S the work done on Amanda's car while speeding up (i) greater than, (ii) less than, or (iii) the same as the work done on Bertha's car while speeding up?

Explanation:

Mass of amanda and bertha car = Ma = Mb

Initial velocity of amanda, ua = 10 m/s

Final velocity of amanda, va = 20 m/s

Initial velocity of bertha, ub = 20 m/s

Final velocity of bertha, vb = 30 m/s

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Ea = Ma × 1/2 × (20^2 - 10^2)

= Ma × 150

= 150 Ma J

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Eb = Mb × 1/2 × (30^2 - 20^2)

= Mb × 250

= 250 Mb J

Option ii. Less than. Thus this is because the workdone by amanda car (150Ma) is less than the workdone in berthas car (250Mb).

B.

Impulse, p = force × time

p = Mass × change in velocity

pa = Ma × va - Ma × ua

= Ma × (20 - 10)

= 10 × Ma

= 10 Ma

p = Mass × change in velocity

pa = Mb × vb - Mb × ub

= Mb × (30 - 20)

= 10 × Mb

= 10 Mb

Option iii. The same as.

The impulse as seen above is the same in amanda car (10Ma) as the same with bertha (10Mb)

an energy transformation flow diagram is shown

where is it ?

where is the photo and the answer choices?