 2.85 x 10^-2 m.
3. Magnetic field of the wave From the theory of electromagnetic waves we know that the magnetic field and electric field of an electromagnetic wave are related by the equation B = E/c, where c is the speed of light in vacuum.
Therefore, we can find the magnetic field of the wave as follows :B = E/c = (375 N/C) / (3 x 10^8 m/s) = 1.25 x 10^-6 T Now, we need to express it using sinusoidal function.
As the wave is traveling in the +x direction, the magnetic field is oriented along the z axis. Hence, the magnetic field can be expressed as follows: Bz = B sin(kx - wt + φ) = (1.25 x 10^-6) sin(2.2 x 10^9 t - 2.85 x 10^-2 x)
4. Energy of one photon in this given EM wave From the theory of electromagnetic waves, we know that the energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency of the wave.
Therefore, we can find the energy of one photon in this given EM wave as follows:E = hf = (6.63 x 10^-34 J s) x (3.5 x 10^8 Hz) = 2.32 x 10^-25 J, The energy of one photon in this given EM wave is 2.32 x 10^-25 J.
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(a) A bullet with a mass of 2.10 g moves at a speed of 1.50 x 10³ m/s. If a tennis ball of mass 57.5 g has the same momentum as the bullet, what is its speed (in m/s)? m/s (b) Which has greater kinetic energy, the ball or the bullet? O Both have the same kinetic energy. The bullet has greater kinetic energy. O The ball has greater kinetic energy. A 7.80-g bullet moving at 540 m/s penetrates a tree trunk to a depth of 6.50 cm. (a) Use work and energy considerations to find the average frictional force that stops the bullet. (Enter the magnitude.) N (b) Assuming the frictional force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving. S A professional golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00480 s. After the collision, the ball leaves the club at a speed of 39.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club? N
(a) A bullet with a mass of 2.10 g moves at a speed of 1.50 x 10³ m/s. If a tennis ball of mass 57.5 g has the same momentum as the bullet, then its speed is 54.79 m/s.
(b) the bullet has a greater kinetic energy than the tennis ball.
(a)The average frictional force that stops the bullet is 223.6 N.
(b) Assuming the frictional force is constant, we can use Newton's second law, F = ma, to find the time it takes for the bullet to come to a stop.
Rearranging
(a) To find the speed of the tennis ball, we can use the conservation of momentum. The momentum of an object is given by the product of its mass and velocity. Since momentum is conserved in a collision, the momentum of the bullet will be equal to the momentum of the tennis ball.
Let's denote the mass of the bullet as m1 (2.10 g) and the speed of the bullet as v1 (1.50 x 10^3 m/s). The mass of the tennis ball is m2 (57.5 g), and we need to find the speed of the tennis ball, denoted as v2.The momentum of the bullet is given by p1 = m1 * v1, and the momentum of the tennis ball is given by p2 = m2 * v2. Since the momenta are equal, we can set up an equation: m1 * v1 = m2 * v2.
Plugging in the values, we have (2.10 g) * (1.50 x 10^3 m/s) = (57.5 g) * v2.
Solving for v2, we find v2 = [(2.10 g) * (1.50 x 10^3 m/s)] / (57.5 g).
Performing the calculation, v2 ≈ 54.79 m/s.
(b) The kinetic energy of an object is given by the formula KE = (1/2) * m * v^2, where m is the mass of the object and v is its velocity.Comparing the kinetic energy of the bullet and the tennis ball, we can calculate the kinetic energy for each using their respective masses and velocities.
For the bullet: KE_bullet = (1/2) * (7.80 g) * (540 m/s)^2. For the tennis ball: KE_tennis_ball = (1/2) * (55.0 g) * (39.0 m/s)^2.Performing the calculations, we find that KE_bullet ≈ 846,540 J and KE_tennis_ball ≈ 48,247 J.Thus, the bullet has a greater kinetic energy than the tennis ball.
(a) To find the average frictional force that stops the bullet, we can use the work-energy principle. The work done by the frictional force is equal to the change in kinetic energy of the bullet.
The initial kinetic energy of the bullet is given by KE_initial = (1/2) * m * v_initial^2, where m is the mass of the bullet and v_initial is its initial velocity. In this case, m = 7.80 g and v_initial = 540 m/s.
The final kinetic energy of the bullet is zero since it comes to a stop. Therefore, the work done by the frictional force is equal to the initial kinetic energy of the bullet.
The work done by the frictional force is given by W = F * d, where F is the average frictional force and d is the distance the bullet penetrates the tree trunk.
Setting W equal to KE_initial, we have F * d = KE_initial.
Rearranging the equation to solve for the average frictional force, we get F = KE_initial / d.
Plugging in the values, F = (0.5 * 7.80 g * (540 m/s)^2) / (6.50 cm).
Converting the units to N and m, F ≈ 223.6 N.
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An elevator shaft is drilled directly through the Earth along its diameter, running from near Buenos Aires to near Shanghai. An elevator car with a physicist inside is dropped through the shaft. Show that the motion of the elevator car is simple harmonic motion and find an expression for the time period of the motion in terms of rho (the density of Earth) and G. From the time period, calculate the shortest time for the physicist to reach the Shanghai end if dropped in the Buenos Aires end at t=0.
For this problem assume that the radius of the Earth is RE=6.37×106 m, that the mass of the Earth is ME=5.972×1024 kg, that the density of the Earth is uniform, and that the Earth is a perfect sphere. (Hint: you will need to have an expression for how g depends on radius r inside the Earth.)
Give your answer to exactly 3 significant figures, in minutes.
G=6.67×10−11 N m2/kg2.
Substituting the given values for G and ρ and performing the calculations, we find the shortest time for the physicist to reach the Shanghai end is approximately 31.2 minutes.
To analyze the motion of the elevator car dropped through the Earth, let's consider the forces acting on it. There are two forces to consider: the gravitational force pulling the car towards the Earth's center and the centrifugal force pushing the car outwards due to the rotation of the Earth.
1. Gravitational Force:
The gravitational force acting on the elevator car can be calculated using Newton's law of gravitation:
F_gravity = G * (m_car * M_Earth) / r^2,
where G is the gravitational constant (6.67×10^−11 N m^2/kg^2), m_car is the mass of the elevator car, M_Earth is the mass of the Earth (5.972×10^24 kg), and r is the distance between the car and the center of the Earth.
2. Centrifugal Force:
The centrifugal force is given by:
F_centrifugal = m_car * ω^2 * r,
where ω is the angular velocity of the Earth's rotation. The angular velocity ω can be calculated as:
ω = 2π / T,
where T is the time period of one complete revolution of the Earth (24 hours or 86400 seconds).
For simple harmonic motion, the net force acting on the elevator car must be proportional to the displacement from the equilibrium position. Therefore, the gravitational force and the centrifugal force must be equal and opposite:
F_gravity = F_centrifugal.
Substituting the equations for the forces, we have:
G * (m_car * M_Earth) / r^2 = m_car * ω^2 * r.
Simplifying the equation, we find:
G * M_Earth / r^2 = ω^2 * r.
Substituting ω = 2π / T, we get:
G * M_Earth / r^2 = (2π / T)^2 * r.
Solving for T, we have:
T^2 = (4π^2 * r^3) / (G * M_Earth).
Now, we need to express r in terms of the density of the Earth (ρ). The volume of a sphere is given by V = (4/3)πr^3, and the mass of the Earth is M_Earth = ρ * V, where ρ is the density of the Earth. Substituting these expressions, we have:
M_Earth = ρ * (4/3)πr^3.
Substituting M_Earth in the equation for T^2, we get:
T^2 = (4π^2 * r^3) / (G * ρ * (4/3)πr^3).
Canceling out common terms, we find:
T^2 = (3π / (G * ρ)).
Finally, solving for T, we have:
T = √((3π / (G * ρ))).
To calculate the shortest time for the physicist to reach the Shanghai end, we divide the time period T by 2 (since the time period represents a complete round trip):
Shortest time = T / 2.
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5. (6 pts) An 0.08 8 piece of space debris travelling at 7500 m/s hits the side of a space station and is brought to a stop A3 cm deep crater is left in the side of the space station from the impact. What was net the force of the impact of the piece of space debris?
The net force of the impact of the space debris can be calculated using the concept of impulse. Given the mass of the debris, the initial velocity, and the depth of the crater, we can determine the force exerted during the impact.
To calculate the net force of the impact, we can use the equation for impulse: Impulse = Change in momentum. The change in momentum is equal to the mass of the debris multiplied by the change in velocity (since the debris comes to a stop).
The force can be found by dividing the impulse by the time it takes for the impact to occur. Since the time is not provided, we can assume that the impact occurs over a very short duration, allowing us to consider it an instantaneous collision.
Therefore, the force of the impact is determined by the impulse, which can be calculated using the given mass, initial velocity, and depth of the crater.
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A hypothetical charge -0.2pc with a mass 65fg moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 74mT and is directed into the page. If the speed of the hypothetical charge is 54km/s/ A. Determine the radius of the circular path. B. Determine the time interval required to complete one revolution.
A hypothetical charge with a charge of -0.2pc and a mass of 65fg is moving in a circular path perpendicular to a uniform magnetic field with a magnitude of 74mT.
The speed of the charge is given as 54km/s. To determine the radius of the circular path, we can use the equation for the centripetal force in a magnetic field. To find the time interval required to complete one revolution, we can use the relationship between the speed, radius, and time.
(a) To determine the radius of the circular path, we can use the equation for the centripetal force in a magnetic field. The centripetal force (F) is given by F = qvB, where q is the charge, v is the velocity, and B is the magnetic field strength.
In this case, the charge is -0.2pc, the velocity is 54km/s, and the magnetic field strength is 74mT.
By rearranging the formula to solve for the radius (r), we get r = mv/(qB), where m is the mass of the charge. Plugging in the given values, we can calculate the radius.
(b) To determine the time interval required to complete one revolution, we can use the relationship between the speed, radius, and time.
The formula for the time required for one revolution is T = 2πr/v, where T is the time, r is the radius, and v is the velocity.
By substituting the calculated radius and the given velocity, we can find the time interval required to complete one revolution.
By following these calculations, we can determine the radius of the circular path and the time interval required to complete one revolution for the hypothetical charge.
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A car manufacturer claims that its product, starting from rest, will travel 0.4 km in 10 s. What is the magnitude of the constant acceleration (m/s2) required for this? Give your answer to one decimal place.
The car manufacturer claims that their product can travel 0.4 km in 10 seconds, starting from rest. we can use the kinematic equation. we find that the magnitude of the constant acceleration needed is 8 m/s².
The magnitude of the constant acceleration required for the car to travel 0.4 km in 10 seconds can be calculated using the kinematic equation:
[tex]\(d = \frac{1}{2}at^2\),[/tex]
where d is the distance traveled, a is the acceleration, and t is the time taken.
Given that d = 0.4km = 0.4 * 1000 m = 400 m and t = 10 s, we can rearrange the equation to solve for a:
[tex]\(a = \frac{2d}{t^2}\).[/tex]
Substituting the values, we have:
[tex]\(a = \frac{2 \times 400}{10^2} = \frac{800}{100} = 8\) m/s^{2}[/tex]
Therefore, the magnitude of the constant acceleration required for the car to travel 0.4 km in 10 seconds is 8 m/s².
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Man has the capability of changing the half life of a radioactive material. True or False?
Which type of force is very short range? electric force magnetic force strong force gravitational force
Correct answer : True
The term
“radioactive decay”
refers to the process in which an unstable atomic nucleus loses energy by emitting radiation such as alpha particles, beta particles, gamma rays, or positrons.
These particles are released until the nucleus becomes stable again by releasing energy.
The decay rate of an unstable substance is measured by the half-life, which is the time it takes for half of the original substance to decay.The half-life of a
radioactive element
cannot be altered. However, the rate at which radioactive decay occurs can be influenced by a variety of factors, including external conditions and the use of certain procedures and techniques to treat the radioactive element. This is accomplished by modifying the atoms of the substance or through manipulation of its physical surroundings.
In the case of
short-range forces
, the strong force is the one that is primarily involved. The strong force holds atomic nuclei together by binding the protons and neutrons within the nucleus. The range of the strong force is restricted to only a few femtometers, which is a very short distance. When two nucleons are near each other, this force is quite strong, but it rapidly weakens as the distance between the particles grows.In summary, Man does have the ability to influence the rate of radioactive decay but not the half-life of a radioactive element. The strong force is a type of force that has a very short range.
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A circular loop of copper wire is placed next to a long, straight wire. The current / in the long, straight wire is increasing. What current does this induce in the circular loop? A. a clockwise current B. a counterclockwise current C. zero current D. either A or B E. any of A, B, or C
A counterclockwise current will be induced in the circular loop of copper wire.
The current in the long, straight wire creates a magnetic field around it. As the current increases, the magnetic field also increases. The changing magnetic field induces an electric field in the circular loop of copper wire. This electric field causes a current to flow in the loop, and the direction of the current is such that it creates a magnetic field that opposes the change in the magnetic field from the long, straight wire. This is known as Lenz's law.
In this case, the current in the long, straight wire is increasing, so the magnetic field is also increasing. The induced current in the circular loop of copper wire will flow in the counterclockwise direction, because this creates a magnetic field that opposes the increasing magnetic field from the long, straight wire.
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Problem#15(Please Show Work 20 Points) What is the peak emf generated by a 0.250 m radius, 500-turn coil that is rotated one-fourth of a revolution in 5.17 ms, originally having its plane perpendicular to a uniform magnetic field? Problem# 16 (Please Show Work 10 points) Verify that the units of AD/A are volts. That is, show that 1T·m²/s=1V_
The peak emf generated by the rotated coil is zero. The units of AD/A are volts (V).
Problem #15:
The peak emf generated by the rotated coil is zero since the magnetic flux through the coil remains constant during rotation.
Problem #16:
We are asked to verify that the units of AD/A are volts.
The unit for magnetic field strength (B) is Tesla (T), and the unit for magnetic flux (Φ) is Weber (Wb).
The unit for magnetic field strength times area (B * A) is T * m².
The unit for time (t) is seconds (s).
To calculate the units of AD/A, we multiply the units of B * A by the units of t⁻¹ (inverse of time).
Therefore, the units of AD/A are (T * m²) * s⁻¹.
Now, we know that 1 Wb = 1 V * s (Volts times seconds).
Therefore, (T * m²) * s⁻¹ = (V * s) * s⁻¹ = V.
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An elevator cabin has a mass of 363.7 kg, and the combined mass of the people inside the cabin is 177.0 kg. The cabin is pulled upward by a cable, in which there is a tension force of 7638 N. What is the acceleration of the elevator?
The acceleration of the elevator is approximately 14.12 m/s².
The mass of an elevator cabin and people inside the cabin is 363.7 + 177.0 = 540.7 kg.
The tension force is 7638 N.
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Fnet = ma
Where:
Fnet = net force acting on the object
m = mass of the object
a = acceleration of the object
Rearranging this equation gives us:
a = Fnet / m
Substituting the given values gives us:
a = 7638 N / 540.7 kg
a ≈ 14.12 m/s²
Therefore, the acceleration of the elevator is approximately 14.12 m/s².
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You have an inclined surface whose angle of inclination is 30°, if you have a coefficient of kinetic friction of 0.2. What will be the acceleration of an object on this surface?
The acceleration of the object on the inclined surface with an angle of inclination of 30° and a coefficient of kinetic friction of 0.2 is approximately 0.75 m/s^2.
To calculate the acceleration of an object on an inclined surface, we can use the following equation:
a = g * sin(theta) - mu * g * cos(theta)
where:
a is the acceleration of the object,
g is the acceleration due to gravity (which is approximately equal to 9.8 m/s^2),
theta is the inclination angle of the surface,
mu is the coefficient of kinetic friction.
theta = 30°,
mu = 0.2.
Substituting these values in the given equation, we have:
a = 9.8 m/s^2 * sin(30°) - 0.2 * 9.8 m/s^2 * cos(30°)
Simplifying this expression, we get:
a ≈ 4.9 m/s^2 * 0.5 - 0.2 * 9.8 m/s^2 * 0.866
a ≈ 2.45 m/s^2 - 1.7 m/s^2
a ≈ 0.75 m/s^2
Therefore, the acceleration of the object on the inclined surface will be approximately 0.75 m/s^2.
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A solenoid that is 97.2 cm long has a cross-sectional area of 24.6 cm2. There are 1320 turns of wire carrying a current of 5.78 A. (a) Calculate the energy density of the magnetic field inside the solenoid
Given; Length of solenoid, l = 97.2 cm = 0.972 m Cross-sectional area of solenoid, A = 24.6 cm² = 0.0246 m²Number of turns of wire, n = 1320Current, I = 5.78 A
Energy density of the magnetic field inside the solenoid is given by; Energy density, u = (1/2)µ₀I²where µ₀ = Permeability of free space = 4π x 10⁻⁷ T m/I After substituting the values of I and µ₀, we get Energy density, u = (1/2) x 4π x 10⁻⁷ x 5.78² u = 1.559 x 10⁻³ J/m³Let's calculate Energy density, u of the magnetic field inside the solenoid. The magnetic energy density is equal to (1/2) µ0 N I² where N is the number of turns per unit length and I is the current density through the solenoid. Thus, the magnetic energy density of the solenoid is given by (1/2) µ0 N I².
However, in the problem, we're only given the number of turns, current, and cross-sectional area of the solenoid, so we have to derive the number of turns per unit length or the length density of the wire. Here, length density of wire = Total length of wire / Cross-sectional area of solenoid Total length of wire = Cross-sectional area of solenoid x Length of solenoid x Number of turns per unit length of wire= A l n Length density of wire, lN = n / L, where L is the length of the wire of the solenoid.
Then, Energy density, u = (1/2) µ₀ lN I²= (1/2) * 4 * π * 10^-7 * n * I² / L= 1.559 x 10^-3 J/m³.
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A 2 M resistor is connected in series with a 2.5 µF capacitor and a 6 V battery of negligible internal resistance. The capacitor is initially uncharged. After a time t = ↑ = RC, find each of the following. (a) the charge on the capacitor 9.48 HC (b) the rate at which the charge is increasing 1.90 X HC/s (c) the current HC/S (d) the power supplied by the battery μW (e) the power dissipated in the resistor μW (f) the rate at which the energy stored in the capacitor is increasing. μW
The rate at which the energy stored in the capacitor is increasing. = μW
We know that;
Charging of a capacitor is given as:q = Q(1 - e- t/RC)
Where, q = charge on capacitor at time t
Q = Final charge on the capacitor
R = Resistance
C = Capacitance
t = time after which the capacitor is charged
On solving this formula, we get;
Q = C X VC X V = Q/C = 6 V / 2.5 µF = 2.4 X 10-6 C
Other data in the question is:
R = 2 MΩC = 2.5 µFV = 6 V(
The charge on the capacitor:
q = Q(1 - e- t/RC)q = 2.4 X 10-6 C (1 - e- 1)q = 9.48 X 10-6 C
The rate at which the charge is increasing:
When t = RC; q = Q(1 - e- 1) = 0.632QdQ/dt = I = V/RI = 6/2 X 106 = 3 X 10-6 Adq/dt = d/dt(Q(1 - e-t/RC))= I (1 - e-t/RC) + Q (1 - e-t/RC) (-1/RC) (d/dt)(t/RC)q = Q(1 - e- t/RC)dq/dt = I (1 - e- t/RC)dq/dt = (3 X 10-6 A)(1 - e- 1) = 1.9 X 10-6 A
the current: Current flowing through the circuit is given by; I = V/R = 6/2 X 106 = 3 X 10-6 A
the power supplied by the battery: Power supplied by the battery can be given as:
P = VI = (6 V)(3 X 10-6 A) = 18 X 10-6 μW
the power dissipated in the resistor: The power dissipated in the resistor can be given as; P = I2 R = (3 X 10-6 A)2 (2 X 106 Ω) = 18 X 10-6 μW
the rate at which the energy stored in the capacitor is increasing: The rate at which the energy stored in the capacitor is increasing is given as;dW/dt = dq/dt X VdW/dt = (1.9 X 10-6 A)(6 V) = 11.4 X 10-6 μW
Given in the question that, a 2 M resistor is connected in series with a 2.5 µF capacitor and a 6 V battery of negligible internal resistance. The capacitor is initially uncharged. We are to find various values based on this. Charging of a capacitor is given as;q = Q(1 - e-t/RC)Where, q = charge on capacitor at time t
Q = Final charge on the capacitor
R = Resistance
C = Capacitance
t = time after which the capacitor is charged
We have;R = 2 MΩC = 2.5 µFV = 6 VTo find Q, we have;Q = C X VQ = 2.4 X 10-6 C
Other values that we need to find are
The charge on the capacitor:q = 2.4 X 10-6 C (1 - e- 1)q = 9.48 X 10-6 C
The rate at which the charge is increasing:dq/dt = I (1 - e- t/RC)dq/dt = (3 X 10-6 A)(1 - e- 1) = 1.9 X 10-6 A
The current: Current flowing through the circuit is given by; I = V/R = 6/2 X 106 = 3 X 10-6 A
The power supplied by the battery: Power supplied by the battery can be given as:
P = VI = (6 V)(3 X 10-6 A) = 18 X 10-6 μW
The power dissipated in the resistor: Power dissipated in the resistor can be given as; P = I2 R = (3 X 10-6 A)2 (2 X 106 Ω) = 18 X 10-6 μW
The rate at which the energy stored in the capacitor is increasing: The rate at which the energy stored in the capacitor is increasing is given as;dW/dt = dq/dt X VdW/dt = (1.9 X 10-6 A)(6 V) = 11.4 X 10-6 μW
On calculating and putting the values in the formulas of various given entities, the values that are calculated are
The charge on the capacitor = 9.48 HC
The rate at which the charge is increasing = 1.90 X HC/s
The current = HC/S
The power supplied by the battery = μW
The power dissipated in the resistor = μW
The rate at which the energy stored in the capacitor is increasing. = μW.
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The turbine of a power plant receives steam from a boiler at
520oC and expels it towards a condenser at 100oC. What is its
maximum possible efficiency?
The Carnot cycle gives the greatest possible efficiency for an engine working between two specified temperatures, provided the cycle is completely reversible. The Carnot cycle is made up of four processes.
The heat energy input and the heat energy output of a steam turbine are determined by the enthalpies of the steam entering and leaving the turbine, respectively. The change in enthalpy of the steam is given by:
Where H1 and H2 are the enthalpies of the steam entering and leaving the turbine, respectively. It is possible to obtain the efficiency of the turbine using the following equation. where W is the work output, and Qin is the heat energy input.
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A planet with mass m, is at a distance r from a star with mass 5m. At what separation distance is the gravitational attraction between the planet and the star equal?
The separation distance at which the gravitational attraction between the planet and the star is equal is equal to the distance r₁ multiplied by the square root of 5. The force of attraction is proportional to the masses and inversely proportional to the square of the distance between the two masses, i.e., the planet and the star.
According to Newton's law of gravitation, the force of gravity between two objects is directly proportional to their masses and inversely proportional to the square of the distance between them. Let the distance between the planet and the star be r₁. The force of gravity between them is given by:
F₁ = G(m)(5m) / r₁²
where G is the gravitational constant.
Subsequently, the force of gravity between them when the distance between them is r₂ is given by:
F₂ = G(m)(5m) / r₂²
We are asked to find the distance between the planet and the star where the gravitational attraction between them is equal.
Therefore, F₁ = F₂.G(m)(5m) / r₁²
= G(m)(5m) / r₂²
Simplifying, r₂ = r₁ √5
The separation distance at which the gravitational attraction between the planet and the star is equal is equal to the distance r₁ multiplied by the square root of 5. The force of attraction is proportional to the masses and inversely proportional to the square of the distance between the two masses, i.e., the planet and the star.
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