Given:
density of air at inlet, 
density of air at inlet, 
Solution:
Now,

(1)
where
A = Area of cross section
= velocity of air at inlet
= velocity of air at outlet
Now, using eqn (1), we get:

= 1.14
% increase in velocity =
=114%
which is 14% more
Therefore % increase in velocity is 14%
If the object's <em>velocity is constant</em> ... (it's speed isn't changing AND it's moving in a straight line) ... then the net force on the object is zero.<em> (D)</em>
Either there are no forces at all acting on the object, OR there are forces on it but they're 'balanced' ... when you add up all of their sizes and directions, they just exactly cancel each other out, and they have the SAME EFFECT on the object as if there were no forces at all.
Force = Work/distance
Force = 150/10
= 15 Newtons
Force = 15 Newtons
Therefore, 15 newtons of force is applied to the body when 150 joules of work
is done in displacing the body through a distance of 10m in the direction of the force.
Answer:
We know from the basic speed distance relation that

Since the car started from rest and it covered the distance between the 2 officer's in 19 minutes we have speed of the car

Which clearly exceeds the limit of 