To answer this question, it helps enormously if you know
the formula for momentum:
Momentum = (mass) x (speed) .
Looking at the formula, you can see that momentum is directly
proportional to speed. So if speed doubles, so does momentum.
If the car's momentum is 20,000 kg-m/s now, then after its speed
doubles, its momentum has also doubled, to 40,000 kg-m/s.
Answer:
Acceleration = 0.0282 m/s^2
Distance = 13.98 * 10^12 m
Explanation:
we will apply the energy theorem
work done = ΔK.E ( change in Kinetic energy ) ---- ( 1 )
<em>where :</em>
work done = p * t
= 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J
( note : convert 1 year to seconds )
and ΔK.E = 1/2 mVf^2 given ; m = 1200 kg and initial V = 0
<u>back to equation 1 </u>
473040000 * 10^6 = 1/2 mv^2
Vf^2 = 2(473040000 * 10^6 ) / 1200
∴ Vf = 887918.92 m/s
<u>i) Determine how fast the rocket is ( acceleration of the rocket )</u>
a = Vf / t
= 887918.92 / ( 1 year )
= 0.0282 m/s^2
<u>ii) determine distance travelled by rocket </u>
Vf^2 - Vi^2 = 2as
Vi = 0
hence ; Vf^2 = 2as
s ( distance ) = Vf^2 / ( 2a )
= ( 887918.92 )^2 / ( 2 * 0.0282 )
= 13.98 * 10^12 m
Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.
Explanation: Hope this helped :)
Answer:
3/7 ω
Explanation:
Initial momentum = final momentum
I(-ω) + (2I)(3ω) + (4I)(-ω/2) = (I + 2I + 4I) ωnet
-Iω + 6Iω - 2Iω = 7I ωnet
3Iω = 7I ωnet
ωnet = 3/7 ω
The final angular velocity will be 3/7 ω counterclockwise.
Clarify what you mean by ratios?
