The momentum of the mass expelled in the opposite direction ... the rocket-engine exhaust, or the ionized matter expelled from an ion drive.
THAT's why every propulsion engine has outlet nozzles designed with super-high-intensity math, to achieve the highest possible velocity for the mass that gets shot out the back ... so that it will carry the maximum possible momentum.
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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BaCI2 stands for Barium Chloride.
Answer:
you can simply answer by derivative = 3.5x^2+25x+250-y=0 you can derivate this eqn 7x +25-1=0 7x=24 yo u can divide you get it
