By applying the third law of motion, which force is greater when you push on a wall? Your force, the wall’s force, your force an
2 answers:
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Answer:
Part a)
Part b)
Explanation:
As we know that mountain climber is at rest so net force on it must be zero
So we will have force balance in X direction
now we will have force balance in Y direction
Part a)
so from above equations we have
Part b)
Now for tension in right string we will have
Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s