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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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3 years ago
Find the area under the standard normal curve to the right of z=0.49. round your answer to four decimal places, if necessary.
VikaD [51]

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

<h3>What is  the standard normal curve?</h3>

The horizontal axis is approached by the standard normal bend as it extends indefinitely both in directions without ever being touched by it. The center of the bell-shaped, z=0 standard normal curve. Between z=3 and z=3, almost the entire area underneath the standard normal curve is located.

<h3>Use of the standard normal curve:</h3>

Use the normal distribution's standard form to calculate probability. Since the standard normal distribution is indeed a probability distribution, the probability that a variable will take on a range of values is indicated by area of the curve between two points. 100% or 1 is the total area beneath the curve.

<h3>According to the given data:</h3>

the region to the left of the standard normal curve,

z=0.49

To the right of,

z = 2.05

So,

The area will be:

= P[z < 0.49] + P[ z >2.05]

= P[z < 0.49] + 1 -  P[ z < 2.05]

= .6879 + 1 - .9798

= 0.7081

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

To know more about standard normal curve visit:

brainly.com/question/12972781

#SPJ4

I understand that the question you are looking for is:

Find the area under the standard normal curve to the left of z = 0.49 and to the right of z = 2.05. Round your answer to four decimal places, if necessary.

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What statement best describes his motion as he jogs around the curved part of the track?
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Answer:

The answer would be A.

Explanation:

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3 years ago
1) A record is spinning at the rate of 25 rpm. If a ladybug is sitting 10 cm from the
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<h2>distance = 523 cm</h2>

Explanation:

( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s

= 5/12 rev/sec

( b ) The definition of frequency is the number of rotations per second .

Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz

( c ) The tangential speed is v = angular velocity x radius of rotation

The angular velocity ω = 2π x n , where n is the number of rotations per second

Thus angular velocity = 2π x 5/12   = 5π/6 rad/sec

The linear velocity = angular velocity x distance from center of record

Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec

Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad

Linear displacement = angular displacement x distance from center of record

= 50π/3 x 10 = 500π/3 = 523 cm

8 0
3 years ago
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