1. if an egg has a mass of .1k and its dropped from a height of 5 meters, calculate its potential energy at the top

2. if all the potential energy at the top is converted to kinetic energy at the bottom, calculate the eggs velocity when it reaches the ground.

3. calculate the momentum of the egg as it hits the ground.

4. if the time of impact is .03 seconds, calculate the force acting on the egg using the impulse formula

SHOW WORK!!!

(a) 5Joules

(b) 10m/s

(c) 1.0kgm/s

(d) 33.3N

Explanation:

(a) P E = mass * gravity * height

= 0.1*10*5 = 5Joules

(b) All PE is converted to KE

Therefore;

KE = 1/2 * mass * velocity^2

5 = 1/2 * 0.1 * v^2

V = sqrt(100)

v = 10m/s

(c) change in momentum = mass * velocity

= 0.1 * 10

= 1.0kgm/s

(d) impulse = change in momentum

Change in momentum = force * time

1.0 = force * 0.03

Force = 1.0/0.03

Force = 33.3N

Therefore, the force acting on the egg is 33.3N

wavelength variability is the right one

heat the gas and measure the pressure