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A 3.5 kW drill transfers 5 000 kJ of kinetic energy during 15 seconds of use. What is the percentage efficiency of the drill?

Sagot :

Answer:

1.04%

Explanation:

Given that,

The power of drill = 3.5 kW = 3500 W

Transferred kinetic energy = 5000 kJ during 15 seconds of use.

We need to find the percentage efficiency of the drill. It can be given by :

[tex]\eta=\dfrac{P_o}{P_i}\times 100[/tex]

Where

Po and Pi are output and input powers.

[tex]P_o=\dfrac{5000\times 10^3}{15}\\\\=3.34\times 10^5\ W[/tex]

So,

[tex]\eta=\dfrac{3500}{3.34\times10^{5}}\times100\\\\=1.04\%[/tex]

So, the percentage efficiency of the drill is 1.04%.

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