The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.
The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.
v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s
Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.
The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.
Combining this together we get:
(1) vx=40m/s and vy=10m/s
Answer:
28716.4740661 N
1.2131147541 m/s
51.2474965841%
Explanation:
m = Mass of plane = 74000 kg
s = Displacement = 3.7 m
f = Frictional force = 14000 N
t = Time taken = 6.1 s
u = Initial velocity = 0
v = Final velocity

Force is given by

The force with which the team pulls the plane is 28716.4740661 N

The speed of the plane is 1.2131147541 m/s
Kinetic energy is given by

Work done is given by

The fraction is given by

The teams 51.2474965841% of the work goes to kinetic energy of the plane.
Answer:
10
Explanation:
i = 5/.5 = 10 Amps. Hope this helps :)
D I think .... don’t be mad if I’m wrong
Answer:
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s