By: Ryan Carpenter
formulas
1)
Calculate
the speed of an 8.0E 4 kg airliner with a kinetic energy of 1.1E9 J
EK = .5mv^2
1.1E9 = .5 (8.0E4)v^2
1.1E9/(.5 (8.0E4)) = v^2
(1.1E9/(.5 (8.0E4)))^.5 = v
v = 165.8312395
v = 170 m/s TOP
2)
What
is the speed of a 0.145 kg baseball if its kinetic energy is 109 J?
EK = .5mv^2
109 = .5 (0.145)v^2
109/(.5 (0.145)) = v^2
(109/(.5 (0.145)))^.5 = v
v = 38.77432496
v = 38.8 m/s TOP
3)
Two
bullets have masses of 3.0 g and 6.0 g respectively. Both are fired with a speed of 40.0 m/s. (a) Which bullet has more kinetic energy?
(b) what is the ratio of their kinetic energies?
Bullet 1
EK = .5mv^2
EK = .5(.03)(40)^2
EK = 24 j
Bullet 2
EK = .5mv^2
EK = .5(.06)(40)^2
EK = 48 j
Thus bullet 2 has a greater kinetic energy with a ration of 48:24 or 2:1. TOP
4) Two
3.0 g bullets are fired with speeds of 40.0 m/s and 80.0 m/s,
respectively. (a) What is their kinetic
energies? (b) Which bullet has more kinetic energy? (c) What is the ration of
their kinetic energies?
Bullet 1
EK = .5mv^2
EK = .5(.03)(40)^2
EK = 24 j
Bullet 2
EK = .5mv^2
EK = .5(.03)(80)^2
EK = 96 j
Bullet 2 has a greater kinetic energy, with ratio of 96:24 or 4:1 TOP
5) A
car has a kinetic energy of 4.32E5 J when traveling at the speed of
23 m/s. What is its mass?
EK = .5mv^2
4.32E5 = .5m(23)^2
4.32E5/.5(23)^2 = m
m = 1633.270321
m = 1600 kg TOP