A negatively charged rod is brought close to an uncharged electroscope. While the rod is close one's finger touches the far side of the metal ball on the electroscope. After the charged rod has been removed the finger is removed. The electroscope is

Yes, if the velocity of an object is nonzero, its acceleration can still be zero. An example is a car moving at a constant speed in a straight line. In this case, the car has a nonzero velocity, but its acceleration is zero because its velocity is not changing over time.

For example, if a car is traveling at a constant speed of 50 miles per hour in a straight line on a flat road, its velocity is nonzero but its acceleration is zero because there is no change in its speed or direction. In other words, acceleration is the rate of change of velocity, so if the velocity is not changing, then the acceleration must be zero. However, it's important to note that if the object's velocity changes, even if momentarily, then its acceleration will not be zero.

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We can use the conservation of momentum principle, which states that the total momentum of a system before a collision is equal to the total momentum of the system after the collision. In this case, the system consists of the child and the skateboard.

Before the collision, the momentum of the child is:

p_ child = m_ child * v_ child
p_ child = 30 kg * 4 m/s
p_ child = 120 kg*m/s

Since the skateboard is stationary, its momentum is zero:
p _skateboard = 0 kg*m/s

The total momentum before the collision is:
p_ before = p_ child + p_ skateboard
p_ before = 120 kg*m/s + 0 kg*m/s
p_ before = 120 kg*m/s

After the collision, the child and the skateboard move together as one system. Let's assume their final velocity is v_ final. The total momentum after the collision is:
p_ after = (m_ child + m_ skateboard) * v_ final

Substituting the values we know:
p_ after = (30 kg + 10 kg) * v_ final
p_ after = 40 kg * v_ final

According to the conservation of momentum principle, p_ before = p_ after, so we can set these two equations equal to each other:

p_ before = p_ after
120 kg*m/s = 40 kg * v_ final

Solving for v_ final:

v_ final = 3 m/s

Therefore, the speed of the child and the skateboard after the collision is approximately 3 m/s.

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a) Relative to your home, the football stadium is south of the west. So, the correct option is (4). b) The straight-line distance from your house to the stadium is approximately 707.1 meters.

a) From your house, you first go 1000 m north (up on the diagram), then 500 m west (left on the diagram), and finally 1500 m south (down on the diagram). The final location (the football stadium) is located to the left (west) and below (south) of your house.

b) To find the straight-line distance from your house to the stadium, we can use the Pythagorean theorem.

The distance you traveled north and south can be combined into one vertical distance of 1000 m - 1500 m = -500 m (negative because you traveled south). The distance you traveled west is 500 m.

So, the straight-line distance from your house to the stadium is the hypotenuse of a right triangle with legs of 500 m and 500 m.

Using the Pythagorean theorem,

distance² = 500² + (-500)²
distance² = 500,000
distance = √500,000
distance ≈ 707.1 meters

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The acceleration of the second disk would be half of the acceleration of the first disk.

This is because the torque applied to both disks is the same, but the moment of inertia of the second disk is four times greater than that of the first disk (since the moment of inertia is proportional to the radius squared). Therefore, the second disk will have a larger moment of inertia, which means it will be harder to accelerate.

However, the same torque will produce the same angular acceleration in both disks. Since the second disk has twice the radius, its linear acceleration will be half that of the first disk.

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When a hydrogen atom electron is excited to an energy of −13.6/4 eV, it enters the fourth excited state (n=4) with a corresponding energy level of -3.4 eV.

There are a total of 16 quantum states with this energy level, corresponding to the different possible combinations of the electron's orbital angular momentum quantum number (l=0,1,2,3) and magnetic quantum number (ml = -l, -l+1, ..., l-1, l) within the n=4 energy level. Therefore, there are 16 different quantum states with an energy of −13.6/4 eV for a hydrogen atom electron.
The number of different quantum states for a hydrogen atom electron excited to an energy of -13.6/4 eV.

There are 4 different quantum states for a hydrogen atom electron with an energy of -13.6/4 eV.
Here's a step-by-step explanation:
1. The energy levels of a hydrogen atom can be determined using the formula E_n = -13.6 eV / n^2, where n is the principal quantum number.
2. In this case, the given energy is -13.6/4 eV. To find the corresponding value of n, set E_n equal to -13.6/4 eV and solve for n:
-13.6/4 = -13.6 / n^2
n^2 = 4
n = 2
3. For a given principal quantum number n, there are n^2 possible quantum states.
4. Therefore, for n = 2, there are 2^2 = 4 different quantum states with the energy of -13.6/4 eV.

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A uniform sphere of radius r and mass m rotates freely about a horizontal axis that is tangent to an equatorial plane of the sphere. The moment of inertia (I) of the sphere about this axis can be calculated using the parallel axis theorem.

Step 1: Find the moment of inertia for a uniform sphere about its center of mass. For a sphere, this value is given by the equation:

I_center = (2/5) * m * r^2

Step 2: Apply the parallel axis theorem. The parallel axis theorem states that the moment of inertia (I) about an axis parallel to and a distance d away from the axis through the center of mass is:

I = I_center + m * d^2

Step 3: In our case, the distance d is equal to the radius r, since the horizontal axis is tangent to the equatorial plane of the sphere. Plug this value and I_center into the parallel axis theorem equation:

I = (2/5) * m * r^2 + m * r^2

Step 4: Simplify the equation:

I = m * r^2 * ((2/5) + 1)

I = m * r^2 * (7/5)

So, the moment of inertia of the uniform sphere about the horizontal axis tangent to the equatorial plane is:

I = (7/5) * m * r^2

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The wave that is being described at a football game is a longitudinal wave. This is because the movement of the wave is parallel to the direction of the wave itself. In other words, as the wave moves through the stands, the people are standing up and sitting down in the same direction as the wave is traveling.

Longitudinal waves are characterized by the compression and rarefaction of the medium through which they are traveling. In the case of the football game wave, the medium is the people in the stands. As the wave passes through them, they are compressed and then allowed to expand, creating the up and down motion that is characteristic of the wave.

It is important to note that the wave at the football game is not a true physical wave, as it is not a disturbance that travels through a medium in a continuous manner. Rather, it is a coordinated movement of individuals that creates the appearance of a wave. However, the wave can still be characterized as a longitudinal wave based on the nature of its motion.

In conclusion, the wave at a football game would be characterized as a longitudinal wave due to the parallel movement of the wave and the medium through which it is traveling.

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rogue waves are best described as a single massive wave that suddenly develops and disappears in the open ocean. Hence option D is correct.

Rogue waves, sometimes referred to as freak waves, monster waves, episodic waves, killer waves, severe waves, and anomalous waves, are abnormally enormous, erratic, and abruptly arising surface waves that can be quite harmful to ships, even huge ones. They differ from tsunamis, which are typically hardly perceptible in deep oceans and are brought on by the movement of water as a result of other events (like earthquakes). Sneaker waves are rogue waves that suddenly erupt near the coast. it is described as single massive wave.

Hence option D is correct.

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As the mass oscillates up and down,

1. Kinetic energy (KE) will change periodically. It will be at its maximum when the mass is at the midpoint of its oscillation, and at its minimum (zero) when the mass reaches the highest and lowest points in its oscillation.

2. Elastic potential energy (EPE) will also change periodically. It will be at its maximum when the mass is at the highest and lowest points in its oscillation (where the spring is most compressed or stretched), and at its minimum (zero) when the mass is at the midpoint of its oscillation.

3. Mechanical energy (ME), which is the sum of kinetic energy and elastic potential energy, will remain constant throughout the oscillation, as long as there are no external forces (e.g., friction) causing energy loss. This is because, as the mass oscillates, the energy is transferred between kinetic energy and elastic potential energy, but the total energy remains the same.

As the mass oscillates up and down, the kinetic energy will continuously alternate between maximum and zero values. At the highest points of the oscillation, the kinetic energy will be zero as the mass momentarily stops moving. At the lowest points of the oscillation, the kinetic energy will be at its maximum as the mass is moving at its highest speed.

The elastic potential energy will also change in a similar fashion, oscillating between maximum and zero values. At the highest points of the oscillation, the elastic potential energy will be at its maximum as the spring is stretched to its greatest extent. At the lowest points of the oscillation, the elastic potential energy will be at its minimum as the spring is relaxed.

The mechanical energy, which is the sum of the kinetic and potential energies, will remain constant as the mass oscillates. This is because energy is conserved in an elastic system, meaning that the total mechanical energy will remain constant even as the kinetic and potential energies alternate.

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The new gravitational force between the two masses is 4 times the original force, or 4F.

To find the new gravitational force between the two masses when the mass M is replaced by a mass of 2M, we'll use the formula for gravitational force and plug in the appropriate values.
The gravitational force (F) between two objects with masses m1 and m2, separated by a distance (R), is given by the formula:
F = G × (m1 × m2) / R²
where G is the gravitational constant.
Initially, we have the masses M and 2M, with a force F:
F = G × (M ×2M) / R²
Now, let's replace the mass M with 2M and find the new force (F_new):
F_new = G × (2M ×2M) / R²
We can simplify this as:
F_new = 4 × G × (M × 2M) / R²
Notice that the expression G × (M × 2M) / R² is equal to the original force F:
F_new = 4 ×F
So the new gravitational force between the two masses is 4 times the original force, or 4F.

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When two objects are traveling in a circular orbit. The centripetal acceleration of object A is twice that of object B.

Centripetal acceleration is the acceleration of the body that travels in a circular motion. Any object that moves in the circular path and its vector is towards to the center is called as Centripetal acceleration. It is obtained by the ratio of the velocity square and the radius of the circle.

From the givens,

Object A moves with the velocity twice that of Object B and the diameter of  the circle moves by object A is twice that of the diameter of circle moves by object B. Acceleration (a) = v² / r.

Object B's acceleration, a =  v² / 2r ( diameter d = 2r)

Object A's acceleration, a = (2v)² / (2r)

                                           =  4/2 (v²/ 2r)

                                           = 2 (v²/ 2r).

Thus, the object A's acceleration is twice that of acceleration of object B.

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The direction of sound travel compares to the shapes of the sound waves.

The direction of sound travel is determined by the propagation of the sound waves through a medium, such as air or water. Sound waves are longitudinal waves, which means that the particles in the medium vibrate parallel to the direction of the wave's travel. However, the direction of sound travel is perpendicular to this vibration, meaning that it travels in a straight line away from the source of the sound. This is why we can hear sound from different directions, even though the waves themselves are moving in a specific direction

In other words, when the sound waves move, they cause the particles to compress and rarefy (move closer together and further apart) in the same direction as the wave's movement.

To summarize, the direction of sound travel is parallel to the shapes of the sound waves, since sound waves are longitudinal waves that cause particles in the medium to vibrate along the same direction as the wave's movement.

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The efficiency of the engine is 83.4% assuming  that the

heat capacity is 21 J/K and the universal gas constant is 0.08206 L · atm/mol/K =8.314 J/mol/K.

Efficiency is described as the often measurable ability to avoid wasting materials, energy, efforts, money, and time while performing a task.

The efficiency of the engine is given by:

E = W/Q

where;

W = the work done in the four steps,

Q = the energy input

Since there at four steps in a cycle:

E = w1+ w2 +w3+ w4/ q1+ q2+q3+q4

We calculate that the work done in the first step (isothermal expansion)

n= 1 mole, T1 = 402 K, V2 = 41.2 L, V1 = 18.5 L

We also solve for Steps 2 and 4 are constant volume processes,

We also calculate  work done in the third step (isothermal expansion) is

where;

n = 1 mol, T3 = 273 K, V4 = 41.2 L, V3 = 18.5 L

We notice that Heat enters the system only during steps (1) and (4).

The internal energy of the gas increases in step 4 but no work is done, while the internal energy is constant change in step 1 but work is done by the gas.

Cv =21 J/K, T3 = 273 K, T4 = 402 K

We Solve  for efficiency, ɛ:

ɛ = 2676.01  +0 +0 + 1879.29/ 2676.01  +0 +0 + 2709 = 83.4%.

Therefore, the efficiency of the engine is 83.4%.

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Answer: estimate could be anywhere from 15 to 30 minutes, or longer.

Explanation: If your vehicle has incurred significant mechanical and exterior damage, an estimate could be anywhere from 15 to 30 minutes, or longer. When the damage is minimal and mechanical issues don't exist, an estimate usually takes 15-20 minutes.

The magnitude of the force on the charge is 240 N.

To calculate the magnitude of the force on the charge in a uniform electric field, you can use the following formula:

F = q * E

where F is the force, q is the charge, and E is the electric field magnitude.

In this case, the uniform electric field has a magnitude of 40 N/C, and the charge is -6 C. Plug these values into the formula:

F = (-6 C) * (40 N/C)

F = -240 N

Therefore, the magnitude of the force on the charge is 240 N. Note that the negative sign indicates that the force is directed downward.

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It is properly calibrated, and there is no external magnetic field present.

A galvanometer is a sensitive instrument used for detecting and measuring small electric currents. It works on the principle of electromagnetic induction and consists of a coil of wire suspended within a magnetic field. When a current is passed through the coil, it experiences a force due to the interaction with the magnetic field, causing the coil to rotate.

In order for a galvanometer to show a reading of zero, the following conditions must be met:

No current is flowing through the galvanometer - When there is no current passing through the galvanometer, there will be no interaction between the magnetic field and the coil, and the coil will remain stationary in its initial position.

The galvanometer is properly calibrated - The galvanometer must be calibrated to ensure that its zero position corresponds to no current passing through it. This calibration process involves adjusting the position of the coil or adding a small compensating magnet to the instrument.

There is no external magnetic field present - Any external magnetic field can cause the coil to rotate, even in the absence of current flowing through it. This can result in a false reading on the galvanometer. To prevent this, the galvanometer should be shielded from any external magnetic fields.

Overall, a galvanometer will show a reading of zero when no current is flowing through it, it is properly calibrated, and there is no external magnetic field present.

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The work required to bring the hollow cylinder from rest to an angular velocity of 0.305 m/s without the hoop slipping is 305.76 joules

To calculate the work required to bring the hollow cylinder from rest to an angular velocity of 0.305 m/s, we need to use the formula,

work = (1/2)I[tex]w^{2}[/tex]

where I is the moment of inertia of the cylinder, and ω is the angular velocity.

From the problem statement, we know that the moment of inertia of the cylinder is [tex]mR^2[/tex], where m is the mass of the cylinder and R is its radius. Plugging in the given values, we get

I = [tex](73 kg)(3.05 m)^2[/tex] = 6665.35 [tex]kg m^2[/tex]

Next, we plug in the given angular velocity,

ω = 0.305 m/s

Now we can calculate the work,

work = [tex](1/2)(6665.35 kg m^2)(0.305 m/s)^2[/tex] = 305.76 joules

Therefore, the work required to bring the hollow cylinder from rest to an angular velocity of 0.305 m/s is 305.76 joules.

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A roller coaster starts from rest at its highest point and then descends on it ( frictionless) track. Its speed is 30 m/s when it reaches ground level. We have to find its speed when its height was half of its starting point.

To solve this problem, we can use the conservation of energy principle.

At the highest point, the roller coaster has only potential energy, which is equal to its initial potential energy when it is at half of its starting point, it will have lost some potential energy and gained an equal amount of kinetic energy. Since there is no friction, the total mechanical energy of the roller coaster is conserved. Therefore, we can set the initial potential energy equal to the final kinetic energy and solve for the final velocity.

Initial potential energy = Final kinetic energy
mgh = [tex]\frac{1}{2} mv^{2}[/tex]

where m is the mass of the roller coaster, g is the acceleration due to gravity, h is the initial height of the roller coaster, and v is the final velocity.

We can simplify this equation by canceling out the mass and rearranging:

gh = [tex]\frac{1}{2} v^{2}[/tex]

When the roller coaster is at half of its starting point, its initial height (h) is divided by 2. Therefore, we can plug in the given values and solve for the final velocity:

g(h/2) = [tex]\frac{1}{2} v^{2}[/tex]
(9.8 m/s^2)(h/2) = [tex]\frac{1}{2} v^{2}[/tex]
(9.8 m/s^2)(50 m/2) = [tex]\frac{1}{2} v^{2}[/tex]
v = 21 m/s

Therefore, the answer is (C) 21 m/s.

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a. To find the period of oscillation of a large simple pendulum with a length of 2m and a mass of 3kg, we can use the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. Plugging in the values, we get T = 2π√(2/9.81) ≈ 2.02 seconds.

b. If the length of the pendulum is doubled to 4m, we can use the same formula to find the new period. Plugging in the new value for L, we get T = 2π√(4/9.81) ≈ 4.04 seconds. So doubling the length of the pendulum results in a doubling of the period.

c. If the length of the pendulum is shortened back to 2m, but the mass is doubled to 6kg, we can again use the same formula to find the new period. Plugging in the new values for L and the mass, we get T = 2π√(2/9.81) ≈ 1.43 seconds. So doubling the mass while keeping the length constant results in a shorter period.

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When a sound wave enters water from air, and the speed of sound in water is 5 times larger than the speed of sound in air, the wave number of the sound wave will increase.

Here's a step-by-step explanation:

1. First, recall that the wave number (k) is defined as 2π divided by the wavelength (λ): k = 2π/λ.
2. The relationship between wavelength, frequency (f), and speed of sound (v) is given by the equation v = fλ.
3. As the sound wave enters water, its frequency (f) remains constant, while the speed of sound (v) increases by a factor of 5.
4. Using the equation v = fλ, you can deduce that if the speed of sound (v) increases by a factor of 5, the wavelength (λ) must also increase by a factor of 5 to maintain the same frequency (f).
5. Now, considering the wave number formula k = 2π/λ, as the wavelength (λ) increases by a factor of 5, the wave number (k) will decrease by a factor of 5.

In conclusion, when the sound wave enters water from air, the wave number of the sound wave will decrease by a factor of 5.

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We can apply the equations for constant acceleration only for Situation 2, where the angular acceleration is constant.

To determine whether we can apply the equations for constant acceleration, we need to check whether the angular acceleration is constant for each situation. We can calculate the angular acceleration as the second derivative of the angular position with respect to time:

α = d^2θ/dt^2

If the angular acceleration is constant, then we can use the equations for constant acceleration. Otherwise, we need to use the more general equations for angular motion with variable acceleration.

Situation 1:

θ(t) = 2t^3 - 3t^2 + 6t - 1

Taking the first derivative:

dθ/dt = 6t^2 - 6t + 6

Taking the second derivative:

d^2θ/dt^2 = 12t - 6

The angular acceleration is not constant, so we cannot use the equations for constant acceleration.

Situation 2:

θ(t) = 4t^2 - 2t + 3

Taking the first derivative:

dθ/dt = 8t - 2

Taking the second derivative:

d^2θ/dt^2 = 8

The angular acceleration is constant, so we can use the equations for constant acceleration.

Situation 3:

θ(t) = 2t^3 - 3t^2 + 6t - cos(t)

Taking the first derivative:

dθ/dt = 6t^2 - 6t + 6 + sin(t)

Taking the second derivative:

d^2θ/dt^2 = 12t - 6 + cos(t)

The angular acceleration is not constant, so we cannot use the equations for constant acceleration.

Situation 4:

θ(t) = 6t + 2sin(t)

Taking the first derivative:

dθ/dt = 6 + 2cos(t)

Taking the second derivative:

d^2θ/dt^2 = -2sin(t)

The angular acceleration is not constant, so we cannot use the equations for constant acceleration.

Therefore, we can apply the equations for constant acceleration only for Situation 2, where the angular acceleration is constant.

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The take-off angle that will give her the shortest flight time is 90 degrees.

When a dancer jumps, their velocity, take-off angle, and flight time are interconnected. At a 90-degree take-off angle, the dancer is jumping straight upward, minimizing horizontal motion and focusing all energy into vertical height. This results in the shortest flight time, as the dancer's time in the air is determined by the time it takes to reach the peak of the jump and then fall back down due to gravity.

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The dart will oscillate with an angular frequency of approximately 56.47 rad/s.

In physics, angular frequency "ω" is a scalar measure of rotation rate. It refers to the angular displacement per unit time or the rate of change of the phase of a sinusoidal waveform, or as the rate of change of the argument of the sine function.

To find the angular frequency with which the dart oscillates, we can use the spring constant, the mass of the dart, and the equation for angular frequency.

Step 1: Identify the spring constant (k) and the mass of the dart (m).
k = 220 N/m
m = 0.069 kg

Step 2: Use the equation for angular frequency (ω) in a spring-mass system.
ω = √(k/m)

Step 3: Plug the values of k and m into the equation and solve for ω.
ω = √(220 N/m / 0.069 kg)

Step 4: Calculate the angular frequency.
ω ≈ 56.47 rad/s

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Answer:

P = 2 π (L / g)^1/2     describes period of a simple pendulum

The period of the pendulum does not depend on the mass (weight)

Pendulum b has the same period (frequency) as pendulum a

The final temperature if you changed the relative amounts of water so that there was the same amount of hot water in the film container but less cool water in the cup would be higher.

1. The hot water has a higher temperature than the cool water.
2. When the two are mixed, the hot water transfers some of its heat energy to the cool water, causing the cool water to increase in temperature.
3. Since there is less cool water, it requires less heat energy to increase its temperature.
4. The hot water will still have more heat energy remaining after transferring some to the cool water.
5. The final mixture will have a higher temperature because the remaining heat energy in the hot water is distributed among less cool water.

So, if you change the relative amounts of water to have less cool water in the cup, the final temperature of the mixture will be higher.

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Answer:

below

Explanation:

When sound travels from a loudspeaker to a microphone, it does so through a medium such as air or water. The sound waves created by the loudspeaker cause the particles in the medium to vibrate, creating a disturbance in the air molecules.The vibrating air molecules then collide with other molecules, causing them to vibrate as well. This creates a chain reaction, with the sound wave traveling through the medium until it reaches the microphone.

As the sound wave reaches the microphone, it causes the diaphragm of the microphone to vibrate. This vibration creates an electrical signal that is then transmitted to a recording device, such as a computer or tape recorder.

The electrical signal is then processed and stored in the recording device, where it can be played back later. The sound wave is thus converted into an electrical signal, which can be manipulated and processed as needed for various applications.

Overall, the process of sound traveling from a loudspeaker to a microphone involves the conversion of sound waves into electrical signals through the vibration of particles in a medium.

James Prescott Joule was the scientist who performed the first experiment demonstrating the equivalence of mechanical energy and heat.

The first experiment that demonstrated the equivalence of mechanical energy and heat was performed by James Prescott Joule in the mid-19th century. Joule's experiment involved measuring the increase in temperature of water as it was stirred by paddles driven by falling weights.

He showed that the amount of mechanical work done by the falling weights was equivalent to the amount of heat generated in the water. This work was instrumental in establishing the principle of conservation of energy, which states that energy cannot be created or destroyed, only converted from one form to another.

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Conservative forces are independent of path, whereas nonconservative forces are path dependent.

Nonconservative forces are forces that do work on an object that depends on the path taken by the object, rather than just its initial and final positions. Examples of nonconservative forces include friction, air resistance, and tension in a rope that is being stretched. When nonconservative forces are present, the equation of deltaE= delta U + delta K still holds, but the change in energy (deltaE) must take into account the work done by the nonconservative forces.
Conservative forces, on the other hand, are forces that do not depend on the path taken by the object, only on its initial and final positions. Examples of conservative forces include gravity and electrostatic forces.

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The torque on the particle about the point with coordinates (-1.00 m, -2.00 m, 5.00 m) is:

τ = (-5.00 Nm)i^ + (15.0 Nm)j^ + (13.0 Nm)k^

The torque τ about a point is given by the cross product of the vector from the point to the particle and the force acting on the particle:

τ = r x F

where r is the vector from the point to the particle, and x denotes the cross product.

We can find the vector r by subtracting the coordinates of the point from the coordinates of the particle:

r = r_particle - r_point

r = (2.00 m, 3.00 m, 4.00 m) - (-1.00 m, -2.00 m, 5.00 m)

r = (3.00 m, 5.00 m, -1.00 m)

Now we can calculate the torque by taking the cross product of r and F:

τ = r x F

τ = (3.00 m, 5.00 m, -1.00 m) x (5.0 N)i^ + (-1.00 N)k^

τ = (5.00 N)(-1.00 m)i^ + (15.0 Nm)j^ + (13.0 Nm)k^

Therefore, the torque on the particle about the point with coordinates (-1.00 m, -2.00 m, 5.00 m) is:

τ = (-5.00 Nm)i^ + (15.0 Nm)j^ + (13.0 Nm)k^

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The Earth and the moon will be stronger at the smaller distance, causing the moon to move faster and complete its orbit in less time.

C less than 27.3 days.

The orbital period of a satellite (like the moon) depends on its distance from the object it's orbiting (like the Earth). According to Kepler's laws of planetary motion, the square of the orbital period is directly proportional to the cube of the semi-major axis of the orbit.

If the moon is brought into a new circular orbit with a smaller radius (i.e., a smaller semi-major axis), its orbital period will decrease. This is because the gravitational force between the Earth and the moon will be stronger at the smaller distance, causing the moon to move faster and complete its orbit in less time.

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