A1000.–kilogram car traveling 20.0 meters per second east experiences an impulse of 2000. newton • seconds west. what is the final velocity of the car after the impulse has been applied?
(1)18.0 m/s east
(2)19.5 m/s east
(3)20.5 m/s west
(4)22.0 m/s west

So the formula is equal to

M1v1 + I = m2v2

Where m is the mass of the object

V is the velocity of the object

I is the impulse applied

(1000 kg) (20 m/s) – 2000 = m2v2

20000 – 2000 = m2v2

18000 = m2v2

Since the car did change its mass, m1 = m2

18000 = (1000)v2

V2 = 18 m/s east 

The Final velocity ( V₂ ) of the car after the impulse has been applied is ; ( A ;  18.0 m/s east

Given data :

Mass of car = 1000 kg

Initial velocity of car = 20.0 m/s  east

Impulse experienced by car = 2000 N.s  west

V₂ ( final velocity ) after impulse is applied = ?

To determine the final velocity Apply the formula below

M₁V₁  +  Impulse = M₂V₂   ( 1 )   note : M₁ = M₂

since Impulse is applied in opposite direction to the movement of the car  

Equation ( 1 ) becomes

1000 ( 20 ) - 2000 = 1000 ( V₂ )

∴ V₂ = ( 20000 - 2000 ) / 1000

       = ( 18,000 ) / 1000

       = 18.0 m/s east  

Hence we can conclude that the final velocity of the car after the Impulse is applied is 18.0 m/s east.

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the ima (higher than 1m) will be greater than the ima with the height of 1m.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j