A 30-caliber rifle with a mass of 2.94 kg fires a 9.73- g bullet with a speed of 666 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
The bullet has a KE of 1/2 *.0973*666^2 joules.
Now the energy released by the explosion is much more...hot gases, gases expanding, sound, fireball, and the energy in kickback.
To calculate the kinetic energy released by the explosion of the gunpowder, we can use the formula:
Kinetic Energy = (1/2) * mass * velocity^2
First, we need to convert the mass of the bullet from grams to kilograms:
9.73 g = 9.73 * 10^-3 kg
Now, we can calculate the kinetic energy:
Kinetic Energy = (1/2) * (mass of bullet) * (velocity of bullet)^2
= (1/2) * (9.73 * 10^-3 kg) * (666 m/s)^2
Calculating the kinetic energy using a calculator:
Kinetic Energy ≈ 1122.551 J
To find the tolerance of 0.1 percent, we need to calculate 0.1 percent of the answer:
0.1/100 * 1122.551 J ≈ 1.122551 J
Therefore, the kinetic energy released by the explosion of the gunpowder is approximately 1122.551 Joules with a tolerance of 1.122551 Joules.
To find the kinetic energy released by the explosion of the gunpowder, we need to use the formula:
Kinetic energy (KE) = 1/2 * mass * velocity^2
First, we need to convert the mass of the bullet from grams to kilograms:
9.73 g = 9.73 / 1000 kg = 0.00973 kg
Now, we can calculate the kinetic energy:
KE = 1/2 * mass * velocity^2
= 1/2 * 0.00973 kg * (666 m/s)^2
= 1/2 * 0.00973 kg * 443556 m^2/s^2
= 0.004865 kg * 443556 m^2/s^2
≈ 2146.4 J
Therefore, the kinetic energy released by the explosion of the gunpowder is approximately 2146.4 Joules.