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Pachacha [2.7K]
2 years ago
6

If you ride quickly down a hill on a bicycle your eardrums are pushed in before they pop back. Why is this?

Physics
1 answer:
Mice21 [21]2 years ago
7 0

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.

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Explanation:

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However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.

Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.

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Eddie

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1 year ago
La) What is meant by the term Basic Quantities.​
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Answer:

A basic quantity is basically the physical quantity that can not be defined in terms of other quantities.

Explanation:

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Some of the names of the basic quantities include:

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Answer:

a)h_{max}=14536.16 m

b)h = 15687.9 m

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Explanation:

a) Using the energy conservation we have:

E_{initial}=E_{final}

we have kinetic energy intially and gravitational potential energy at the maximum height.

\frac{1}{2}mv^{2}=mgh_{max}

h_{max}=\frac{v^{2}}{2g}

h_{max}=\frac{43^{2}}{2*0.0636}

h_{max}=14536.16 m  

b)  We can use the equation of the gravitational force

F=G\frac{mM}{R^{2}}   (1)

We have that:

F = ma    (2)

at the surface G will be:

G=\frac{gR^{2}}{M}

Now the equation of an object at a distance x from the surface.

is:

F=\frac{mgR^{2}}{(R+x)^{2}}

m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}

Using that dv/dt is vdx/dt and integrating in both sides we have:

v_{0}=\sqrt{\frac{2gRh}{R+h}}

h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}

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c) The difference is:

So the percent difference will be:

PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%

PD=7.62\%

The estimate is low.

I hope it helps you!

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Answer:

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