A 12.0-g bullet is fired horizontally into a 114-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 146 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 75.5 cm, what was the speed of the bullet at impact with the block

The speed of the bullet-block is .

Explanation:

The expression of the conservation of energy to the given system is as follows;

Here, are the initial kinetic energy and final kinetic energy and .

Here, m is the mass of the bullet, M is the mass of the wooden block, k is the spring constant, x is the distance, v is the speed of the bullet and V is the speed of the bullet with the block.

Calculate the speed of the bullet-block.

Convert the distance from cm to m.

x=75.5 cm

x=0.755 m

Put m= 12 g, M=114 g, v=0, x= 0.755 m and in the above expression according to the given values in the problem.

Therefore, the speed of the bullet-block is .

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge