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vodomira [7]
2 years ago
7

An ear squeeze occurs when: Select one: The pressure inside the middle ear space is greater than ambient (surrounding) pressure.

The pressure inside the middle ear space is less than ambient (surrounding) pressure. Divers forget to wear ear plugs. None of the above.
Physics
1 answer:
matrenka [14]2 years ago
6 0

B. An ear squeeze occurs when the pressure inside the middle ear space is less than ambient (surrounding) pressure.

<h3>When external ear squeeze occurs</h3>

External ear squeeze occurs when gas trapped in the external ear canal remains at atmospheric pressure while the external water pressure increases during descent.

Thus, we can conclude that, an ear squeeze occurs when the pressure inside the middle ear space is less than ambient (surrounding) pressure.

Learn more about ear squeeze here: brainly.com/question/11430998

#SPJ1

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NEED ASAP PLEASE
miskamm [114]

Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.

6 0
3 years ago
A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches
Luden [163]
Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.
We have been given a value of dr/dt and are required to find dh/dt
Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:
128 = πr²h
h = 128/πr²
Now we differentiate both sides:
dh/dr = -256/πr³
Applying the chain rule:
dh/dt = dh/dr x dr/dt
dh/dt = (-256/πr³) x -0.05
dh/dt = 64/5πr³; substituting the value of r
dh/dt = 64/5π(1.5)³
dh/dt = 1.21 in/sec
4 0
3 years ago
A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
3 years ago
If I travel 300 m east, then 400 m west, what is my distance &amp;<br> displacement?
ad-work [718]

Answer:100m west

Explanation:

6 0
3 years ago
A car dropped from a height of 44 meters fall to a height of zero meters. How fast will the car be traveling as it hits the grou
FinnZ [79.3K]
Vi=0m/s
Vf=?
A=9.81
D=44
T=not needed

Vf^2=Vi^2+2ad
Vf=2ad square rooted
Vf=2(9.81)(44) square root it
Vf=29.3m/s
6 0
3 years ago
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