Question

1. what is the kinetic energy of a 2025.7 kg car that travels at 23.2 m/s?

Explanation:

Step1: Recall the kinetic energy formula

The formula for kinetic energy (\(KE\)) is \(KE=\frac{1}{2}mv^{2}\), where \(m\) is the mass of the object and \(v\) is its velocity.

Step2: Substitute the given values

Here, \(m = 2025.7\space kg\) and \(v=23.2\space m/s\). Substitute these values into the formula:
\(KE=\frac{1}{2}\times2025.7\times(23.2)^{2}\)
First, calculate \((23.2)^{2}=23.2\times23.2 = 538.24\)
Then, \(\frac{1}{2}\times2025.7 = 1012.85\)
Now, multiply these two results: \(1012.85\times538.24\)
\(1012.85\times538.24 = 1012.85\times(500 + 38+0.24)=1012.85\times500+1012.85\times38 + 1012.85\times0.24\)
\(1012.85\times500 = 506425\)
\(1012.85\times38=(1000 + 12.85)\times38=1000\times38+12.85\times38 = 38000+488.3 = 38488.3\)
\(1012.85\times0.24=(1000 + 12.85)\times0.24=1000\times0.24+12.85\times0.24 = 240+3.084 = 243.084\)
Now, sum these three results: \(506425+38488.3 + 243.084=506425+38731.384 = 545156.384\) (approximate value, more accurately calculated directly as \(1012.85\times538.24 = 545156.384\))

Answer:

The kinetic energy of the car is approximately \(545156.38\space J\) (or \(5.45\times 10^{5}\space J\) when rounded to a reasonable number of significant figures).