Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air $y$ at time $t$ is given by

$y=100\,\mathrm m-\dfrac g2t^2$

Let $d$ be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

$-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}$

7 0

4 years ago

An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?

snow_lady [41]

The wire vibrates back and forth between the poles of the magnet.
The frequency of the vibration is the frequency of the AC supply.

7 0

2 years ago

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

shusha [124]

Answer:

$v=d\sqrt{\frac{k}{m}}$

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

$U_{0}+K_{0}=U_{f}+K_{f}$

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

$U_{0}=K_{f}$