Answer:
The impulse on the ball delivered by the floor is 2.52 kg-m/s.
Explanation:
Given that,
Mass of the ball, m = 0.22 kg
It is dropped from an initial height of 1.80 m. It rebounds back after colliding with the floor to a final height of 1.50 m. Initial velocity and final velocity can be calculated using conservation of energy as :
Final velocity,
As the ball rebounds, v = -5.47 m/s
We need to find the impulse on the ball delivered by the floor. We know that impulse is equal to the change in momentum as follows :
So, the impulse on the ball delivered by the floor is 2.52 kg-m/s.
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Answer:
<em>The difference in pressure between the external air pressure, and the internal air pressure of the middle ear.</em>
Explanation:
First of all, we should note that pressure decreases with height and increases with depth. The air within the middle ear (between the ear drum and the Eustachian tube) adjusts itself to respond to the atmospheric pressure, or when we yawn. At a high altitude like on the hill, the air pressure in the middle ear, is fairly low (this is to balance the low air pressure at this height). While riding down the hill quickly, there is little time for the air pressure in the ear to readjust itself to the increasing external air pressure, causing the external air to push into the ear drum. Along the way, the air within the middle ear is adjusted by the opening of the Eustachian tube, allowing more air into the space in the middle ear to balance the external air pressure. This readjustment causes the ear to pop.
