Acar travels along a straight line at a constant speed of 60.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 30.0 mi/h. (a) what is the constant speed with which the car moved during the second distance d

velocity during second d = 20.0 mi/h

Explanation:

Total distance travelled is 2d, with an average velocity of 30.0 mi/h you can express the time travelled in terms of d:

distance = velocity * time

time = distance / velocity

time = 2d/30.0

The time needed for the first d at 60.0 is:

time = d/60.0

The time in the second d you can get it by substracting both times (total time - time for the first d)

second d time = 2d/30.0 - d/60.0

= 4d/60.0 - d/60.0

= 3d/60.0

and with the time (3d/60.0) and the distance travelled (d) you can get the velocity:

velocity = distance / time

velocity = d / (3d/60.0)

= 60.0/3 = 20.0 mi/h

it may be wrong but i think its a

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