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
The higher the depth, the higher the mass of the liquid that creates pressure
Answer:
Final speed of security car v = 65 m/s
Explanation:
Given:
Speed of race car u1 = 35m/s
Speed of security car u2 = 5 m/s
Acceleration = 5 m/s²
Find:
Final speed of security car v
Computation:
Assume, they chase S meter
So
S = u1t + [1/2]at²
S = 35t
S = u2t + [1/2]at²
so,
35t = 5t + [1/2](5)t²
t = 12 s
So
v = u + at
v = 5 + 5(12)
Final speed of security car v = 65 m/s
Answer:
C
Explanation:
If the theory were to be proved you you need to repeat the experiment over and over again so that way you can prove that it is true wuth the same results.
