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3 years ago

a pulley system has an efficiency of 74.2%. If you perform 200 J of work how much useful work does the pulley perform

Physics

1 answer:

oksian1 [2.3K]3 years ago

3 0

Answer:

148.4 J

Explanation:

We are given that the efficiency of a pulley system is 74.2% and we are to calculate the useful work done if we perform 200 J of work

We know that efficiency is basically the measure of how efficient a process is.

The formula of efficiency is given by:

Efficiency = useful energy output / total energy input

So substituting the given values in the above formula:

0.742 = Useful energy / 200 J

Useful energy = 148.4 J

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Answer: D (doubled)

According to ohm's law

I = V/R ,

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V is directly proportional to the current, and it is given that R is constant ;

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4 years ago

You observe a distant galaxy. you find that a spectral line normally found in the visible part of the spectrum is shifted toward

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3 years ago

the gravitational force that earth exerts on the moon equals 2.03 x 10^20N. The moons mass is 7.35 x 10^22kg. What is the accele

Ghella [55]

The first time we read this, we think that we'll have to go look up a bunch of stuff about the gravitational constant, centripetal force, orbital mechanics, the Moon's period of revolution, and who knows whut owl !

But you know whut ? The only thing we need is Newton's 2nd law:

F = m A

Divide each side by m :

Acceleration = Force / mass

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8 0

3 years ago

A 70.0-kg person throws a 0.0420-kg snowball forward with a ground speed of 35.0 m/s. A second person, with a mass of 57.0 kg, c

ollegr [7]

Answer:

so throwers new velocity = 2.18032m/s

so catchers new velocity = 0.02577m/s

Explanation:

Directly by conservation of momentum we can write

$m_1u_1+m_2u_2= m_1v_1+m_2v_2$

let x be the thrower's new velocity

(70+0.042)×2.2 + 57×0 = 70× x +0.042×35 +57×0

x = 2.18032m/s

so the velocity of 70 kg man = 2.18032m/s

so throwers new velocity = 2.18032m/s

now again by conservation of momentum

0.042×35 = (57+0.042) ×y

y = 0.02577m/s

so catchers new velocity = 0.02577m/s

3 0

3 years ago

Two force A and B at a point at right angles. If their resultant is 50N and their sum is 70N,

kodGreya [7K]

Taking a wild guess here, but it sounds like you're asked to find |A| and |B| given the magnitude of the resultant A + B, i.e. |A + B| = 50 N, and the sum of the individual magnitudes, |A| + |B| = 70 N.

Recall that the dot product of a vector with itself is equal to the square of that vector's magnitude,

A • A = |A|²

Then

|A + B|² = (A + B) • (A + B)

|A + B|² = (A • A) + 2 (A • B) + (B • B)

|A + B|² = |A|² + 2 (A • B) + |B|²

Since A and B are perpendicular to one another, their dot product is

A • B = 0

So it follows that

|A + B|² = |A|² + |B|²

2500 N² = |A|² + |B|²

Substitute |B| = 70 N - |A| and solve for |A| :

2500 N² = |A|² + (70 N - |A|)²

2500 N² = |A|² + 4900 N² - (140 N) |A| + |A|²

2 |A|² - (140 N) |A| + 2400 N² = 0

|A|² - (70 N) |A| + 1200 N² = 0

|A|² - (70 N) |A| = - 1200 N²

|A|² - (70 N) |A| + 1225 N² = - 1200 N² + 1225 N²

(|A| - 35 N)² = 25 N²

|A| - 35 N = ± 5 N

|A| = 35 N ± 5 N

so that |A| = 30 N or |A| = 40 N. If we fix |A| to be one of these, then |B| will have the other value.

So, the magnitudes are |A| = 30 N and |B| = 40 N.

3 0

3 years ago