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Tatiana [17]
3 years ago
7

The pressure at the bottom of a lake is 300 000 N/m2 more than at the surface of the lake. The lake is fresh water (density = 10

00 kg/m3).
Physics
1 answer:
MakcuM [25]3 years ago
5 0

Answer:

h = 30.61 m

Explanation:

In this problem they indicate the pressure at the bottom of the lake and the density of the water, for which they must ask what is the depth of the lake.

The pressure is given by the expression

           P = P_{atm} + rho g h

           P - P_{atm} = rho g h

       

The gauge pressure value is P- P_{atm} = 3 105 N / m²

          h = \frac{P - P_{atm}}{ \rho \ g}

let's calculate

          h = \frac{3 \ 10^5}{1000 \ 9.8}

          h = 30.61 m

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Answer:

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We know that :

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\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

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t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

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t'' = 20.56 years

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