The front 1.20 m of a 1,550-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a car traveling 24.0 m/s stops uniformly in 1.20 m, how long does the collision last

t = 0.1 s

Explanation:

Given:

- The distance crushed (Stopping-Distance) s = 1.20 m

- The mass of the car m = 1,550 kg

-  The initial velocity vi = 24.0 m/s

Find:

- How long does the collision last t?

Solution:

- The stopping distance s is the average velocity v times time t as follows:

                                s = t*( vf + vi ) / 2

Where,

             vf = 0 m/s ( Stopped )

                                s = t*( vi ) / 2

                                t = 2*s / vi

                                t = 2*1.20 / 24

                                t = 0.1 s

t=0.1seconds.

Explanation:

The mass of the car m = 1,550 kg  

We know the stopping distance, 1.20m,

we know the final velocity, 0m/s (its stopped),

the starting velocity, 24m/s.

d=t(v2+v1)/2

Rearange to solve for t, and remove the v2^2 as its zero

t=2d/v1  

t=2(1.20m)/(24m/s)

t=0.1seconds.

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27

spare some change and give it to the poor

explanation:

plz : ((