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3 years ago

While working out, a man performed 2525 J of work in 19 seconds. What was his power?

Physics

1 answer:

PIT_PIT [208]3 years ago

8 0

Answer:

$power = \frac{work \: done}{time \: passed} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

power=2525J/19S

=132.8947368421JS^-1

=132.9W

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3 years ago

A major difference between AC and DC electricity is that A) the current in AC electricity varies in magnitude and direction. B)

galina1969 [7]

Answer:

A) the current in AC electricity varies in magnitude and direction.

C) the voltage in AC electricity varies in magnitude and direction.

Explanation:

In DC current and voltage the direction of current will not change with time and it always remains the same.

So here in DC voltage and DC current the magnitude may change with time but the direction will always remain same

While in AC voltage and AC current the direction of AC will change with time

periodically.

So here magnitude and direction both will change in AC current and AC voltage.

so the correct answer is

A) the current in AC electricity varies in magnitude and direction.

C) the voltage in AC electricity varies in magnitude and direction.

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3 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

VMariaS [17]

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3 years ago

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi

DedPeter [7]

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

$\frac{dC}{dt}=-15cm/s$

Magnetic field,B=0.9 T

We have to find the magnitude of the emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux= $\phi=BA=B(\pi r^2)$

Circumference,C= $2\pi r$

$r=\frac{C}{2\pi}$

$r=\frac{168}{2\pi}$ cm

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s$

$\int dr=-\int \frac{15}{2\pi}dt$

$r=-\frac{15}{2\pi}t+C$

When t=0

$r=\frac{168}{2\pi}$

$\frac{168}{2\pi}=C$

$r=-\frac{15}{2\pi}t+\frac{168}{2\pi}$

E= $-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}$

$E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}$

t=8 s

B=0.9

$E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})$

$E=103.1 V$

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3 years ago

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3 years ago