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A gasoline-powered engine produces a peak output of 10 hp. Express the given power P in the SI.
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Solution:

Approach:

As listed in equations for derived units in the Uses, hp refers to horsepower, Then the Conversions for power.

$ \begin{aligned}\\ &1 h p_1=0.7457 \mathrm{kw} . \\\\ &1 h p=\frac{550(\mathrm{ft} \cdot 1 \mathrm{~b})}{\mathrm{s}}\\ \end{aligned}\\ $

Solution:

Here kijo prefix in, $1000=10^3$

Applying the Conversion factor to the engine's power rating, we have.

$ \begin{aligned}\\ P &=(10 \mathrm{hR})\left(0.7457 \frac{\mathrm{kw}}{\mathrm{AR}}\right) \\\\ &=7.457(\mathrm{hR})\left(\frac{\mathrm{kw}}{\mathrm{hp}}\right) \\\\ &=7.457 \mathrm{kw} \\ \end{aligned} \\ $

Discussion:-

In terms of the derived unit watt, the motor produces $\mathrm{cm}$ output of $7457 \mathrm{w}$.

However, since this numerical value is greater then 1000, the prefix "kiJo" is used,

$ P=7.457 \mathrm{kw}\\ $

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