Two airplanes leave an airport at the same time. The velocity of the first airplane is 660 m/h at a heading of 28.6 ◦ . The velo
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Answer:
C. 590 mph
Explanation:
Given:
- velocity of jet,
- direction of velocity of jet, east relative to the ground
- velocity of Cessna,
- direction of velocity of Cessna, 60° north of west
Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.
Refer the attached schematic.
So,
&
Now the vector of relative velocity of Cessna with respect to jet:
Now the magnitude of this velocity:
is the relative velocity of Cessna with respect to the jet.
Answer:
F = 4000 N
Explanation:
given,
mass of rocket (M)= 5000 Kg
10 Kg gas burns at speed (m)= 4000 m/s
time = 10 s
average force = ?
at the end the rocket is at rest
by conservation of momentum
M v + m v' = 0
5000 x v - 10 x 4000 = 0
5000 v = 40000
v = 8 m/s
speed of rocket = 8 m/s
now,
we know
change in momentum = F x Δ t
F = 4000 N
Hence, the average force applied to the rocket is equal to F = 4000 N