Answer:
0.955286 j
Explanation:
A 500.0 kg module is attached to a 440.0 kg shuttle craft, which moves at 1050. m/s relative to the stationary main spaceship. Then a small explosion sends the module backward with speed 100.0 m/s relative to the new speed of the shuttle craft. As measured by someone on the main spaceship, by what fraction did the kinetic energy of the module and shuttle craft, Ki, increase because of the explosion?
M=500 kg, m=440 kg
V=1000 m/s, v = 100 m/s
Let relative speed =Vs
Momentum rule says
(M+m)V=mVs+M(Vs-v)
940(1000)=500(Vs-100)+440Vs
940000=500Vs-50000+440Vs
940Vs=940000+50000
940Vs=990000
Vs= 990000/940=1053.19 m/s
So, the module speed = Vs-v=1053.19-100=953.19 m/s
Fractional increase in KE is given by;
Total KE after explosion / He before explosion
=500(953.19)2+ 400(1053.19)2/ 940(1000)2= 0.955286
Answer:

Explanation:
As we know that the that the Astronaut dropped a stone on the surface of the planet takes 0.420 s to fall a distance of d = 1.90 m
so we will have



now we know that the acceleration due to gravity is given as

so we will have


The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.
The Coulomb force between the two balls is:

where

is the Coulomb's constant,

is the intensity of the two charges, and

is the distance between them.
Substituting these numbers into the equation, we get

The force is repulsive, because the charges have same sign and so they repel each other.
11/23/2012 - 2.2 mag, 5.0mi depth 1.0875 mi from <span>Gloucester Township, NJ
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