What is the impulse on a 45,000 kg airplane when it changes its velocity from 242 m/s to 258 m/s?*

=720000kgm/s

Explanation:

The impuse is equivalent to the rate of change of momentum which is expressed below as

Impulse= mv2 - mv1

=m(V2 - V1)

V1= initial velocity= 242 m/s

V2=final velocity= 258 m/s?

m= mass of the plane =45,000 kg

Then if we input the values, we have,

= 45000(258-242)

=720000kgm/s

Hence,the impulse on a 45,000 kg airplane is 720000kgm/s

1/0.14nm = 10^10/1.4 a lot