g A car with a mass of 1,120.0 kg accelerates from 0 to 80.0 km/h in 10.0 s. The driver applies the brakes when the car is moving at 80.0 km/h, and the car comes to rest after traveling 44.0 m. What is the net force (in N) on the car during its deceleration

8761.5 N

Explanation: Given that a car with a mass of 1,120.0 kg accelerates from 0 to 80.0 km/h in 10.0 s. The driver applies the brakes when the car is moving at 80.0 km/h, and the car comes to rest after traveling 44.0 m. What is the net force (in N) on the car during its deceleration

Mass m = 1120kg

First change km/h to m/s

(80 × 1000) ÷ 3600

Velocity = 22.2 m/s

Acceleration = ( 22.2 - 0 ) ÷ 10

Acceleration = 2.2 m/s^2

Force F1 = mass × acceleration

Substitutes all the parameters into the formula

F1 = 1120 × 2.2

F1 = 2489 N

For the car to come to rest, it will be decelerating.

Using 3rd law of motion

V^2 = U^2 + 2as

V = 0 since it is coming to rest.

Substitute all the parameters into the formula

0 = 22.2^2 - 2 × a × 44

88a = 492.84

a = 492.84/88

a = 5.6 m/s^2

The force F2 = 1120 × 5.6

F2 = 6272.5N

Net force = 6272.5 + 2489

Net force = 8761.5 N

Therefore, the net force is 8761.5 N

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer