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.
Answer:0
Explanation:
Given
two particles of mass m and 3m
They move towards each other with velocity v
As the collision is Elastic therefore value of coefficient of restitution is 1
if 
and 
are velocity of approach of m and 3 m
and 
and 
are velocity after collision respectively
therefore
conserving momentum
Therefore velocity of heavier mass is zero
Answer:
Explanation:
Conservation of angular momentum
The MGR originally has momentum
L = 100(3.0) = 300 kg•m²/s²
The child can be thought of as a point mass with I = mr²
When she jumps onto the rim of the MGR
300 = (100 + 22(2.0²)ω
ω = 300 / 188 = 1.5957... 1.6 rad/s
As she moves toward the center of the MGR, her moment of inertia goes to zero as her radius goes to zero.
The angular velocity when she reaches the center will again be 3.0 rad/s
the force acting on that object
