Complete Question
A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?
Answer:

Explanation:
From the question we are told that:
Angular velocity 
Mass 
Diameter d 
Off Time 
Oscillation at Power off 
Generally the equation for Angular displacement is mathematically given by




Generally the equation for Time to come to rest is mathematically given by



Therefore Angular displacement is


Answer:
B. twice as much kinetic energy
Explanation:
Lets take the mass of the first marble =2 m
the mass of the second marble = m
We know that velocity of particle does not depends on their mass that is the velocity of both mass will be same after dropping from the roof.
We know that kinetic energy of a mass is given as

Kinetic energy for heavier mass

Kinetic energy for light mass

KE=2 KE '
Form above two equation we can say that ,the kinetic energy for the heavier mass is twice the lighter mass.
Therefore the answer will be B.
Answer:
M' = μ₀n₁n₂πr₂²
Explanation:
Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.
So, M = N₂Ф₂₁/i₁
substituting the values of the variables into the equation, we have
M = N₂Ф₂₁/i₁
M = N₂B₁A₂/i₁
M = n₂lμ₀n₁i₁πr₂²/i₁
M = lμ₀n₁n₂πr₂²
So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²
M' = μ₀n₁n₂πr₂²
True
The half-life isn’t applicable to a first order reaction because it does not rely on the concentration of reactant present. However the 2nd order reaction is dependent on the concentration of the reactant present.
The relationship between the half life and the reactant is an inverse one.
The half life is usually reduced or shortened with an increase in the concentration and vice versa.
Answer:
The sun
Explanation:
In this system the energy of the sun heats the water in the pipe, producing a high pressured steam, which is used for moving a turbine and producing electricity, is a transformation of energy from solar to thermal, then to mechanical to electrical.