Answer:
When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.
Explanation:
Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.
Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.
When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.
Answer:
The answer is going to be C.
Explanation:
Trust me. Im an expert in physics
Answer:
11:1
Explanation:
At constant acceleration, an object's position is:
y = y₀ + v₀ t + ½ at²
Given y₀ = 0, v₀ = u, and a = -g:
y = u t − ½g t²
After 6 seconds, the ball reaches the maximum height (v = 0).
v = at + v₀
0 = (-g)(6) + u
u = 6g
Substituting:
y = 6g t − ½g t²
The displacement between t=0 and t=1 is:
Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]
Δy = 6g − ½g
Δy = 5½g
The displacement between t=6 and t=7 is:
Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]
Δy = (42g − 24½g) − (36g − 18g)
Δy = 17½g − 18g
Δy = -½g
So the ratio of the distances traveled is:
(5½g) / (½g)
11 / 1
The ratio is 11:1.
Answer:
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Explanation:
Answer:
A.The spring constant for B is one quarter of the spring constant for A.
Explanation:
If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.
As 
, and the 2 springs have the same mass
So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.
