At what speed would a 3.00 x 10^4 kg airplane have to fly and with a momentum of 1.60 x 10^9 kg.m/s
1 answer:
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Answer:
25 m/s
Explanation:
This question can be solved using equation of motion
where
v is the final velocity
u is the initial velocity
s is the distance covered while moving from initial to final velocity
a is the acceleration
_____________________________________________
Given
box moved for distance of 62.5 m
Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.
thus in this problem
a = -5.0 m/s2
V = 0 as body came to rest due to friction deceleration
u the initial velocity we have to find
the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.
so if we find speed of box, we will be able get sped of truck as well.
using equation of motion
Thus, initial speed with the truck was travelling was 25 m/s.