Final answer:
The kinetic energy of a 10g bullet moving at 70 m/s is calculated to be 24.5 Joules using the kinetic energy formula KE = \( \frac{1}{2}mv^2 \).
Step-by-step explanation:
To determine the kinetic energy of a bullet, we can use the formula for kinetic energy which is KE = \( \frac{1}{2}mv^2 \), where m is the mass of the bullet and v is its velocity. In this case, the mass of the bullet is 10 g (which is 0.01 kg when converted to kilograms) and its velocity is 70 m/s. Therefore, the calculation will be:
KE = \( \frac{1}{2} \times 0.01 \text{kg} \times (70 \text{m/s})^2 \)
KE = \( \frac{1}{2} \times 0.01 \times 4900 \)
KE = \( 0.005 \times 4900 \)
KE = 24.5 J
So, the kinetic energy of the bullet before it stops is 24.5 Joules.
To determine the kinetic energy of the bullet, we can use the equation:
KE = ½mv^2
where KE is the kinetic energy, m is the mass of the bullet, and v is the velocity of the bullet.
Given that the mass of the bullet is 10g (0.01kg) and the velocity is 70m/s, we can substitute these values into the equation:
KE = ½ · 0.01kg · (70m/s)^2
Calculating this expression, we get:
KE = 0.005kg · 4900m^2/s^2 = 24.5J
Therefore, the kinetic energy of the bullet is 24.5J.