What is the gravitational potential energy of uranus due to the sun? assume muranus = 8.68×10^25 kg , msun = 2.0 × 1030 kg , and the orbital distance is r = 2.88×10^9 km .

If a star remains at a constant size and temperature for a long period of time, it is most likely to be true that the star generates about as much energy as it radiates.

If a star remains at a constant size and temperature for an extended period, it suggests that the star is in a state of equilibrium. In such a state, the energy generated by the star's internal processes, such as nuclear fusion, is balanced by the energy radiated into space. This equilibrium is crucial for maintaining the star's stability and preventing it from expanding or contracting over time. If the star were to generate more energy than it radiates, it would accumulate excess energy and eventually experience an imbalance, causing changes in size, temperature, or both. Likewise, if the star generated less energy than it radiates, it would gradually deplete its internal energy reserves. Therefore, the most likely scenario is that the star generates about as much energy as it radiates, maintaining a steady state.

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The gravitational potential energy of a 79 kg person standing atop Mt. Everest at an altitude of 8,848 m is approximately 6.12 x 10^7 J.

The gravitational potential energy (GPE) of an object is given by the formula GPE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above some reference point. In this case, we are given that the person has a mass of 79 kg and is standing atop Mt. Everest at an altitude of 8,848 m above sea level, which we can use as our reference point (i.e., y=0).

We can find the acceleration due to gravity at this altitude using the formula g' = (GM)/(r+h)^2, where G is the gravitational constant, M is the mass of the Earth, r is the radius of the Earth, and h is the height of the person above the Earth's surface. Plugging in the appropriate values, we get g' ≈ 9.760 m/s^2.

Using this value of g', we can now calculate the GPE of the person using the formula GPE = mgh. Plugging in the values we have, we get GPE ≈ (79 kg)(9.760 m/s^2)(8,848 m) ≈ 6.12 x 10^7 J. Therefore, the gravitational potential energy of the person is approximately 6.12 x 10^7 J.

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The energy released when 100 kg of deuterium and 150 kg of tritium are consumed in one year in a fusion reactor is approximately 8.4 x 10^16 joules.

Fusion reactions release energy according to Einstein's mass-energy equivalence formula (E=mc^2), where m is the mass difference between the reactants and products, and c is the speed of light. Deuterium-tritium fusion is one of the most promising reactions for practical fusion power. It releases more energy per reaction compared to other fusion reactions, making it an attractive choice for future fusion reactors. The energy released when 100 kg of deuterium and 150 kg of tritium are consumed in one year in a fusion reactor is approximately 8.4 x 10^16 joules.

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The power intensity of radiation per unit wavelength emitted at a wavelength of 500.0 nm by a blackbody at a temperature of 10,000 K is 3.63 × 10⁻² W⋅m⁻²⋅nm⁻¹.

To determine the power intensity of radiation per unit wavelength emitted by a blackbody at a given temperature and wavelength, we can use Planck's law, which gives the spectral radiance of a blackbody as a function of its temperature and wavelength;

B(λ, T)=(2hc²/λ⁵) × 1/(exp(hc/λkT) - 1)

where; B(λ, T) is the spectral radiance of the blackbody at wavelength λ and temperature T

h is Planck's constant

c is the speed of light

k is the Boltzmann constant

To obtain the power intensity per unit wavelength, we need to multiply the spectral radiance by the wavelength and divide by the speed of light;

I(λ, T) = B(λ, T) × λ / c

Substituting λ = 500.0 nm

= 5.00 × 10⁻⁷ m and T

= 10,000 K, we get;

B(500.0 nm, 10,000 K) = (2hc²/λ⁵) × 1/(exp(hc/λkT) - 1)

= 1.09 × 10⁸ W⋅m⁻²⋅sr⁻¹⋅m⁻¹

I(500.0 nm, 10,000 K)

= B(500.0 nm, 10,000 K) × λ / c

= 3.63 × 10⁻² W⋅m⁻²⋅nm⁻¹

Therefore, the power intensity is 3.63 × 10⁻² W⋅m⁻²⋅nm⁻¹.

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In an oscillating RLC circuit with R = 2.1 Ω, L = 2.0 mH, and C = 200 µF, you are asked to determine the angular frequency of the oscillations (in rad/s).

To calculate the angular frequency (ω), we will use the formula for the resonance frequency (f) of an RLC circuit, which is given by:

f = 1 / (2π * √(L * C))

Where L is the inductance (2.0 mH) and C is the capacitance (200 µF). First, convert the given values into their base units:

L = 2.0 mH = 2.0 * 10^(-3) H

C = 200 µF = 200 * 10^(-6) F

Now, plug the values into the formula:

f = 1 / (2π * √((2.0 * 10^(-3) H) * (200 * 10^(-6) F)))

f ≈ 1 / (2π * √(4 * 10^(-9)))

f ≈ 1 / (2π * 2 * 10^(-4.5))

f ≈ 795.77 Hz

To find the angular frequency (ω), we use the relationship between angular frequency and frequency:

ω = 2π * f

ω = 2π * 795.77 Hz

ω ≈ 5000 rad/s

In conclusion, the angular frequency of the oscillations in the given oscillating RLC circuit is approximately 5000 rad/s.

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Explanation:

The change in an object's momentum is equal to the final momentum minus the initial momentum.

The momentum of an object is given by the product of its mass and velocity:

Initial momentum = mass * initial velocity

Final momentum = mass * final velocity

Given:

Mass of the tennis ball = 5.69x10^-2 kg

Initial velocity = 13 m/s

Final velocity = -18 m/s (opposite direction)

Let's calculate the initial momentum and final momentum:

Initial momentum = (5.69x10^-2 kg) * (13 m/s)

Final momentum = (5.69x10^-2 kg) * (-18 m/s)

Now, let's calculate the change in momentum:

Change in momentum = Final momentum - Initial momentum

Plugging in the values:

Change in momentum = [(5.69x10^-2 kg) * (-18 m/s)] - [(5.69x10^-2 kg) * (13 m/s)]

Performing the calculation will give you the change in the ball's momentum.

Hope I helped

(a) The motor has done 25.12 J of work on the rotor when it has rotated through four revolutions.

(b) The rotor's angular velocity is approximately 167.55 rpm when it has rotated through four revolutions.

To calculate the work done by the motor on the rotor, we use the formula W = τΔθ, where W is the work done, τ is the torque exerted by the motor, and Δθ is the angle through which the rotor has rotated. Since the rotor has rotated through four revolutions, Δθ = 8π radians. Thus, W = 0.8 N-m × 8π rad = 25.12 J.

To calculate the angular velocity of the rotor, we use the formula ω = Δθ/Δt, where ω is the angular velocity, Δθ is the angle through which the rotor has rotated, and Δt is the time taken to rotate through that angle. Since the rotor has rotated through four revolutions, Δθ = 8π radians. The time taken can be calculated from the formula Δθ = ωt. Rearranging this formula, we get t = Δθ/ω. Substituting the values, we get t = 8π/ω. We know that one revolution is equal to 2π radians, so four revolutions is equal to 8π radians. Therefore, t = 4/ω. Substituting this value of t in the formula for ω, we get ω = Δθ/t = (8π)/(4/ω) = 2ωπ. Solving for ω, we get approximately 167.55 rpm.

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The magnitude of the electrostatic force between the two charges is approximately 5.29 x 10^-5 N. The force between these two charges is attractive.

(a) To find the magnitude of the electrostatic force between the two charges, we can use Coulomb's Law which states

that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, this can be expressed as:

F = k * (q1 * q2) / r^2

Where F is the electrostatic force, k is Coulomb's constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges in Coulombs, and r is the distance between the charges in meters.

Plugging in the given values, we get:

F = 9 x 10^9 * [(7.66 x 10^-9) * (4.54 x 10^-9)] / (1.93)^2
F ≈ 5.29 x 10^-5 N

Therefore, the magnitude of the electrostatic force between the two charges is approximately 5.29 x 10^-5 N.

(b) To determine if the force is attractive or repulsive, we need to look at the signs of the charges. In this case, one charge is positive (7.66 nc) and the other is negative (4.54 nc). Opposite charges attract each other, while like charges repel each other. Therefore, the force between these two charges is attractive.

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The capacitance of the two wires is approximately 6.07 pF (pico-Farads).

To find the capacitance of the two wires, we'll use the formula for capacitance, which is:

Capacitance (C) = Charge (Q) / Potential Difference (V)

Given the information in your question, we have:

Charge (Q) = 85 pC (pico-Coulombs) = 85 x 10^(-12) C (Coulombs)
Potential Difference (V) = 14 V

Now we can calculate the capacitance:

C = (85 x 10^(-12) C) / (14 V)

C ≈ 6.07 x 10^(-12) F (Farads)

So, the capacitance of the two wires is approximately 6.07 pF (pico-Farads).

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The change in entropy of the universe is 5.26 J/K.

To find the change in entropy of the universe, we need to use the formula ΔS_univ = ΔS_hot + ΔS_cold, where ΔS_hot is the change in entropy of the hot reservoir and ΔS_cold is the change in entropy of the cold reservoir.

The change in entropy of the hot reservoir can be calculated using the formula ΔS_hot = Q_hot/T_hot, where Q_hot is the amount of heat transferred from the hot reservoir and T_hot is the temperature of the hot reservoir. Substituting the given values, we get:

ΔS_hot = 1050 J/576 K = 1.82 J/K

Similarly, the change in entropy of the cold reservoir can be calculated using the formula ΔS_cold = -Q_cold/T_cold, where Q_cold is the amount of heat absorbed by the cold reservoir and T_cold is the temperature of the cold reservoir. Since the heat is being transferred irreversibly from the hot reservoir to the cold reservoir, we know that Q_cold = -1050 J. Substituting the given values, we get:

ΔS_cold = -(-1050 J)/305 K = 3.44 J/K

Now we can calculate the change in entropy of the universe:

ΔS_univ = ΔS_hot + ΔS_cold
ΔS_univ = 1.82 J/K + 3.44 J/K
ΔS_univ = 5.26 J/K

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The frequency of the photon emitted when the electron in a hydrogen atom drops from energy level E6 to E3 is 4.56 x 10¹⁴ Hz. The emitted photon falls in the ultraviolet range of the electromagnetic spectrum.

The energy change of an electron in a hydrogen atom dropping from energy level n=6 to n=3 is given by:

ΔE = E6 - E3 = -13.6 eV[(1/3²) - (1/6²)] = -1.89 eV

The frequency of the emitted photon can be calculated using the Planck-Einstein equation:

E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10⁻³⁴ J s), and f is the frequency of the photon.

Converting the energy change to joules:

ΔE = -1.89 eV x 1.6 x 10⁻¹⁹ J/eV = -3.02 x 10⁻¹⁹ J

Solving for f:

f = E/h = (-3.02 x 10⁻¹⁹ J) / (6.626 x 10⁻³⁴ J s) = 4.56 x 10¹⁴ Hz

The frequency of the emitted photon is 4.56 x 10¹⁴ Hz, which corresponds to the range of the electromagnetic spectrum known as ultraviolet (UV).

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The mass of the sample of 238/92U is 0.401 kg.

First, we can use the decay constant (λ) formula to calculate the decay rate:

[tex]λ = ln(2)/t1/2 = ln(2)/(4.468×10^9 yr) ≈ 1.549 × 10^-10 /s[/tex]

Then, we can use the decay rate formula to find the number of atoms (N) in the sample:

[tex]N = (decay rate) / λ = 450 / (1.549 × 10^-10 /s) ≈ 2.906 × 10^12 atoms[/tex]

Finally, we can use the atomic mass of 238/92U (which is approximately 238 g/mol) to calculate the mass of the sample:

mass = N × (atomic mass) = 2.906 × 10^12 atoms × (238 g/mol / 6.022 × 10^23 atoms/mol) ≈ 0.401 kg

Therefore, the mass of the sample is 0.401 kg (to three significant figures).

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The new volume of the balloon when heated to 48.0°C is approximately 1.95 L (option a).

To calculate the new volume of the balloon when it is heated, we can use the formula derived from Charles's Law, which states that for a given amount of gas at constant pressure, the volume is directly proportional to its temperature in Kelvin.

The formula is V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.

First, we need to convert the temperatures from Celsius to Kelvin:

T1 = 24.0°C + 273.15 = 297.15 K
T2 = 48.0°C + 273.15 = 321.15 K

Now, we can plug in the values into the formula:

V1 = 1.80 L (given)
T1 = 297.15 K
T2 = 321.15 K

1.80 L / 297.15 K = V2 / 321.15 K

Solving for V2, we get:

V2 = (1.80 L * 321.15 K) / 297.15 K
V2 ≈ 1.95 L

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The discovery of Jupiter-sized planets much closer than 1 au from their parent stars was surprising to astronomers because according to the current understanding of planetary formation, such large gas giants should not be able to form so close to their stars due to the intense heat and radiation.

Additionally, the detection of these planets using the radial velocity method was difficult as the wobble of the star caused by the planet's gravitational pull is smaller when the planet is closer to the star. Therefore, the discovery of these "hot Jupiters" challenged astronomers' assumptions about planetary formation and the conditions required for the existence of extrasolar planets.

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The maximum diffraction order seen is m = 3.

To find the maximum diffraction order seen for a 500 lines per mm diffraction grating illuminated by light of      wavelength 620 nm, follow these steps:

Step 1: Convert lines per mm to lines per meter.
500 lines/mm = 500,000 lines/m

Step 2: Calculate the grating spacing (d) using the formula:
d = 1 / (lines per meter)
d = 1 / 500,000
d = 2 x 10^-6 m

Step 3: Use the diffraction formula to find the maximum order (m):
m * λ = d * sinθ

Since we want to find the maximum diffraction order, the angle θ will be at its maximum (90 degrees).

Therefore, sinθ = sin(90°) = 1.

Step 4: Solve for m:
m = (d * sinθ) / λ
m = (2 x 10^-6 * 1) / (620 x 10^-9)
m = 3.2258

Since the diffraction order must be an integer, the maximum diffraction order seen is m = 3.

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You would absorb 8.5 ×[tex]10^{6}[/tex]J of solar energy in a 60-minute sunbath.

The amount of solar energy you absorb in a 60-minute sunbath can be estimated as follows:

Calculate the area of the beach blanket you occupy:

Area = length x width = (1.7 m) x (0.3 m) = 0.51 [tex]m^{2}[/tex]

Calculate the fraction of solar energy that reaches the surface of the Earth:

Fraction reaching Earth's surface = 60% = 0.6

Calculate the fraction of solar energy that you absorb:

Fraction absorbed = 50% = 0.5

Calculate the solar energy that you absorb per unit area:

Energy absorbed per unit area = (intensity of solar radiation at the top of Earth's atmosphere) x (fraction reaching Earth's surface) x (fraction absorbed)

Energy absorbed per unit area = (1,370 W/[tex]m^{2}[/tex]) x (0.6) x (0.5) = 411 W/[tex]m^{2}[/tex]

Calculate the solar energy you absorb in a 60-minute sunbath:

Energy absorbed = (energy absorbed per unit area) x (area of beach blanket) x (time)

Energy absorbed = (411 W/[tex]m^{2}[/tex]) x (0.51 [tex]m^{2}[/tex]) x (60 min x 60 s/min) = 8,466,120 J

Therefore, you would absorb approximately 8.5 ×[tex]10^{6}[/tex] J of solar energy in a 60-minute sunbath. Note that this is an order-of-magnitude estimate and the actual value may be different due to various factors such as the actual solar radiation intensity, the actual fraction of solar energy reaching Earth's surface, and the actual fraction of solar energy absorbed by your body, among others.

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The electric potential at the origin of the xy-coordinate system due to two positive point charges, both of magnitude 4.0x10^6C, situated along the x-axis at x=-2.0m and x=2.0m is 0.

To calculate the electric potential at the origin, we first need to find the electric potential due to each charge separately. Using the formula for electric potential due to a point charge V=kQ/r, where k is Coulomb's constant, Q is the magnitude of the charge, and r is the distance from the charge to the point where the potential is being calculated, we can calculate the electric potential due to each charge as follows:

V1=k(4.0x10^6)/2.0=2.16x10^7V
V2=k(4.0x10^6)/2.0=2.16x10^7V

Since the charges are of the same magnitude and opposite signs, their electric potentials cancel out at the origin, resulting in a net electric potential of 0. Therefore, the correct answer is 0.

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The frequency of the wave is 2.50 Hz.

The velocity of a wave can be calculated using the equation v = λf, where v is the velocity of the wave, λ is the wavelength, and f is the frequency. We can rearrange this equation to solve for the frequency as follows:

v = λf => f = v/λ

The velocity of the wave can be calculated using the distance traveled and the time taken as follows:

v = d/t = 60.0 m / 12.0 s = 5.00 m/s

Substituting the given values for the wavelength and velocity, we get:

f = v/λ = 5.00 m/s / 3.00 m = 1.67 Hz

However, this is the frequency of one complete wavelength. To find the frequency of the entire wave, we need to divide by the number of wavelengths that pass a point in one second. The number of wavelengths that pass a point in one second is equal to the velocity of the wave divided by the wavelength, which is:

Number of wavelengths per second = v/λ = 5.00 m/s / 3.00 m = 1.67 Hz

Therefore, the frequency of the wave is 1.67 Hz / wavelength = 2.50 Hz (to two significant figures).

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An elementary particle travels 60 km through the atmosphere at a speed of 0.9996c. According to the particle, the thickness of the atmosphere is 32.4 km.

According to the particle, the length of the atmosphere it travels through is shortened due to time dilation and length contraction effects predicted by special relativity.

The proper length of the atmosphere (i.e., the length measured by a stationary observer on Earth) is L = 60 km.

The length contracted distance, as measured by the particle, is given by

L' = L / γ

Where γ is the Lorentz factor

γ = 1 / [tex]\sqrt{(1- v^{2} /c^{2} )[/tex]

Where v is the velocity of the particle and c is the speed of light.

Substituting the given values into the above equation, we get

γ = 1 / [tex]\sqrt{(1- (0.9996c)^{2} / c^{2} )[/tex]

γ = 1.854

Therefore, the length of the atmosphere as measured by the particle is

L' = L / γ

L' = 60 km / 1.854

L' ≈ 32.4 km

Therefore, according to the particle, the thickness of the atmosphere is 32.4 km.

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The maximum charge on the capacitor during the oscillations is equal to i/ω.

At time t=0, the current in the oscillating lc circuit with inductance L and capacitance C is at its maximum value of i. As the circuit oscillates, the charge on the capacitor varies periodically, resulting in a back-and-forth flow of energy between the inductor and the capacitor. During each oscillation, the maximum charge on the capacitor occurs when the current is at its zero crossing.

To determine the maximum charge on the capacitor, we can use the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor. At the point where the current is at its zero crossing, the voltage across the capacitor is at its maximum value, which is given by V = i/(ωC), where ω = 1/√(LC) is the angular frequency of the oscillation. Substituting this into the equation for Q, we get:

Qmax = CVmax = C(i/(ωC)) = i/ω

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To determine electrical energy from a plot of power versus time, you can integrate the power over time. The integral of power with respect to time gives you the total energy consumed or produced.

Here are the steps to determine electrical energy from a plot of power versus time:

1. Obtain a plot of power (P) as a function of time (t).

2. Identify the area under the power-time curve.

3. Calculate the integral of power with respect to time (∫P dt) over the desired time interval.

4. Evaluate the integral to find the numerical value of energy.

If the power values in the plot are constant over different time intervals, you can simply calculate the energy for each interval by multiplying the constant power by the duration of that interval.

Keep in mind that the unit of power should be consistent throughout the calculation (e.g., watts, kilowatts), and the resulting energy value will be in units such as joules or watt-hours (Wh) depending on the units used for power and time.

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The complete expression for the magnetic field vector B of the electromagnetic wave is given by B = Bo cos(kx - wt) i, where Bo is the amplitude of the magnetic field vector.

The complete expression for the magnetic field vector B of the electromagnetic wave is given by B = Bo cos(kx - wt) i, where Bo represents the amplitude of the magnetic field vector. The magnetic field vector B is perpendicular to both the electric field vector and the direction of wave propagation.

In the given expression, cos(kx - wt) represents the time and space dependence of the wave. The term cos(kx - wt) indicates the phase of the wave and determines how the magnetic field varies as a function of position (x) and time (t). The quantity k represents the wave number, which is related to the wavelength of the wave.

The presence of the unit vector i indicates that the magnetic field vector is directed along the x-axis. This means that the magnetic field oscillates in a direction perpendicular to both the direction of wave propagation (positive x direction) and the y-axis.

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The minimum frequency of light needed to release electrons from the metal is approximately 9.67 x [tex]10^1^4[/tex]Hz.

The minimum frequency of light required to cause electrons to be emitted from a metal can be found by using equation:

E = hf

where E is the energy of a single photon, h is Planck's constant (6.626 x [tex]10^-^3^4[/tex] J·s), and f is the frequency of light.

The work function, denoted by φ, is the minimum energy required to remove an electron from the metal. In this case, the work function is given as 4.00 eV.

We need to convert the work function from electron volts (eV) to joules (J) since Planck's constant is in joules. The conversion factor is 1 eV = 1.602 x [tex]10^-^1^9[/tex]J.

Therefore, the work function in joules is:

φ = 4.00 eV × (1.602 x [tex]10^-^1^9[/tex] J/eV) = 6.408 x[tex]10^-^1^9[/tex]J

We can equate the energy of a single photon to the work function

E = φ

hf = φ

From this equation, we can solve for the frequency f:

f = φ / h

Substituting the values:

f = (6.408 x [tex]10^-^1^9[/tex]J) / (6.626 x [tex]10^-^3^4[/tex]J·s)

f ≈ 9.67 x 1[tex]0^1^4[/tex]Hz

Therefore, the minimum frequency of light required to cause electrons to be emitted from the metal is approximately 9.67 x[tex]10^1^4[/tex] Hz.

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The skateboarder needs to regularly push with their foot when skateboarding along a level surface because of the presence of frictional forces that oppose motion. When the skateboarder gives themselves a push, they increase their forward velocity.

The skateboarder needs to regularly push with their foot when skateboarding along a level surface because of the presence of frictional forces that oppose motion. When the skateboarder gives themselves a push, they increase their forward velocity. However, over time, the velocity decreases due to the force of friction between the skateboard's wheels and the ground, which acts in the opposite direction to the skateboard's motion. During a push, the skateboarder exerts a force on the skateboard that propels it forward. Between pushes, the skateboard moves at a constant velocity due to the balanced forces acting upon it. However, as frictional forces act on the skateboard, it slows down until the next push is required. The net force acting on the skateboarder is unbalanced, as the force of friction acting against the skateboard's motion is greater than the force of the skateboarder's push. The resulting net force causes the skateboarder to slow down over time. Thus, by pushing themselves, the skateboarder overcomes the force of friction and maintains their forward motion.

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All the answers as as follows:

1. An electromagnetic wave is a type of wave that is produced by the motion of electric and magnetic fields. The waves consist of oscillating electric and magnetic fields, which are perpendicular to each other and to the direction of the wave's propagation. These waves travel through a vacuum at the speed of light and have different properties depending on their wavelength and frequency.
2. Polarized light refers to light that has a specific polarization direction. This means that the electric field of the light waves oscillates in a single plane, rather than in all possible directions. Polarized light can be produced by passing unpolarized light through a polarizing filter, which only allows waves with a certain polarization direction to pass through.
3. Wave optics is needed to describe polarized light because the polarization direction of light is determined by the properties of the wave itself. This means that the behavior of polarized light can only be explained using wave optics, which is a branch of physics that deals with the wave-like behavior of light.
4. The rotary motion probe is used in Activity 1-3 to measure the angle of rotation of a polarizing filter. The probe is attached to the filter and is used to record the angle of rotation as the filter is turned.
5. To find the mathematical relationship between the data from Activity 1-3, you will need to plot the light intensity vs. the angle of rotation of the polarizing filter. This will produce a graph that shows how the intensity of the light changes as the filter is rotated. You can then use this graph to determine the mathematical relationship between the two variables, which will allow you to make predictions about how the intensity of the light will change for different angles of rotation.

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7. The gravitational force between the two sumo wrestlers is approximately 8.52 N when they are embraced and their centers are 1.2 m apart.

8. The mass of the planet is approximately 4.44 x 10²⁴ kg.

9. (a) The ratio of the orbital radii of the planets is 4:1.

(b) the ratio of the periods of the two planets is 8:1.

7. To estimate the gravitational force between two sumo wrestlers, we can use Newton's law of gravitation:

F = G * m1 * m2 / r²

where F is the gravitational force, G is the gravitational constant (6.67 x 10⁻¹¹ N*m²/kg²), m1 and m2 are the masses of the two objects, and r is the distance between their centers.

Substituting the given values, we get:

F = (6.67 x 10⁻¹¹ N*m²/kg²) * (220 kg) * (240 kg) / (1.2 m)²

Solving for F, we get:

F = 8.52 N

Therefore, the gravitational force between the two sumo wrestlers is approximately 8.52 N when they are embraced and their centers are 1.2 m apart.

8. To determine the mass of a planet from the acceleration due to gravity on its surface, we can use Newton's law of gravitation and the formula for the acceleration due to gravity:

F = G * m * M / r²

a = G * M / r²

where F is the gravitational force between the planet and an object with mass m, M is the mass of the planet, r is the radius of the planet, G is the gravitational constant, and a is the acceleration due to gravity on the planet's surface.

Substituting the given values, we get:

a = 18 m/s²

r = 1.2 x 10⁷ m

G = 6.67 x 10⁻¹¹ N*m²/kg²

Substituting the formula for F into the formula for a, we get:

a = G * M / r²

Solving for M, we get:

M = a * r² / G

Substituting the given values, we get:

M = (18 m/s²) * (1.2 x 10⁷ m)²/ (6.67 x 10⁻¹¹N*m²/kg²)

Solving for M, we get:

M = 4.44 x 10²⁴ kg

Therefore, the mass of the planet is approximately 4.44 x 10²⁴ kg.

9. (a)Two planets in circular orbits around a star are subject to the gravitational force of the star. The gravitational force provides the centripetal force that keeps the planets in their circular orbits. The speed of each planet is related to its distance from the star by the formula:

v = √(G * M / r)

where v is the speed of the planet, G is the gravitational constant, M is the mass of the star, and r is the distance between the planet and the star.

Since both planets are in circular orbits around the same star, the gravitational force acting on each planet is the same. Therefore, we can equate the centripetal force to the gravitational force:

m * v² / r = G * M * m / r²

where M is the mass of the star.

Solving for the ratio of the radii, we get:

r2 / r1 = (v2 / v1)²

r2 / r1 = 4

Therefore, the ratio of the orbital radii of the two planets is 4:1.

(b) The ratio of their periods can be found using Kepler's third law:

T² / r³ = 4 * pi² / (G * M)

where T is the period of the planet and r is its orbital radius.

Since both planets are orbiting the same star, we can take the ratio of their periods:

(T2 / T1)² = (r2 / r1)³

(T2 / T1)²= 64

(T2 / T1) = 8

Therefore, the ratio of the periods of the two planets is 8:1.

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Because angular momentum must be conserved, as a gas cloud contracts due to gravity, it will spin faster. The correct answer is (b)

This is due to the conservation of angular momentum, which states that the product of the angular velocity and the moment of inertia of an object must remain constant if there is no net external torque acting on it.

As the cloud contracts, its moment of inertia decreases, so in order to conserve angular momentum, the angular velocity (spin rate) of the cloud must increase.

This is similar to what happens when an ice skater pulls in their arms while spinning - they spin faster to conserve their angular momentum. Therefore, the correct answer is (b) spin faster.

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b. spin faster.

This is because as the gas cloud contracts due to gravity, its radius decreases, which means its moment of inertia decreases. In order for angular momentum to be conserved, the cloud must spin faster to compensate for the decrease in moment of inertia.

As a gas cloud contracts due to gravity, it needs to conserve angular momentum. To do this, the cloud will spin faster. This is because angular momentum (L) is given by the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. As the cloud contracts, its moment of inertia (I) decreases, so to maintain constant angular momentum (L), the angular velocity (ω) must increase, causing the gas cloud to spin faster.

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Yes, the speed of asteroid 1 relative to asteroid 2 would be v=0.22c.

To find the relative speed of asteroid 1 and asteroid 2, we can use the formula for relative velocity:

v(relative) = v(1) - v(2)

where v(1) is the velocity of asteroid 1 and v(2) is the velocity of asteroid 2.

Given that the speeds relative to Earth are 0.78c for asteroid 1 and 0.58c for asteroid 2, we can convert these to their velocities relative to the speed of light (c):

v(1) = 0.78c
v(2) = 0.58c

Substituting these values into the formula for relative velocity, we get:

v(relative) = 0.78c - 0.58c
v(relative) = 0.20c

Therefore, the speed of asteroid 1 relative to asteroid 2 is v=0.20c, which is equivalent to v=0.22 times the speed of light.

Yes, you are correct that the relative speed of asteroid 1 and asteroid 2 would be v=0.22c.

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The peak current is  1.14 A

The phase angle is 17.7 degrees

The power lost is  130 W

The capacitive reactance is;

Xc = 1/2πfc

Xc = 1/2 * 3.14 * 60 * 30 * 10^-6

Xc = 88.5 ohm

XL = 2πfL

= 2 * 3.14 *60 * 0.15

= 56.5 ohm

Impedance;

Z = √R^2 + (XL - XC)^2

Z = √(100)^2 + (56.5 - 88.5)^2

Z = 105 ohm

I = V/Z

= 120V/105 Ohm

= 1.14 A

The phase angle is;

Tan-1 (XL - XC)/R

= Tan-1 (-32/100)

= 17.7 degrees

The average power loss is;

IV cosφ

= 1.14 * 120 8 Cos 17.7

= 130 W

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The maximum charge on the capacitor is 0.185 μC and the charge on the capacitor at an instant when the current in the inductor is 0.500 mA will be 0.109 μC.

(a)  We can calculate [tex]Q_{max}[/tex] by using the equation [tex]Q_{max} =C*V_{max}[/tex].

Given C = 3.20μF

And we know, [tex]V_{max} = I_{max} * XL[/tex]

Here, Inductive reactance(XL) = 2πfL, where f is the resonant frequency.

We know,[tex]f=\frac{1}{2\pi \sqrt{LC} }[/tex]

So, f = 1 / [2π√(85.0 mH * 3.20 μF)] = 1.28 kHz

∴ Inductive reactance(XL) = 2πfL

                                           = 2π * 1.28 kHz * 85.0 mH = 68.3 Ω

Now, [tex]V_{max} = I_{max} * XL[/tex]

∴ Vmax = 0.850 mA * 68.3 Ω = 58.05 mV

Finally, the maximum charge on the capacitor can be calculated as,

Qmax = C * Vmax

          = 3.20 μF * 58.05 mV

          = 0.185 μC

Therefore, the maximum charge on the capacitor is 0.185 μC.

(b)  When the current in the inductor has a magnitude of 0.500 mA, the voltage across the inductor will be,

V = I * XL

  = 0.500 mA * 68.3 Ω

  = 34.15 mV

Now the charge at required instant i.e., when I = 0.500 mA

Q = C * V

   = 3.20 μF * 34.15 mV

   = 0.109 μC.

Therefore, the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 mA is 0.109 μC.

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