A 30-caliber rifle with a mass of 3.54 kg fires a 7.33- g bullet with a speed of 596 m/s with respect to the ground. What kinetic energy is released by the explosion of the gunpowder that fires the bullet? (Note there is a tolerance 0.1 percent for the answer)

To find the kinetic energy released by the explosion of the gunpowder, we need to calculate the kinetic energy of the bullet.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 1/2 * mass * velocity^2

First, let's convert the mass of the bullet from grams to kilograms:
7.33 g = 7.33 * 10^(-3) kg

Now, we can plug in the values to calculate the kinetic energy of the bullet:

KE = 1/2 * 7.33 * 10^(-3) kg * (596 m/s)^2

Simplifying the equation, we get:

KE = 1/2 * 7.33 * 10^(-3) kg * 596^2 m^2/s^2

KE = 1/2 * 7.33 * 10^(-3) * 596^2 J

Calculating further, we have:

KE ≈ 1/2 * 7.33 * 0.00596 * 358816 J (rounding to 6 decimal places)

KE ≈ 1/2 * 0.02706088 * 358816 J

KE ≈ 9670.95303616 J

Rounded to 0.1 percent tolerance, the final answer is approximately 9670.95 J.

Therefore, the kinetic energy released by the explosion of the gunpowder is approximately 9670.95 J.