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olasank [31]
3 years ago
10

Find the total power developed in the circuit. Express your answer using three significant figures and include the appropriate u

nits.
Physics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:

37 W

Explanation:

Power is the time rate of dissipation or absorbing energy. The power supplied or absorbed by an element is the product of the current flowing through the element and the voltage across the element. Power is measured in watts. If the power is positive then it is absorbed by the element and if it is negative then it is supplied by the element.

Power = voltage * current

For element A: Power = 36 V * -4 A = -144 W

For element B: Power = -20 V * -4 A = 80 W

For element C: Power = -24 V * 4 A = -94 W

For element D: Power = -80 V * -1.5 A = 120 W

For element E: Power = 30 V * 2.5 A = 75 W

The total power developed in the circuit = sum of power through the element = (-144 W) + 80 W + (-94 W) + 120 W + 75 W = 37 W

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Answer:

a = 9.81[m/s^2]; v = 18683.5[m/s]

Explanation:

The stament of the problem is:

Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).

a. What acceleration does he experience due to the earth's pull?

b. What tangential velocity must he possess in order that he orbit safely (in m/s)?

First we need to convert all the initial data to units of the SI

Rs = 310 [mil] = 498897 [m]

RT= 400 [mil] = 643738 [m]

r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"

F=G*\frac{M*m}{r^{2} } \\where:\\

G = universal gravitational constant

M = mass of the earth [kg]

m = mass of the man [kg]

r = distance from the center of the earth to the man [m]

a)

The acceleration he is experimenting is the same acceleration given by the gravity, therefore:

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

b)

To find the tangential velocity, we must determinate the force exerted by the earth.

Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.

G = 6.67*10^{-11} [\frac{N*m^{2} }{kg^{2} } ]\\M=5.98*10^{24}[kg]\\ m=100 [kg]\\Replacing:\\F = G*\frac{M*m}{r^{2} }

F = 6.67*10^{-11}*\frac{5.98*10^{24}*100 }{1142635^{2} } \\ F= 30550 [N]

And this force will be equal to the following expression:

F = m*\frac{v^{2} }{r} \\where:\\v= tangential velocity [m/s]\\v=\sqrt{\frac{F*r}{m} } \\v=\sqrt{\frac{30550*1142635}{100} } \\v=18683.5[m/s]

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What does a ‘0’ mean in binary code?
posledela

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Feeling a bit better about not getting the Nobel Prize, Meitner drove home from the rink. If her 1500kg car accelerated from res
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Answer:

<em>The force exerted by the car engine was 3000 N</em>

Explanation:

<u>Mechanical Force</u>

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On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

v_f=v_o+a.t

Where:

vf is the final speed

vo is the initial speed

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The question describes how Meitner drove home taking her car from rest to a speed of 20 m/s in 10 seconds. This provides us the following data:

vf=20 m/s, v0=0 (rest), t = 10 seconds.

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\displaystyle a=\frac{v_f-v_o}{t}

\displaystyle a=\frac{20\ m/s-0}{10\ s}=2\ m/s^2

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