Force applied on the car due to engine is given as
towards right
Also there is a force on the car towards left due to air drag
towards left
now the net force on the car will be given as

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.
So we can say



So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.
Answer:
96w
Explanation:
p=Iv..where v=12 and I=8.0
Explanation:
It is given that,
A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.
In 2 seconds, distance covered by the mass is 12 cm.
In 1 seconds, distance covered by the mass is 6 cm
So, in 16 seconds, distance covered by the mass is 96 cm
So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.