Atwo stage rocket leaves its launch pad moving vertically with an average acceleration of 4 m/s2. at 10 s after launch the first stage of the rocket (now without fuel) is released. the second stage now had an acceleration of 6 m/s2

a) how high is the rocket when the first stage seperates?

b)how fast is the rocket moving upon first stage seperation?

c) what will be the maximum height attained by the first stage after seperation?

d) what will be the distance between the first and second stages 2 s after separation

a) 200m

b) 40 m/s

c) 81.55m

d) 31.62m

Explanation:

Solution

a)

y = y0 + u×t+⅟2×a×t2 =

y0 = 0

u = 0

y = unknown

a = 4m/s2

t = time = 10 seconds

y = 0.5×4×102 = 200m

b) v = u + at

v = 0 + 4×10 = 40 m/s

c) v2 = u2 - 2×g×y

at maximum height  v = 0

we have

402 = 2×9.81×y

Y =81.55m

d)

for the stage 2 we haace

 y = y0 + u×t+⅟2×a×t2 =

y = 0 + 4×2+0.5×6×22 = 92m

for the stage one we have

y = 0+40×2-0.5×9.81×4= 60.38m

distance between the first and second stage  2s aftee separation = 92-60.38 = 31.62m

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation:

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j