An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the angular momentum and kinetic energy of the skater affected?
Angular Momentum Kinetic Energy
(A) Increases Increases
(B) Increases Remains Constant
(C) Remains Constant Increases
(D) Remains Constant Remains Constant
(E) Decreases Remains Constant

Answers

Answer 1

An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled closer to the body, the angular momentum of the skater will remain constant while the kinetic energy of the skater increases. The correct option is C.

The angular momentum of the skater is given by

[tex]L = I\omega[/tex],

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

When the skater pulls their arms in, their moment of inertia decreases due to the decreased distance between their body and the axis of rotation.

According to the conservation of angular momentum, the product of the moment of inertia and angular velocity must remain constant. Therefore, if the moment of inertia decreases, the angular velocity must increase to keep the angular momentum constant.

The kinetic energy of the skater is given by

[tex]K = (1/2)I\omega^2[/tex]

As the moment of inertia decreases and the angular velocity increases, the kinetic energy of the skater also increases because it is proportional to the square of the angular velocity.

Therefore, the correct answer is: (C) Remains Constant Increases. The angular momentum remains constant, while the kinetic energy increases due to the increased angular velocity.

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Related Questions

water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. the density of water is 1 000 kg/m3. determine its average velocity. multiple choice question. 20 m/s 200 m/s 0.02 m/s 2 m/s 0.2 m/s

Answers

Option D: 2 m/s is the average velocity of the water flowing through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s.

According to the question:

cross-sectional area of the pipe = 0.002m²

Mass flowrate = 4 kg/s

Density of water = 1000 kg/m³

We are asked to find, average velocity =?

Average velocity is the net or total displacement covered by a body in a given time. The mass flow rate divided by the pipe's cross-sectional area and density ratio is the formula for calculating a fluid's average velocity.

As a result, the water's average flow rate through the pipe is provided by:

v = m / (ρ × A)

where, v is the average velocity, m is the mass flow rate, ρ is the density of water, and A is the cross-sectional area of the pipe. Substituting the values in the above equation we get:

v = 4 / (1000 × 0.002)

v = 2m/s

Therefore, the average velocity of water flowing through a pipe of cross-sectional area of 0.002m² is 2m/s.

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Correct question is:

Water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. The density of water is 1 000 kg/m3. Determine its average velocity. Multiple choice question.

20 m/s

200 m/s

0.02 m/s

2 m/s

0.2 m/s

a 10.0-mf capacitor is fully charged across a 12.0-v bat- tery. the capacitor is then disconnected from the battery and connected across an initially uncharged capacitor with capacitance c. the resulting voltage across each capacitor is 3.00 v. what is the value of c?

Answers

The value of  uncharged capacitor in series with a 10.0-microfarad capacitor, given that each capacitor has a voltage of 3.00 volts, can be calculated using the formula for equivalent capacitance in series and  formula for charge on a capacitor. The value of c is approximately 4.00 microfarads.

To determine the value of c, which is [tex]1/Ceq = 1/C1 + 1/C2[/tex] . Initially, the 10.0-microfarad capacitor has a charge of [tex]Q = CV = (10.0 * 10^{-6 }F) * 12.0 V = 1.20 * 10^{-4} C[/tex].

When it is connected in series with uncharged capacitor with capacitance c,  charge is shared between the two capacitors. The charge on  10.0-microfarad capacitor is also equal to the charge on  uncharged capacitor, which is given by [tex]Q = (3.00 V) * C[/tex] .

Equating the two expressions for Q and solving for c, we get [tex]C = Q/3.00[/tex] [tex]V = (1.20 * 10^{-4 C}) / (3.00 V) = 4.00 * 10^{-5 F}[/tex]. Therefore,  value of c is approximately 4.00 microfarads.

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