To calculate the kinetic energy (KE) of the bowling ball when it hits the pin, we can use the formula:
\[ KE = \frac{1}{2}mv^2 \]
where:
- \( m \) is the mass of the object (in kilograms),
- \( v \) is the velocity of the object (in meters per second).
Given:
- \( m = 6 \) kg,
- \( v = 7.6 \) m/s.
Now we can plug in these values into the formula:
\[ KE = \frac{1}{2} \times 6 , \text{kg} \times (7.6 , \text{m/s})^2 \]
Calculating \( (7.6)^2 \):
\[ (7.6)^2 = 57.76 , \text{m}^2/\text{s}^2 \]
Now substitute back into the equation:
\[ KE = \frac{1}{2} \times 6 \times 57.76 \]
\[ KE = 3 \times 57.76 \]
\[ KE = 173.28 , \text{J} \]
So the kinetic energy of the bowling ball when it hits the pin is 173.28 J.
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