select the correct answer.\nthe input work done on a machine is 9.63×10³ joules, and the output work is…

select the correct answer.\nthe input work done on a machine is 9.63×10³ joules, and the output work is 3.0×10³ joules. what is the percentage efficiency of the machine?\na. 0.31 percent\nb. 25.6 percent\nc. 31.1 percent\nd. 42.8 percent\ne. 50.0 percent

select the correct answer.\nthe input work done on a machine is 9.63×10³ joules, and the output work is 3.0×10³ joules. what is the percentage efficiency of the machine?\na. 0.31 percent\nb. 25.6 percent\nc. 31.1 percent\nd. 42.8 percent\ne. 50.0 percent

Answer

Explanation:

Step1: Recall efficiency formula

The formula for the efficiency $\eta$ of a machine is $\eta=\frac{W_{out}}{W_{in}}\times100%$, where $W_{out}$ is the output work and $W_{in}$ is the input work.

Step2: Substitute given values

Given $W_{in} = 9.63\times10^{3}$ joules and $W_{out}=3.0\times 10^{3}$ joules. Then $\eta=\frac{3.0\times 10^{3}}{9.63\times 10^{3}}\times100%$.

Step3: Calculate the value

First, $\frac{3.0\times 10^{3}}{9.63\times 10^{3}}=\frac{3.0}{9.63}\approx0.312$. Then $0.312\times100% = 31.2%$, which is approximately $31.1%$ (due to rounding - differences in options).

Answer:

C. 31.1 percent