A 9.90-g bullet has a speed of 1.33 km/s.
(a) What is its kinetic energy in joules?
(b) What is the bullet's kinetic energy if its speed is halved?
(c) What is the bullet's kinetic energy if its speed is doubled?
(a)
(1/2) M V^2.
Make sure M is in kg and V is in m/s.
(b) (1/4) f what it is in (a)
(c) What do you think? What happens to V^2 if you double V?
why can’t y’all just answer the question :)
To calculate the kinetic energy of an object, we can use the equation:
Kinetic Energy = (1/2) * mass * velocity^2
where mass is given in grams, velocity in meters per second, and the resulting kinetic energy is in joules.
Given:
Mass of the bullet = 9.90 g = 0.00990 kg
Velocity of the bullet = 1.33 km/s = 1330 m/s
(a) To find the bullet's kinetic energy:
Using the formula mentioned earlier:
Kinetic Energy = (1/2) * mass * velocity^2
= (1/2) * 0.00990 kg * (1330 m/s)^2
Calculating this:
Kinetic Energy = (1/2) * 0.00990 kg * (1330 m/s)^2
= (1/2) * 0.00990 kg * 1768900 m^2/s^2
= 9301335 J
Therefore, the bullet's kinetic energy is 9,301,335 joules.
(b) If the bullet's speed is halved, we can find its kinetic energy by dividing the original kinetic energy by 4. This is due to the fact that kinetic energy is proportional to the square of the velocity.
New velocity = 1330 m/s / 2 = 665 m/s
New kinetic energy = (1/2) * 0.00990 kg * (665 m/s)^2
= 2,322,568.25 J (rounded)
Therefore, if the bullet's speed is halved, its kinetic energy would be approximately 2,322,568.25 joules.
(c) If the bullet's speed is doubled, we can find its new kinetic energy using the same formula.
New velocity = 1330 m/s * 2 = 2660 m/s
New kinetic energy = (1/2) * 0.00990 kg * (2660 m/s)^2
= 37,205,340 J
Therefore, if the bullet's speed is doubled, its kinetic energy would be 37,205,340 joules.
To calculate the kinetic energy of a moving object, you can use the equation:
Kinetic Energy (KE) = (1/2) * mass * velocity^2
(a) What is its kinetic energy in joules?
Given:
Mass (m) = 9.90 g = 0.00990 kg (convert grams to kilograms)
Velocity (v) = 1.33 km/s = 1,330 m/s (convert kilometers per second to meters per second)
Substituting these values into the kinetic energy equation:
KE = (1/2) * 0.00990 kg * (1,330 m/s)^2
KE = 0.5 * 0.00990 kg * (1,330 m/s)^2
KE = 0.5 * 0.00990 * 1,768,900
KE ≈ 8726.53 J
Therefore, the kinetic energy of the 9.90-g bullet is approximately 8726.53 Joules.
(b) What is the bullet's kinetic energy if its speed is halved?
If the speed is halved, the new velocity (v') is half of the original velocity (v).
v' = v / 2 = 1,330 m/s / 2 = 665 m/s
Using the kinetic energy equation again:
KE' = (1/2) * 0.00990 kg * (665 m/s)^2
KE' = 0.5 * 0.00990 kg * (665 m/s)^2
KE' = 0.5 * 0.00990 * 221,225
KE' ≈ 1,095.02 J
Therefore, if the speed is halved, the bullet's kinetic energy would be approximately 1,095.02 Joules.
(c) What is the bullet's kinetic energy if its speed is doubled?
If the speed is doubled, the new velocity (v'') is twice the original velocity (v).
v'' = 2 * v = 2 * 1,330 m/s = 2,660 m/s
Using the kinetic energy equation:
KE'' = (1/2) * 0.00990 kg * (2,660 m/s)^2
KE'' = 0.5 * 0.00990 kg * (2,660 m/s)^2
KE'' = 0.5 * 0.00990 * 7,075,600
KE'' ≈ 35,091 J
Therefore, if the speed is doubled, the bullet's kinetic energy would be approximately 35,091 Joules.