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:
(c). The two blocks end in a tie
Explanation:
the reason being the absence of any resistance offered to both of the blocks.
if the slope of the hill is for instance 60 deg.
then the acceleration in absence of any resistance is a= 9.81sin(60)
since the acceleration is same then both of the blocks will reach at the same instant
All black surfaces is the correct answer
