3 years ago

Which statement describes how work and power are similar?

Physics

1 answer:

Minchanka [31]3 years ago

3 0

A) you mist know force and distance to calculate both.

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The coil of an ac generator has 50 loops and a cross-sectional area of LaTeX: 0.2~m^20.2 m 2 . What is the maximum emf that can

lisov135 [29]

Answer:

Maximum emf in the coil is 100 volt

Explanation:

We have given number of loops in the coil N = 50

Cross sectional area $A=0.2m^2$

Angular speed is given $\omega =5rad/sec$

Magnetic field is given $B=0.2T$

We have to find the maximum emf in the coil

Maximum emf in the coil is given by $e=NBA\omega$

$e=50\times 2\times 0.2\times 5=100volt$

So maximum emf induced in the coil is 100 volt

7 0

3 years ago

The sun delivers an average power of 1.499 w/m2 to the top of neptune's atmosphere. find the magnitudes of vector e max and vect

sesenic [268]

See attachment for answer:

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3 years ago

A velocity vs. time graph starts at 0 and ends at 10 m/s, stretching over a time- span of 15 s with constant acceleration. What

Alex777 [14]

Answer:

75m

Explanation:

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3 years ago

A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th

solniwko [45]

Answer:

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

$a_t = R\alpha$

2) Centripetal acceleration

$a_c = \omega^2 R$

here we know that

$\omega = \alpha t$

$a_c = (\alpha t)^2 R$

now we know that net linear acceleration is given as

$a = \sqrt{a_c^2 + a_t^2}$

so we have

$a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}$

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

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3 years ago

A driver in a car traveling at a speed of 21.8m/s sees a cat 101 m away on the road. How long will it take for the car to accele

kobusy [5.1K]

The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:

$v^2 = v_0^2 + 2ad\ \to\ a = \frac{-v_0^2}{2d} = -\frac{21.8^2\ m^2/s^2}{2\cdot 99\ m} = -2.4\frac{m}{s^2}$

We can clear time in the speed equation:

$v = v_0 + at\ \to\ t = \frac{-v_0}{a} = \frac{-21.8\ m/s}{-2.4\ m/s^2} = \bf 9.08\ s$

If you find some mistake in my English, please tell me know.

3 0

3 years ago

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