Answer:
Average speed of Elain = 60 km/h
Explanation:
Total Distance covered by Jack = 360km
Average Speed of Jack = 80 km/h
Time taken by Jack to complete his journey = Distance / Average speed = 360 km / 80 km/h
Time taken by Jack to complete his journey = 4.5 hours
As it is given the both Jack and Elain travelled the same amount of distance:
Total distance travelled by Elain = 360 km
It is given that Elain took 1.5 hourse more than Jack to cover the distance, so Time taken by Elain to cover the distance is = 4.5 hours + 1.5 hours = 6 hours
Average speed of Elain = Distance/ time = 360 km / 6 hours
Average speed of Elain = 60 km/h
Density<span> is the </span>mass<span> of an object </span>divided<span> by its </span>volume<span>. So the answer would be Yes. Hope it helps! (:</span>

Explanation:
Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as
(1)
Assuming that the velocity remains constant then

Solving for
we get

Before we plug in the given values, we need to convert them first to their appropriate units:
The thrust <em>F</em><em> </em> is

The exhaust rate dm/dt is


Therefore, the velocity at which the exhaust gases exit the engines is


Answer:
its the teal/second from the left one
Explanation: