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Juliette [100K]
1 year ago
8

An electrical short cuts off all power to a submersible diving vehicle when it is a distance of 28 m below the surface of the oc

ean. The crew must push out a hatch of area 0.70 m2 and weight 200 N on the bottom to escape.If the pressure inside is 1.0 atm, what downward force must the crew exert on the hatch to open it?
Physics
1 answer:
skelet666 [1.2K]1 year ago
6 0

Answer:

F=126339.5N

Explanation:

to find the necessary force to escape we must make a free-body diagram on the hatch, taking into account that we will match the forces that go down with those that go up, taking into account the above we propose the following equation,

Fw=W+Fi+F

where

Fw=   force or weight produced by the water column above the submarine.

to fint Fw we can use the following ecuation

Fw=h. γ. A

h=distance

γ= specific weight for seawater = 10074N / m ^ 3

A=Area

Fw=28x10074x0.7=197467N

w is the weight of the hatch = 200N

Fi is the internal force of the submarine produced by the pressure = 1atm = 101325Pa for this we can use the following formula

Fi=PA=101325x0.7=70927.5N

finally the force that is needed to open the hatch is given by the initial equation

Fw=W+Fi+F

F=Fw-W+Fi

F=197467N-200N-70927.5N

F=126339.5N

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In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6
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Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

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diameter of pipe outlet, d₂ = 15 cm

                                      r₂= 7.5 cm

head upto water level is to rise = 60 + 5

                                          = 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q  

 π r₁² v₁ = π r₂² v₂  = 0.015

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{7.5^2}{5^2} v_2

 v_1= 2.25 v_2

 v_2 = \dfrac{0.015}{\pi r_2^2}

 v_2 = \dfrac{0.015}{\pi 0.075^2}

    v₂ = 0.848 m/s

    v₁ = 1.908 m/s

Applying Bernoulli's equation

 P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h

 P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65

 P_p= 635539.32 Pa

 P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

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2 years ago
A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that
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Answer:

(mv^2/R)/(mg)=1/2

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In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move
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Answer:

To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.

To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.

Explanation:

In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.

On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.  

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rodikova [14]

Answer:

See the explanation below.

Explanation:

12) When an object is falling, how does the objects velocity change? what formula do you use?

The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].

The formula is:

v=v_{0}+g*t

where:

vo = initial velocity = 0

g = gravity = 9.81[m/s^2]

t = time [s]

13)

what is a falling speed at 6s, 9s, 112s?

v = 0 + (9.81*6) = 58.86[m/s]

v = 0 + (9.81*9) = 88.29 [m/s]

v = 0 + (9*112) = 1098.72 [m/s]

14)

If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?

The formula is the same:

v=v_{o}+g*t

65 = 0 + 9.81*t

t = 65/9.81

t = 6.62[s]

15)

What formula is used to determine the distance an object is falling ?

y = y_{o}+v_{o}*t + 0.5*9.81*t^{2}

where:

y = distance [m]

yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero

vo = initial velocity, if it is free fall, then = 0

t = time [s]

g = gravity = 9.81[m/s^2]

This equation will be reduce to:

y =   0.5*g*t^{2}

16)

using the times given in problem 13. Determine the distance fallen for each.

y = 0.5*9,81*(6)^2 = 176.58 [m]

y = 0.5*9,81*(9)^2 = 397.3 [m]

y = 0.5*9,81*(112)^2 = 61528.3 [m]

17)

If an object has fallen a distance of 87.3 [m]. How long was it falling?

87.3 = 0.5*9.81*t^2

t=\sqrt{\frac{87.3}{0.5*9.81} }\\ t=4.21[s]

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