An ice skater spins at 2.5 rev/s when his arms are extended. He draws his arms in and spins at 10.0 rev/s. By what factor does his moment of inertia change in the process?

Answers

Answer 1

Answer:

The moment of inertia decreased by a factor of 4

Explanation:

Given;

initial angular velocity of the ice skater, ω₁ = 2.5 rev/s

final angular velocity of the  ice skater, ω₂ = 10.0 rev/s

During this process we assume that angular momentum is conserved;

I₁ω₁ = I₂ω₂

Where;

I₁ is the initial moment of inertia

I₂ is the final moment of inertia

[tex]I_2 = \frac{I_1 \omega_1}{\omega_2} = \frac{I_1*2.5}{10} \\\\I_2 = 0.25I_1 = \frac{1}{4}I_1[/tex]

Therefore, the moment of inertia decreased by a factor of 4


Related Questions

You have a large flashlight that takes 4 D-cell batteries, each with a voltage of 1.5 volts. If the current in the flashlight is 2.0 amps, what is the
resistance?
A 12 ohms
B 0.75 ohms
C 3.0 ohms
D.0.5 ohms
E. 6.0 ohms

Answers

Answer:

The answer is "C. 3.0 ohms". 

Explanation:

Ohm's law states the following:,

V = IR

where:

Voltage (V) = 1.5 volts × 4 batteries = 6.0 volts

Current (I) = 2.0 amps

Resistance (R) = ? ohms

To solve for Resistance (R) the equation must be rearranged this way:

 R = V / I

Then, the variables must be replaced with the known values:

 R = 6.0 volts / 2.0 amp

R = 3.0 ohms

The answer is  C. 3.0 ohms. 

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P.D: I got it right on PLATO

Need help on circuit question

Answers

Both would be my answer.
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