Answer:
V= 6.974 m/s
Explanation:
Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N
Force of kinetic friction acting parallel and up roof = 18.0 N
Fnet force acting on tool box acting parallel and down roof
Fnet= 55.4 - 18.0
Fnet=37.4 N
acceleration of tool box down roof
a = 37.4(9.81)/88.0
a= 4.169 m/s²
d = 4.90 m
t = √2d/a
t= √2(4.90)/4.169
t= 1.662 s
V = at
V= 4.169(1.662)
V= 6.974 m/s
<u>Answer:</u>
The final velocity of the two railroad cars is 1.09 m/s
<u>Explanation:</u>
Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

where
V= Final velocity
M1= mass of the first object in kgs = 12000
M2= mas of the second object in kgs = 10000
V1= initial velocity of the first object in m/s = 2m/s
V2= initial velocity of the second object in m/s = 0 (given at rest)
Substituting the given values in the formula we get
V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

Which is the final velocity of the two railroad cars
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