Question

1. an engine was built to rotate at 120,000 rotations per second. what is the period of rotation for this engine? 6.0·10^{4}sec 1.2·10^{5}sec 8.3·10^{-6}sec 1.2·10^{6}sec clear all

Explanation:

Step1: Recall the formula for period

The formula for period \(T\) (time per rotation) when given the frequency \(f\) (rotations per second) is \(T=\frac{1}{f}\).

Step2: Substitute the given frequency

Given \(f = 120000=1.2\times10^{5}\) rotations per second. Then \(T=\frac{1}{1.2\times 10^{5}}\).

Step3: Simplify the expression

\(T=\frac{1}{1.2\times 10^{5}}=\frac{1}{1.2}\times10^{- 5}\approx0.83\times10^{-5}=8.3\times10^{-6}\) seconds.

Answer:

\(8.3\times 10^{-6}\text{ sec}\) (the third option)