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Soloha48 [4]
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?
Physics
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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Oksana_A [137]

Answer:

m=1.01\times 10^6\ kg

Explanation:

Given that,

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Area of the square surface, A = 10² = 100 m²

We need to find the mass of vertical column of air. We know that, pressure is equal to the force acting per unit area. So,

P=\dfrac{mg}{A}\\\\m=\dfrac{PA}{g}\\\\m=\dfrac{101325\times 10^2}{10}\\\\m=1.01\times 10^6\ kg

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3 years ago
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistan
Rama09 [41]

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

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(b) Given in the x direction:

v₀ = 1.10 m/s

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Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

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vᵧ = aᵧt + v₀ᵧ

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v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

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