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:
t = 0.96 s is the time it takes for the angle to reduce
Explanation:
In the launch of projectiles, the velocity is broken down into its x and y components, the velocity in the x-axis is constant, as there is no acceleration, instead the velocity in the axis and is reduced by the effect of the acceleration of gravity.
We can find Vox for the initial conditions
Voy = Vo sin θ
Voy = 29 sin 36
VoY = 17 m/s
Vox = Vo cos θ
Vox = 29 cos 36
Vox= 23.5 m/s
Vx = Vox
The velocity on the x axis is constant
By trigonometry, we find the firing angle
tan θ = Voy/ Vx
Vy = Vx tan θ
Vy = 23.5 tan 18
Vy = 7.64 m/s
Now that we have the vertical speed we can find the time
Vy = I'm going - g t
t = (Vy -Voy) / g
t = (17 - 7.64) /9.8
t = 0.96 s
Be the time it takes for the angle to reduce
Answer:
electricity
If a rod is charged it is because of the electrical force acting on it
