1. A baseball that has been thrown is travelling at 40m/s. If the ball has a mass of 0.142kg, how much kinetic energy
does it have?
2. If a similar baseball was hit and travelled at 52m/s and
had 196J of kinetic energy, what is the mass of that
baseball?
3. A 4kg shot put is thrown and it has 288J of kinetic energy.
What is the mass of the shot put?

See explanations below

Explanation:

1) KE = 1/2mv²

mass m = 0.142kg

v = 40m/s

KE = 1/2 * 0.142 * 40^2

KE = 1/2 * 0.142 * 1600

KE = 800*0.142

KE = 113.6Joules

2) KE = 1/mv^2

KE = 196

mass m = ?

velocity v = 52m/s

Substitute

196 = 1/2*m*52²

196 = 1352m

m = 196/1352

m = 0.145kg

Hence the mass of the baseball is 0.145kg

3) KE = 1/2mV^2

288 = 1/2 * 4v²

288 = 2v²\

v² = 288/2

v² = 144

v = 12m/s

The velocity of the shotput is 12m/s

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.