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Nat2105 [25]
1 year ago
8

A machine has an efficiency of 20 percent. Find the input work if the output work is 140 Joules

Physics
2 answers:
Lapatulllka [165]1 year ago
6 0

Answer:

X= 700 Joules

Explanation:

The question asked about the efficiency of the work done.

The formula for efficiency is: Efficiency = (Useful output / input work) * 100%

The useful output given in the question is 140J, the question asked for input work. Let X be the input work. It is also given that the efficiency is 20%.

Using the formula of efficiency,

20 = (140/X) * 100

So, we simply solve the above equation.

X= 140*100/20

X= 700 Joules

vazorg [7]1 year ago
4 0

Answer:

The input work = 700 Joules

Explanation:

Given Data:

Efficiency = 20%

Output work = 140 joules

Input work = ?

The input work done can be calculated using the efficiency formula;

Efficiency = output work done/input work done * 100

Substituting, we have

20 = 140/Input work * 100

20 = 14000/input work

20*input work = 14000

input work = 14000/20

               = 700 joules

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aniked [119]

Answer:

a. N = 2.49W b.  0.40

Explanation:

a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?

Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s

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b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?

The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.

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Answer:

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