Acar slows from 22m/s to 3.0m/a at a constant rate of 2.1m/s. how many seconds are required before the car is traveling at 3.0m/s?

t = 9.05 s

Explanation:

Given,

The initial velocity of the car, u = 22 m/s

The final velocity of the car, v = 3.0 m/s

The rate of change of speed of the car is 2.1 m/s,

                                      ∴ the acceleration, a = -2.1 m/s²

The acceleration and velocity is given by the relation

                                           a = (v - u)/t     m/s²

Therefore,

                                           t = (v-u)/a  s

Substituting the values in the above equation

                                          t = (3.0 - 22)/(-2.1)

                                            = 9.05 s

Hence, the time required before the car is traveling at 3.0m/s is t = 9.05 s

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

there are no atoms