The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N
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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Answer:
F =
.
Explanation:
Gravitational force between two objects of masses
kept at a distance r is given by the formula
F = 
Here ,
= 2m
= 
Thus , F = 
F =
.
Lift force exerted by the air on the rotors=143244 N
Explanation:
we use Newtons second law
F- (M+m)g=(M+m)a
F= lift force
m= mass of helicopter= 13000 Kg
M= mass of car= 2000 lb=907.2 kg
a= acceleration= 0.5 m/s²
g= acceleration due to gravity
F- (M+m)g=(M+m)a
F=(M+m)(a+g)
F=(13000+907.2)(0.5+9.8)
F=143244 N