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3 years ago

8

An exploration submarine should be able to descend 1200 m down in the ocean. If the ocean density is 1020 kg/m3, what is the max

imum pressure on the submarine hull?

1 answer:

motikmotik3 years ago

8 0

Answer:

11995200 N/m²

Explanation:

Pressure: The is the ratio of force to the surface area in contact. The S.I unit of pressure is N/m².

Generally pressure in fluid can be expressed as

P = ρgh......................... Equation 1

Where P = maximum pressure on the submarine hull, ρ = Density of ocean, h = depth of ocean, g = acceleration due to gravity.

Given: h = 1200 m, ρ = 1020 kg/m³

Constant: g = 9.8 m/s²

Substitute into equation 1

P = 1200(1020)(9.8)

P = 11995200 N/m²

Hence the maximum pressure on the submarine hull = 11995200 N/m²

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Answer:

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Answer:b) atoms

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3 years ago

Read 2 more answers

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aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

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It is given that,

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Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

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Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

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The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

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So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

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Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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Answer:

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taurus [48]

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