Question
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=12mv2
(1 point)
Responses
346.65 J
346.65 J
173.28 J
173.28 J
693.12 J
693.12 J
22.8 J
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=12mv2
(1 point)
Responses
346.65 J
346.65 J
173.28 J
173.28 J
693.12 J
693.12 J
22.8 J
Answers
GPT-4o mini
To calculate the kinetic energy (KE) of the bowling ball, we can use the formula:
\[ KE = \frac{1}{2} mv^2 \]
Where:
- \( m \) is the mass of the bowling ball (6 kg)
- \( v \) is the velocity of the bowling ball (7.6 m/s)
Now, plug in the values:
\[ KE = \frac{1}{2} \times 6 \text{ kg} \times (7.6 \text{ m/s})^2 \]
First, calculate \( (7.6 \text{ m/s})^2 \):
\[ (7.6)^2 = 57.76 \text{ m}^2/\text{s}^2 \]
Now substitute back into the kinetic energy formula:
\[ KE = \frac{1}{2} \times 6 \times 57.76 \]
\[ KE = 3 \times 57.76 \]
\[ KE = 173.28 \text{ J} \]
Thus, the kinetic energy of the bowling ball when it hits the pin is **173.28 J**.
So the correct response is:
**173.28 J**
\[ KE = \frac{1}{2} mv^2 \]
Where:
- \( m \) is the mass of the bowling ball (6 kg)
- \( v \) is the velocity of the bowling ball (7.6 m/s)
Now, plug in the values:
\[ KE = \frac{1}{2} \times 6 \text{ kg} \times (7.6 \text{ m/s})^2 \]
First, calculate \( (7.6 \text{ m/s})^2 \):
\[ (7.6)^2 = 57.76 \text{ m}^2/\text{s}^2 \]
Now substitute back into the kinetic energy formula:
\[ KE = \frac{1}{2} \times 6 \times 57.76 \]
\[ KE = 3 \times 57.76 \]
\[ KE = 173.28 \text{ J} \]
Thus, the kinetic energy of the bowling ball when it hits the pin is **173.28 J**.
So the correct response is:
**173.28 J**