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3 years ago

Using 15 percent as machine energy efficiency, what is the actual work output if the total work input is 7500 kilojoules?

Physics

1 answer:

g100num [7]3 years ago

7 0

Answer:

The actual work output is 1,125 kilojoules.

Explanation:

Energy efficiency is defined as the efficient use of energy. An appliance, process, or facility is energy efficient when it consumes less than the average amount of energy to perform an activity. Then, energy efficiency is the ratio between the amount of energy used in an activity and the amount expected to be carried out.

Efficiency is calculated as:

efficiency = output / input

Where output is the amount of mechanical work (in watts) or energy consumed by the process (in joules), and input (input) is the amount of work or energy that is used as input to carry out the process.

In this case:

Efficiency always has a value between 0 and 1. In this case, efficiency=0.15
output=?
input= 7500 kilojoules

Replacing:

0.15=output/7500 kilojoules

Solving:

Output=0.15* 7500 kilojoules

output=1,125 kilojoules

The actual work output is 1,125 kilojoules.

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El valor más preciso de la masa de un electron es 9.11*10^-11kg ¿ Cuánto aumentaría la masa de un cuerpo que se carga con -1 c d

chubhunter [2.5K]

Answer:

$m = 569.375\times 10^{6}\,kg$

Explanation:

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3 years ago

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Stolb23 [73]

Answer:

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Explanation:

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3 years ago

Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.

SVEN [57.7K]

The strength of electric field E is 17 N / C.

Explanation:

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

E = F / q

where E represents the electric field strength,

F represents the force in newton,

q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

force F = 5.0 N

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= 5.0 / 0.30

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