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3 years ago

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An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t

he local atmospheric pressure) in kPa inside of the diving chamber required to keep water from entering the chamber. (Assume SG=1.03)

1 answer:

Fofino [41]3 years ago

8 0

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

$P(z)=P_{0}+\rho _wgh$

Now the gauge pressure is given by

$P(z)-P_{0}=\rho _wgh$

Applying values we get

$P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa$

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So we find Mu.

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2 years ago

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Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

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Part A)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in upward direction = 1800 N

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Part B)

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$T$ - 7000 - 1800 = 0

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3 years ago

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Answer:

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Explanation:

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where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

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Substituting equation (5) into (4), we obtain the following;

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$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

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Answer:

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3 years ago