A helicopter of mass 1.40×103kg1.40×103kg is descending vertically at 3.00 m/sm/s. The pilot increases the upward thrust provided by the main rotor so that the vertical speed decreases to 0.450 m/sm/s as the helicopter descends 3.00 mm. What is the initial energy of the helicopter?

KE=6300 J

Explanation:

Given that

mass of the helicopter , m = 1.4 x 10³ kg

Initial velocity , u = 3 m/s

Velocity after increasing thrust , v = 0.45 m/s

Distance , d= 3 mm

The initial kinetic energy KE is given as follows

Now by putting the values in the above equation we get

KE=6300 J

Therefore the initial energy of the helicopter will be 6300  J.

50 inches should be the resultant vector

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.