3 years ago

20.

Physics

2 answers:

sveticcg [70]3 years ago

4 0

Answer:

I believe the answer is A and D.

I am unsure of C.

MissTica3 years ago

3 0

The answer should be D

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$1.20(\pi 18^2) = L_2(\pi 5^2)$

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So here the net pressure on the smaller area is given as

$P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)$

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$P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)$

$P_{ex} = 2.49\times 10^5 Pascal$

now the force required on the other side is given as

$F = P_ex (area)$

$F = (2.49 \times 10^5)(\pi (0.05)^2)$

$F = 1958.4 N$

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