Answer:
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Explanation:
Answer:
Final velocity of the car will be -9.28 m/sec
Explanation:
We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec
Acceleration of the car
in negative direction so acceleration will be 
From first equation of motion we know that
v = u+at
So 
So final velocity will be -9.28 m/sec
Answer:
ΔF=125.22 %
Explanation:
We know that drag force on the car given as

=Drag coefficient
A=Projected area
v=Velocity
ρ=Density
All other quantity are constant so we can say that drag force and velocity can be given as

Now by putting the values



Percentage Change in the drag force



ΔF=125.22 %
Therefore force will increase by 125.22 %.
Answer:
gap junctions,tight junctions,and desmosomes
Answer:
60 N.C
Explanation:
The box will move in the direction of the push and pull with a force of 60 N. C.