someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (the block of wood is sitting on a frictionless surface, so it moves freely when the bullet hits it). the wood block is initially at rest. the bullet is traveling 300 m/s when it hits the wood block and sticks inside it. now the bullet and the wood block move together as one object. how fast are they traveling?

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

                              mV + MV = mu + MU

                               V(m+M) = mu

                                 V = mu/(m+M)

Substituting the values in the above equation,

                                V = 0.04 Kg x 300 m/s  / (0.04 Kg+ 0.5 Kg)

                                    = 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

wind moving at higher speeds will transfer more energy to the water, resulting in stronger waves.