Rank the following situations according to the magnitude of the impulse of the net force, from largest value to smallest value. in each situation a 1000-kg automobile is moving along a straight east–west road. the automobile is initially : (i) moving east at 25 m> s and comes to a stop in 10 s; (ii) moving east at 25 m> s and comes to a stop in 5 s; (iii) at rest, and a 2000-n net force toward the east is applied to it for 10 s; (iv) moving east at 25 m> s, and a 2000-n net force toward the west is applied to it for 10 s; (v) moving east at 25 m> s; over a 30-s period, the automobile reverses direction and ends up moving west at 25 m> s.

V

I and II

III and IV

Explanation:

The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.

Taking always east as positive direction, and labelling

u the initial velocity

v the final velocity

m = 1000 kg the mass (which is always equal)

We find:

(i)

u = 25 m/s

v = 0

(II)

u = 25 m/s

v = 0

(III)

In this case,

F = 2000 N is the force

is the time

So the magnitude of the impulse is

(IV)

F = 2000 N is the force

is the time

So the magnitude of the impulse is

(V)

u = 25 m/s

v = -25 m/s

So the ranking from largest to smallest is:

V

I and II

III and IV

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.