A700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. what is the car’s change in momentum due to this collision? a. 2.3x 10^4 kg·m/s towards the wall b. 2.3x 10^4 kg·m/s away from the wall c. 1.7 x 10^4 kg·m/s away from the wall d. 1.7 x 10^4 kg·m/s towards the wall impulse

C. 1.7 × 10⁴ kg.m/s away from the wall

Explanation:

Change in momentum: This is the product of mass and change in velocity of a body. The Unit of change in momentum is kg.m/s.

Change in momentum of the car = m(v₂ - v₁) Equation 1

Where m = mass of the car, v₁ = initial velocity of the car, v₂ = final velocity of the car.

Given: m = 700 kg, v₁ = 29 m/s, v₂ = 4.5 m/s

Substituting these values into equation 1

Change in momentum of the car = 700(4.5-29)

Change in momentum of the car = 700(-24.5)

Change in momentum of the car = 700(-24.5)

Change in momentum of the car = -17150 kg.m/s

Therefore the Change in momentum of the car is 1.7 × 10⁴ kg.m/s away from the way wall.

The right option is C. 1.7 × 10⁴ kg.m/s away from the wall

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

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