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1 year ago

How does the output work compare to the input work for a machine that has efficiency

Physics

1 answer:

Leona [35]1 year ago

8 0

(a) For a machine with 100% efficiency, the output work is equal to input work.

(b) For a machine with less than 100% efficiency, the output work is less than the input work.

<h3>What is efficiency of machine?</h3>

The efficiency of a machine indicates how well its input work is converted to useful output work.

It is a major factor determines the usefulness of a machine and is the fraction or percentage of the output energy divided by the input energy.

Mathematically, the formula for efficiency of a machine is given as;

E = output work / input work x 100%

For an ideal machine, the output energy (work) is equal to the input energy (work), and the efficiency of this ideal machine will be 100%.

Thus, a machine with 100% efficiency has not friction in the moving parts and the output work is equal to input work

In a real situation, friction always exists and the efficiency of a machine cannot be 100%.

A machine with less than 100% efficiency, some input work or energy in the machine is used to overcome friction and the output work is less than the input work.

Learn more about efficiency of a machine here: brainly.com/question/27325529

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4 years ago

A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of

lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W= ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7 N

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