Answer:
4.2 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v
315 g m/s − 525 g m/s = -525 g m/s + (75 g) v
315 g m/s = (75 g) v
v = 4.2 m/s
A) Acceleration= (282m/s-0m/s)/5.0s= 56.4m/s^2
b) Deceleration=(0m/s-282m/s)/1.40s=201.4m/s^2 (here we reject negative because deceleration already take into account negative)
c) for the first one the multiple will be 5.76g
for the second one it will be 20.6g
Answer: Option B.
Since here the truck is moving on a circular track, it will experience centripetal force.
F(centripetal) = m × acc
or

where r is the radius of the track.
m is the mass of truck
v is the speed of the truck.
Given: v = <span>13 m/s
m = </span><span>1,600 kg
</span>F = 3300 Newton
To find = radius of track=?


r = 81.94 m
Therefore, radius of track is 81.94 m
P always P because P is an awkward Dorian letter that can always be trusted
Induced motion or movement