What average force is required to stop a 1400 kg car in 6.0 s if the car is traveling at 90 km/h ? express your answer to two significant figures and include the appropriate units. enter positive value if the direction of the force is in the direction of the initial velocity and negative value if the direction of the force is in the direction opposite to the initial velocity.

F=5833.3 N N

Explanation:

Newton's second law applied to the car

F= m*a Formula (1)

F: Force in Newtons (N)

m : mass in kg

a: acceleration ( m/s²)

kinematics car

vf= v₀ + a*t  Formula (2)

vf : final velocity (m/s)

v₀ : final velocity  (m/s)

a : acceleration  ( m/s²)

t : time t

Equivalences

1 km= 1000m

1 h = 3600 s

Data

m= 1000kg

v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s

vf= 0

t= 6 s

Problem Development

We calculate the acceleration replacing the data in the formula (2) :

0 =  25 + a*6

a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)

We calculate the force is required to stop the car replacing the data  in the formula (1)

-F = 1400 kg*(-4.16 m/s²)

F=5833.3 N

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.

-4 on