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maria [59]
3 years ago
6

Atmospheric pressure is greater at the base of a mountain than at

Physics
1 answer:
aliina [53]3 years ago
8 0

At the top of the mountain, when he tightens the cap onto the bottole, there is some water and some air inside the bottle.  Then he brings the bottle down to the base of the mountain.

The pressure on the outside of the bottle is greater  than it was when he put the cap on.  If anything could get out of the bottlde, it would. But it can't . . . the cap is on too tight. So all the water and all the air has to stay inside, and anything that can get squished into a smaller space has to get squished into a smaller space.

The water is pretty much unsquishable.

Biut the air in there can be <em>COMPRESSED</em>.  The air gets squished into a smaller space, and the bottle wrinkles in slightly.

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A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N
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Answer:

<h2>f₀ = 158.12 Hertz</h2>

Explanation:

The fundamental frequency of the string  f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

V = \sqrt{\frac{T}{\mu} } where T is the tension in the string and  \mu is the density of the string

Given T = 600N and \mu = 0.015 g/cm  = 0.0015kg/m

V = \sqrt{\frac{600}{0.0015} }\\ \\V =  \sqrt{400,000}\\ \\V = 632.46m/s

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

5 0
3 years ago
A Man Moved first a Distance of 1000 m in 25 second and 2.5 km in 50 second along a in straight line?​
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Average speed = 46.67 m/s

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Given that the time taken in covering first 1000 m = 25 seconds.

The time taken in covering next 2.5 km = 50 seconds.

Total distance covered = 1000 m + 2500 m = 3500 m

Total time taken = 25+50=75 seconds

Average speed = Total distance covered / total time taken

= 3500/75 = 46.67 m/s

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