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KIM [24]
3 years ago
12

A computer is connected across a 110 V power supply. The computer dissipates 123 W of power in the form of electromagnetic radia

tion and heat. Calculate the resistance of the computer.
Physics
2 answers:
Reika [66]3 years ago
4 0

Answer:

98.37 ohms.

Explanation:

Power, P = I × V

Using ohms law,

V = I × R

P = V^2/R

Given:

V = 110 V

P = 123 W

R =110^2/123

= 98.37 ohms.

nexus9112 [7]3 years ago
3 0

Answer:

81.3ohms

Explanation:

Resistance is known to provide opposition to the flow of electric current in an electric circuit.

Power dissipated by the computer is expressed as;

Power = current (I) × Voltage(V)

P = IV... (1)

Note that from ohms law, V = IR

I = V/R ... (2)

Substituting equation 2 into 1, we will have;

P = (V/R)×V

P = V²/R.. (3)

Given source voltage = 100V, Power dissipated = 123W

To get resistance R of the computer, we will substitute the given value into equation 3 to have

123 = 100²/R

R = 100²/123

R = 10,000/123

R = 81.3ohms

The resistance of the computer is 81.3ohms

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Oduvanchick [21]

Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

(b) 27.013 days

(c) No, their accelerations would not be constant.

Explanation:

In the given question, we have:

Mass of female astronaut M_{1} = 60.0 kg

Mass of male astronaut M_{2} = 74.0 kg

Distance between them (S) = 23.0 m

(a) The free-body diagram is shown in the attached figure.

Using the equations below, the initial accelerations of the two astronauts can be calculated:

Force of gravity (F) = \frac{G*M_{1}*M_{2}}{S^{2} }

G = 6.67*10^-11 \frac{m^{3} }{kg*s^{2} }

F = (6.67*10^-11 *60*74)/23^2 = 5.598*10^-10 N

For the female astronaut, her initial acceleration = F/M_{1} = 5.598*10^-10/60 = 9.33*10^-12 m/s^2

For the male astronaut, his acceleration = F/M_{2} = 5.598*10^-10/74 = 7.56*10^-12 m/s^2

(b) Since the different between their mass is not much, we can deduce that:

a_{average} = \frac{a_{1}+a_{2}}{2} = (9.33*10^-12 + 7.56*10^-12)/2 = 8.445*10^-12 m/s^2

Using the equation below, we can calculate the the time:

S = ut + 1/2 (at^2)   where u = 0

23 = 1/2 (8.445*10^-12)*t^2

t^2 = 5.447*10^12

t = 2333883.044 s = 27.013 days

(c) No, their accelerations will not be constant. It will increase because their radii would be decreasing.

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A muscle inserts 1.5 cm from the joint axis and exerts 300 N of force at an angle of pull of 60 degrees. How much torque is prod
Amanda [17]

Answer:

torque = 3.897 N-m

Explanation:

given data

force = 300 N

angle = 60 degree

distance = 15 cm

to find out

torque

solution

we will apply here torque formula that is given below

torque = force × sinθ × distance    ...................1

put here all these value in equation 1

we get torque

torque  = force × sinθ × distance

torque  = 300 × sin60 × 1.5 ×10^{-2}

torque  = 300 × 0.8660 × 1.5 ×10^{-2}

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torque  = 389.711 ×10^{-2}

torque = 3.897 N-m

8 0
3 years ago
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