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Nadusha1986 [10]
2 years ago
9

A tall tube is evacuated, and its stopcock closed. The open end of the tube is immersed into a container of water (density 103 k

g/m3) that is open to the atmosphere (pressure 105 N/m2). When the stopcock is opened, how far up the tube will the water rise?
Physics
1 answer:
Anna71 [15]2 years ago
7 0

Answer:

The water will rise to a height, H = 10.20 m

Given:

Atmospheric Pressure, P = 10^{5} N/m^{2}

density of water, \rho_{w} = 10^{3} kg/m^{3}

Solution:

The water in the tube will rise to the point where the weight is balanced by the force as a result of the atmospheric pressure.

Therefore,

Pressure, P = \rho_{w} gH

where

g = acceleration due to gravity = 9.8 m/s^{2}

H = height of water in the tube

Now,

H =\frac{P}{\rho_{w} g}

Putting the values in the above formula:

H =\frac{10^{5}}{10^{3}\times 9.8}

H = 10.20 m

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The graph of the problem is missing, find it in attachment.

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P = 18 W

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2 years ago
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