The moment of inertia of the flywheel is 2.63 kg-
It is given that,
The maximum energy stored on the flywheel is given as
E=3.7MJ= 3.7×
J
Angular velocity of the flywheel is 16000
= 1675.51
So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :
E = 

By rearranging the equation:
I = 
I = 2.63 kg-
Thus the moment of inertia of the flywheel is 2.63 kg-
.
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Answer:
Potential energy of book = 7.5 J
Explanation:
Given:
Weight of book = 5 N
Height of shelf = 1.5 meter
Find:
Potential energy of book
Computation:
Weight = Mass x Acceleration of gravity
Mass x Acceleration of gravity = 5 N
Potential energy = Mass x Acceleration of gravity x Height
Potential energy of book = Mass x Acceleration of gravity x Height
We know that;
Mass x Acceleration of gravity = 5 N
So,
Potential energy of book = 5 x 1.5
Potential energy of book = 7.5 J
It's 12.1 m/s, assuming that's the launch velocity that's given.
For projectile motion, velocity's y-component is parabolic/quadratic. It's x-component is constant, so you don't need to know it.
Answer:
See explanation below
Explanation:
The equation to use for this is the following:
dU = q + w
As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.
Therefore the change in the energy would be:
dU = -2.59x10^4 - 6.46^4
dU = -9.05x10^4 kJ
Answer:
B. 0.552
Explanation:
To find the resistance in the circuit above, u simply divide the current in the circuit by the voltage to get the resistance.