Me . see attached for the questions with the graph

Answer

5) b-c

6)    a-b and

       e-f

7) f-g

9) a-b = 0 m/s

    c-d = 0.6667 m/s

    e-f = 0 m/s

    f-g = -3 m/s

10) b-c ⇒ The cart is acceleration.

   e-f ⇒ The cart is moving backwards with a constant velocity.

Explanation

Answer

5) b-c

In the section b-c the cart is accelerating because the slope of the graph is changing. The gradient that represent velocity is increasing.

6) a-b and e-f

At this sections the distance is not changing at all. This can only mean that the cart is not moving. It is at rest.

7) f-g

At this section the slope is negative meaning the cart is moving back to where it came from.

9) a-b = 0 m/s

At a-b the cart is not moving. So the velocity is zero.

    c-d = 0.66667 m/s

Velocity = distance / time

               =(50-40)/(40-25)

                = 10/15

                 = 0.6667  m/s

    e-f = 0 m/s

At e-f the cart is not moving. So the velocity is zero.

    f-g = -3 m/s

Velocity = distance / time

               = (60-30)/(65-75)

                = 30/-10

                = - 3 m/s

10) b-c ⇒ The cart is acceleration.

   e-f ⇒ The cart is moving backwards with a constant velocity.

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.