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Delicious77 [7]
4 years ago
15

What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d

epth of 9400 m in sea water? the density of sea water is 1025 kg/m3?
Physics
1 answer:
Butoxors [25]4 years ago
3 0
Calculate the pressure due to sea water as density*depth.
That is, 
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is  101.3 kPa
Total pressure is  94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
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What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs
Ann [662]

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

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The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

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The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

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<u>brainly.com/question/14226368</u>

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7 0
2 years ago
a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy
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Answer:

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F_{net}= 85.41\ N

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4 years ago
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