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algol13
2 years ago
15

A generator produces 38 mwmw of power and sends it to town at an rms voltage of 78 kvkv. part a what is the rms current in the t

ransmission lines?
Physics
1 answer:
4vir4ik [10]2 years ago
7 0

The rms current in the transmission lines is I = 487.18 A.

The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force  is used to represent the source. it is the rectangular root of the time average of the voltage squared.

Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.

Electric power is  by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values

power = 38 M watt

rms voltage = 78 K v

power = IV

I = power/V

I = (38 * 1000000)/78*1000

I = 487.18 A.

Learn more about rms current here:-brainly.com/question/20913680

#SPJ4

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54 is the correct answer to this question

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List 5 possible effects of not adhering to standards of measurement
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Explanation:

Without the metric system, we'd have a different International System of Units, the metric system is important because 1mm is 0.1cm, 1 cm is 0.01m, with the imperial system the conversion is tedious. The most important feature of the metric system is its base in scientific fact and repeatable standards of measurement

3 0
3 years ago
A wall clock has a second hand 22.0 cmcm long. For related problem-solving tips and strategies, you may want to view a Video Tut
olga55 [171]

Answer:

Explanation:

The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression

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angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.

ω = 2π / T

= 2π / 60

= .1047

angular acceleration =  .1047² x .22

= 2.41 x 10⁻³ rad / s² .

8 0
2 years ago
a0.155 kg arrow is shot from ground level, upward at 31.4 m/s. what is its kinetic energy (ke) when it is 30.0 m above the groun
swat32

The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J

Assuming air friction is negligible,

a = - 9.8 m / s²

u = 31.4 m / s

s = 30 m

v² = u² + 2 a s

v² = 31.4² + ( 2 * - 9.8 * 30 )

v² = 985.96 - 588

v² = 397.96 m / s

KE = 1 / 2 m v²

KE = 1 / 2 * 0.155 * 397.96

KE = 0.0775 * 397.96

KE = 30.85 J

Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J

To know more about kinetic energy

brainly.com/question/24360064

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5 0
1 year ago
If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the
Doss [256]

Answer:

0.84 cm

Explanation:

u = Object distance =  0.35 cm

v = Image distance = -0.6 cm (near point is considered as image distance and negative due to sign convention)

f = Focal length

From lens equation

\frac{1}{f_2}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f_2}=\frac{1}{0.35}+\frac{1}{-0.6}\\\Rightarrow \frac{1}{f_2}=\frac{25}{21}\\\Rightarrow f_2=\frac{21}{25}=0.84\ cm

Focal length of the lens is 0.84 cm

3 0
3 years ago
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