Anne releases a stone from a height of 2 meters. She measures the kinetic energy of the stone at 9.8 joules at the exact point i
2 answers:
The correct answer to the question is A) 0.5 Kg.
CALCULATION:
Before going to answer this question, first we have to understand the law of conservation of energy.
As per law of conservation of energy, the energy can neither be created nor be destroyed. It can only change from one form to another form and the total energy of universe is always constant.
As per the question, the stone was initially at a height of 2m from the ground.The energy possessed by the stone at this height is gravitational potential energy.
When the stone is released, the stone moves downward under gravity. It gains kinetic energy. When it just hits the ground, whole of its potential energy must have been converted into kinetic energy.
Hence, P.E at height 2m = K.E of the stone just before hitting the ground.
The potential energy of a body at a height h from the surface of earth is calculated as-
Potential energy P.E = mgh
Here, g is known as the acceleration due to gravity and its value is .
It is given that the height of the stone h = 2m.
The kinetic energy of the stone just before reaching ground K.E = 9.8 J.
We know- P.E of stone at height 2m = K.E of stone when it hits the ground.
⇒ mgh = 9.8 Joules
⇒ m =
=
= [ans]
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Complete answer
A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100rad/s2. After making 2844 revolutions, its angular speed is 140rad/s
(a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?
Answer:
a) 126.59 radians per second
b) 134.1 seconds
Explanation:
We can use the rotational kinematic equations for constant angular acceleration.
a) For a) let’s use:
(1)
with the initial angular velocity,
the final angular velocity,
the angular acceleration and
the revolutions on radians (2844 revolutions = 17869.38 radians). Solving (1) for initial velocity:
b) Knowing those values, we can use now the kinematic equation
with t the time, solving for t: