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A machine has an efficiency of 80%. How much work must be done on the machine so to make it do 50,000 J of output work?

Respuesta :

Answer;

62,500 Joules

Explanation;

  • Efficiency is a measure of how much work or energy is conserved in a process. It can be stated as the percentage of the output work divided by the input work minus the work lost from to friction and heat.

Efficiency = (output work/ input work) × 100%

        80 % = 50,000 / input work  × 100%

             0.8 = 50,000 /input work

Input work = 50,000/0.8

                  = 62500 Joules

Thus; the work input is 62,500 joules.

The input work for the machine is 62500 Joules.

What is Efficiency?

The efficiency of an object is defined as a dimensionless number that measures the effectiveness of the object in transforming the power input to the device to power output.

The mathematical expression of the efficiency is given below.

[tex]E = \dfrac {O}{I} \times 100[/tex]

Where E is the efficiency, O is the output work and I is the input work.

Given that the efficiency of the machine is 80% and output work is 50,000 J. The input work can be calculated as given below.

[tex]80 = \dfrac { 50000}{I }\times 100[/tex]

[tex]I = \dfrac {50000}{0.8}[/tex]

[tex]I = 62500 \;\rm J[/tex]

Hence we can conclude that the input work for the machine is 62500 Joules.

To know more about the efficiency, follow the link given below.

https://brainly.com/question/6672666.

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