Question

a bowling ball is traveling at 7.6 meters per second when it hits a pin. if the bowling ball has a mass of 6 kilograms, how much kinetic energy does it have when it hits the pin?

$ke = \frac{1}{2}mv^{2}$

(1 point)

22.8 j

173.28 j

346.65 j

693.12 j

Explanation:

Step1: Substitute values into formula

Given \(m = 6\space kg\), \(v=7.6\space m/s\), substitute into \(KE=\frac{1}{2}mv^{2}\).
\(KE=\frac{1}{2}\times6\times(7.6)^{2}\)

Step2: Calculate \((7.6)^{2}\)

\((7.6)^{2}=7.6\times7.6 = 57.76\)
So \(KE=\frac{1}{2}\times6\times57.76\)

Step3: Calculate \(\frac{1}{2}\times6\)

\(\frac{1}{2}\times6 = 3\)
Now \(KE=3\times57.76\)

Step4: Final multiplication

\(3\times57.76 = 173.28\)

Answer:

173.28 J