We can solve the problem by using the law of conservation of energy.
When the rocket starts its motion from the Earth surface, its mechanical energy is sum of kinetic energy and gravitational potential energy:

where
m is the rocket's mass

is the rocket initial speed

is the gravitational constant

is the Earth's mass

is the distance of the rocket from the Earth's center (so, it corresponds to the Earth's radius)
The mechanical energy of the rocket when it is very far from the Earth is just kinetic energy (because the gravitational potential at infinite distance from Earth is taken to be zero):

where

is the final speed of the rocket.
By equalizing the initial energy and the final energy, we can find the final velocity:

<span>Plate tectonics suggest that the lithosphere floats and moves on the asthenosphere.</span>
Answer:
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I hope it's helpful!
Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37