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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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How much force do you push down on the Earth with? (Hint: 1kg is about 2 pounds).​
Gennadij [26K]

Answer: 100lbs

Explanation: The Earth pushes you down at 100lbs, so you push down the earth by 100lbs which is enough to keep you firmly attached to the ground and not allow you to jump more than a couple of feet.

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A box-shaped metal can has dimensions 5 in. by 19 in. by 4 in. high. All of the air inside the can is removed with a vacuum pump
GuDViN [60]

Answer:

The force is  F  =  1397 lb

Explanation:

From the question we are told that

    The length of the box is  l  =  19 \ in

    The width of the box is  w =  5 \ in

     The height is  h  =  4\ in

The pressure experience on one of the sides is mathematically represented as

     p = \frac{F}{A}

Where A is the area of the box which is mathematically evaluated as

    A =  l * w

substituting values

     A =  5 *19

      A = 95 \ in^2

This pressure is equivalent to the atmospheric pressure which has a constant value of  p = 14.7 pi

This implies that

        14.7  = \frac{F}{95}

=>   F  =  14.7 *95

=>    F  =  1397 lb

       

5 0
3 years ago
A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo
Shalnov [3]

Answer:

The Hydrostatic force is   F  =  137.2 kN

The location of pressure center is  Z  = 1.333 \ m  

Explanation:

From the question we are told that

   The height of the gate is  h =  3 \ m

     The weight of the gate is  w =  7 \  m

      The height of the water is  h_w  =  2 \ m

       The density of water is \rho_w  =  1000 \ kg/m^3

Note used h_w for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water  is mathematically represented as

         A =  h_w  * w

substituting values

         A =  2*  7

         A =  14  \ m^2

The hydrostatic force is mathematically represented as

          F  =  \rho_w * g * h_f * A

Where

            h_f =h-  h_w

           h_f =3 -2

           h_f = 1\ m  

So  

              F  =  1000 * 9.8 * 1 * 14

            F  =  137.2 kN

The center of pressure is mathematically represented as

        Z  =  h_f + \frac{I_g}{h_f * A}

Where I_g is the moment of inertia of the gate which mathematically represented as

            I_g =  \frac{w * h_w^2}{12}

The h_w is the height of gate immersed in water

            I_g =  \frac{7  * 2^2 }{12}

             I_g = 4.667\ kg  m^2

Thus  

        Z  = 1  + \frac{4.66}{1 * 14}

        Z  = 1.333 \ m

3 0
3 years ago
Can someone help me?!!!!!
SOVA2 [1]

Answer:magnitude -5; angle 160°

Explanation:

Vector A is described as having magnitude 5 and angle -20°.

To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.

5 @ -20° = 5 @ 340°

5 @ -20° = -5 @ 160°

The third one is the answer.

8 0
3 years ago
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