5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on low-friction wheels and as the bullet struck the target, the target with
bullet embedded bullet sped off with a velocity of 6.0 ms.
Calculate the velocity of the bullet just before it hit the stationary
target.

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas