The change in velocity is +4 m/s to the right (or -4 m/s to the left).
The object's mass is irrelevant.
Answer:
96046 Ns.
Explanation:
We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.
40 km/h in the east
V₁ = 40 i
V₂ = 50j
momentum p₁ = mV₁
= 1500 X 40 i
= 60000 i
Momentum p₂ = mV₂
= 1500 X 50j
= 75000 j
Change in momentum
p₂ - p₁
75000j - 60000i
Magnitude of change
= 
= 96046 Ns.
(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:

(c) The total charge is

. To get the charge on each piece, we should divide this value by 8, the number of pieces:

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):

If we approximate piece 6 as a single charge, the electric field is given by

where

and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have

poiting towards the center of piece 6, since the charge is negative.
(e) missing details on this question.
Since there is no decimal point in the number given above, the counting for the number of the significant figures will start from the left. Then, the first zero from the left is insignificant. Therefore, in this number there are 6 significant figures.
Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00

At x = 1.00 m

= 4J
Kinetic energy = (1/2)mv²

= 18J
Total energy will be =
4J + 18J = 22J
At x = 5

= -0.24J
Kinetic energy =

= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2

= 3.33 m/s
b) magnitude of force when x = 5.0m



At x = 5.0 m


= 0.016N