This can be solved using momentum balance, since momentum is conserved, the momentum at point 1 is equal to the momentum of point 2. momentum = mass x velocity
m1v1 = m2v2
(0.03kg x 900 m/s ) = 320(v2)
v2 = 27 / 320
v2 = 0.084 m/s is the speed of the astronaut
Answer:
T=26.03 N
Explanation:
Given that
Distance between poles = 12 m
Mass of block m= 4 kg
Sag distance = 5 m
Lets take tension in the clothesline is T.
The component of tension in vertical direction will be T cosθ.
By force balancing
2 T cosθ = 40
here 
θ=39.80°
2 T cos39.8 = 40
T=26.03 N
I think it is False because as the Gad relajases fuel it doesn’t move as much anymore
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
