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Vinil7 [7]
3 years ago
5

A person drives to the top of a mountain. On the way up, the person’s ears fail to “pop,” or equalize the pressure of the inner

ear with the outside atmosphere. The pressure of the atmosphere drops from 1.010 x 10^5 Pa at the bottom of the mountain to 0.998 x 10^5 Pa at the top. Each eardrum has a radius of 0.40 cm. What is the pressure difference between the inner and outer ear at the top of the mountain?
a.

1.2 x 10^3 Pa

b.

1.1 x 10^5 Pa


c.

1.0 x 10^2 Pa


d.

1.2 x 10^2 Pa
Physics
1 answer:
KATRIN_1 [288]3 years ago
3 0
The pressure on the inner ear is calculated by subtracting the pressure of the atmosphere of the bottom from the top, so calculating this will give us: 1.010x10^5 - .998x10^5 = 1200Pa outward which would be letter A.
And so the net force would be now calculated as P*A = 1200Pa*π*(0.40x10^-2m)^2 = 0.0603N
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Answer: n=4

Explanation:

We have the following expression for the volume flow rate Q of a hypodermic needle:

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Viscosity of the liquid \eta=\frac{M}{LT}

We need to find the value of n whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}  (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}  (3)

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}  (4)

As we can see n must be 4 if we want the exponent to be 3:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}  (5)

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8 0
4 years ago
Two billiard balls with the same mass undergo a perfectly elastic head-on collision. if one ball's initial speed was 2 m/s, and
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<span>The ball with an initial velocity of 2 m/s rebounds at 3.6 m/s
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  I'll use the following variables
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   For conservation of momentum, we can create this equation:
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a0 + b0 = a1 + b1

   For conservation of energy, we can create this equation:
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Answer:

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Part B.

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Part C.

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Part H

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