Answer:
the moment of inertia with the arms extended is Io and when the arms are lowered the moment
I₀/I > 1 ⇒ w > w₀
Explanation:
The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,
L₀ = L_f
I₀ w₀ = I w
w =
w₀
where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms
I₀ = I_cylinder + 2 m r²
where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.
If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be
I <I₀
I₀/I > 1 ⇒ w > w₀
therefore the angular velocity (rotations) must increase
in this way the skater can adjust his spin speed to the musician.
Temperature doesn't do anything, the boiling point of stuff decreases. If you put water in a vacuum chainber then it will start to boil
Answer:

Explanation:
Since the hoop is rolling on the floor so its total kinetic energy is given as

now for pure rolling condition we will have

also we have

now we will have


now by work energy theorem we can say



now solve for final speed

Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
<u>Explanation:</u>
Given,
Temperature, T = 0°C
Initial mass, Mi = 62kg
Speed, s = 5.48m/s
Distance, x = 26.8m
Friction is present.
Mass of ice melted = ?
We know,
The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice
and

Therefore, 
KE = 930.94 Joules
Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.
Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.
Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g