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Ugo [173]
3 years ago
15

A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assume the density of

sea water is 1.03 × 103 kg/m3. show answer No Attempt Calculate the force, in newtons, needed to open the hatch from the inside, given it is circular and 0.25 m in diameter. The air pressure inside the submarine is 1.00 atm.
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer:

F = 49.55 x 10 ³ N

Explanation:

Given :

ρ = 1.03 x 10³ kg / m³ ,  h= 25 m , g = 9.8 m / s² , r = 0.25 m

So solve to determine force

A = π * r² = π * (0.25 m)²

A = 196.34 x 10 ⁻³ m²

The pressure of the water

P = F / A

Determine the force using the pressure into the area

F = P * A

P = ρ * g * h

Replacing numeric

P = 1.03 x 10³ kg / m³ * 9.8 m / s² * 25 m

P = 252.35 x 10³ N / m²

F = P * A  ⇒ F = 252.35 x 10³ N / m²  * 196.34 x 10 ⁻³ m²  

F = 49.55 x 10 ³ N

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Which of the following is the best description of a gaseous substance?
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a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resi
777dan777 [17]

Answer:

<h2>d₂ = 3d</h2><h2>The diameter of the second wire is 3 times that of the initial wire.</h2>

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

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