A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in
1 answer:
Answer:
96046 Ns.
Explanation:
We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.
40 km/h in the east
V₁ = 40 i
V₂ = 50j
momentum p₁ = mV₁
= 1500 X 40 i
= 60000 i
Momentum p₂ = mV₂
= 1500 X 50j
= 75000 j
Change in momentum
p₂ - p₁
75000j - 60000i
Magnitude of change
=
= 96046 Ns.
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Answer:
distance difference would a) increase
speed difference would f) stay the same
Explanation:
Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.
Their equations of motion for distance and velocities are
Their difference in distance are therefore:
(As
So as time progress t increases, Δs would also increases, their distance becomes wider with time.
Similarly for their velocity difference
Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.
This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.