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icang [17]
2 years ago
13

A submarine dives to a depth of 100-m beneath the surface of the Pacific Ocean. The density of sea water is 1030 kg/m3. The subm

arine has a hatch with an area of 2 m2 located on the top of the submarine. No need to show work. a) Calculate the gauge pressure applied on the submarine at the depth of 100 m. b) Calculate the absolute pressure applied on the submarine at the depth of 100 m. Assume the atmospheric pressure in the air above the ocean is one atmosphere. c) Calculate how much force is required in order to open the hatch from the inside of submarine. Assume that the pressure inside the submarine is one atmosphere.
Physics
1 answer:
kozerog [31]2 years ago
6 0

Answer:

A) 1010430 pa

B) 1111755 pa

C) 2020860 N

Explanation:

Guage pressure = pgh

= 1030 kg/m3 x 9.81 m/s2 x 100 m

= 1010430 pa

Absolute pressure is Guage pressure + atmospheric pressure.

= 1010430 + 101325 = 1111755 pa

If the pressure inside submarine is 1 atm, then net pressure will be

1111755 - 101325 = 1010430 pa

Force required to open hatch against this pressure will be,

F = pghA

pgh = 1010430 pa

F = 1010430 pa x 2 m^2

F = 2020860 N

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3 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
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Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

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By force balancing

2 T cosθ = 40

here tan\theta =\dfrac{5}{6}

θ=39.80°

2 T cos39.8 = 40

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