Answer:
The time is
Explanation:
From the question we are told that
The height of the platform where the ball was dropped from is s = 215 m
The acceleration due to gravity is
Generally from kinematic equations
Here u is the initial velocity of the ball and the value is u = 0 m/s given that the ball was at rest before it was dropped
So
=>
=>
=>
Answer:
Work done, W = 2675.4 J
Given:
mass, m = 70.0 kg
height, H = 3.90 m
Solution:
According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:
Work done, W = mgH
where
g = acceleration due to gravity =
The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
<h3>What is the law of conservation of momentum?</h3>According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of hockey player 1= 91.0-kg
(m₂) is the mass of hockey player 2= 91.0-kg
(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.
(u₂) is the velocity before the collision of hockey player 2=?
a)
Momentum before the collision;
Momentum before the collision = 1237.6 kg m/s².
b)
The velocity of the two hockey players after the collision from the law of conservation of the momentum as:
Momentum before collision = Momentum after the collision
1237.6 kg m/s² = (m₁+m₂)V
1237.6 kg m/s² =(2 ×91.0-kg )V
V=6.8 m/sec.
Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
To learn more about the law of conservation of momentum refer;
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