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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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Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 ×  10^{14} Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

E=h\nu_{o}    ..................1

here E is the energy of electron per atom E=\frac{x}{N}  and h is plank constant i.e. 6.626\times10^{-34} Js  and x is  binding energy

and here N is the Avogadro constant = 6.023\times10^{23}

so E will \frac{x}{6.023\times10^{23}}  

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2 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
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r = \frac{10nm}{\sqrt2} = 7.07 nm

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now the net potential is given as

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now potential energy of alpha particle at this position

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here we know that

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r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

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2 years ago
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Answer:

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<h3>Question:</h3>

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

<h3>Answer:</h3>

1.6nT [in the negative z direction]

<h2>Explanation:</h2>

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I = 20A

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B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{25.0004})

B = (10⁻⁷ x 20 x 0.02) / (25.0)

B = 1.6 x 10⁻⁹T

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2 years ago
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