What best describes an impulse acting on an object? the net force on an object divided by the time of impact the velocity of an
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Answer:
The ball is dropped at a height of 9.71 m above the top of the window.
Explanation:
<u>Given:</u>
- Height of the window=1.5 m
- Time taken by ball to cover the window height=0.15
Now using equation of motion in one dimension we have
Let u be the velocity of the ball when it reaches the top of the window
then
Now u is the final velocity of the ball with respect to the top of the building
so let t be the time taken for it to reach the top of the window with this velocity
Let h be the height above the top of the window
Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
