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bearhunter [10]
3 years ago
15

A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.

Physics
1 answer:
charle [14.2K]3 years ago
6 0

Answer:

<em>1.</em> <em>39068.07 N</em>

<em>2. 19534.036 N</em>

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

<em>height below tank surface = 10 - 0.01 = 9.99</em>

pressure at this depth =  1000 x 9.81 x 9.99 = 98001.9 Pa

<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>

surface area of plug = πr^{2} = 3.142 x 0.25^{2} = 0.196 m^{2}

<em>force required to lift left plug = pressure x area</em>

F =  199326.9 x 0.196 = <em>39068.07 N</em>

<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>

A = 0.196/2 = 0.098 m^{2}

force F required to lift right plug =  199326.9  x 0.098 =<em> 19534.036 N</em>

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sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

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\displaystyle d = \frac{at^2}{2}

Substituting the value of the acceleration:

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\displaystyle a'=\frac{4F}{m}

And the distance is now:

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Thus:

d' = 4d

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3 0
2 years ago
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