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Maru [420]
3 years ago
11

If a machine has an efficiency of 94% and you apply 574J of work, how much work do you get out of the machine

Physics
1 answer:
Mariana [72]3 years ago
7 0
<h3>Answer:</h3>

539.56 Joules

<h3>Explanation:</h3>
  • Efficiency of a machine is the ratio of work output to work input expressed as a percentage.
  • Efficiency = (work output/work input) × 100%
  • Efficiency of a machine is not 100% because so energy is lost due to friction of the moving parts and also as heat.

In this case;

Efficiency = 94%

Work input = 574 Joules

Therefore, Assuming work output is x

94% = (x/574 J) × 100%

0.94 = (x/574 J)

<h3>x = 539.56 J</h3>

Thus, you get work of 539.56 J from the machine

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Answer:

Explanation:

Given that,

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Diameter of the coil d= 11cm = 0.11m

Then, radius = d/2 = 0.11/2 =0.055m

r = 0.055m

Then, the area is given as

A =πr²

A = π × 0.055²

A = 9.503 × 10^-3 m²

Magnetic Field B = 0.35T

Magnetic field reduce to zero in 0.1s, t = 0.1s

so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).

E.M.F is given as

ε = —N • dΦ/dt

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ε = —N • dBA/dt

ε = —NBA/t

Then, its magnitude is

ε = NBA/t

Inserting the values of N, B, A and t

ε = 40×0.35×9.503×10^-3/0.1

ε = 1.33 V

Then, using the relationship between Electric field and electric potential

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E = ε/d

E = 1.33/0.11

E = 12.09 V/m

7 0
3 years ago
Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co
nalin [4]

Answer:

t = 1.02 s

Explanation:

The computation of the time required is shown below:

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= 1.962 m/s^2

And, final velocity = 0

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here

V = 0 = final velocity

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3 years ago
What is the gravitational potential energy of a 65.7 kg person standing on the roof of a 135 meter building?
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When we plug values into the equation, we get following:
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8 0
3 years ago
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