Right now there are 6 astronauts in orbit traveling at nearly 8000 m/s. The othe

Question

Right now there are 6 astronauts in orbit traveling at nearly 8000 m/s. The other nearly 7 billion people on Earth are moving with an average speed somewhat under 0.5 m/s. This means about 20% of all the kinetic energy possessed by all the peoples of the world belongs to these 6 energetic astronauts. A bullet (mass = 15 grams, speed = 300 m/s) imbeds itself in a stationary block of wood (m = l500 grams). What is the final speed of the wood assuming no loss of material for either substance? How high up a frictionless incline could the mass climb? How far could the mass skid on a level table with a static mu of 0.7 and a kinetic mu of 0.5?

Explanation / Answer

question 28 is not clear
I am allowed to answer only 1 question at a time
So i will answer 29

m1 = 15 g
v1i = 300 m/s
v1f = v m/s

m2 = 1500 g
v2i = 0 m/s
v2f = v m/s

use conservation of momentum:
m1*v1i + m2*v2i = m1*v1f + m2*v2f
15*300 + 1500*0 = (15+1500)*v
4500 = 1515*v
v= 2.97 m/s
Answer: 2.97 m/s
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use conservation of energy
0.5*m*v^2 = m*g*h
0.5*v^2 = g*h
0.5*2.97^2 = 9.81*h
h = 0.45 m
Answer: 0.45 m
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use work energy theorem
initial kinetic energy = work done by friction
0.5*m*v^2 = miuk*m*g*d
0.5*v^2 = miuk*g*d
0.5*(2.97)^2 = 0.5*9.81*d
d=0.9
Answer: 0.9 m

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