Identify the arrows that show the movement of carbon from the biosphere to the atmosphere in the carbon cycle.
1 answer:
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Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b =
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’=
h’ =
h’ = 1/9 h
Answer:
<u><em>The plank moves 0.2m from it's original position</em></u>
Explanation:
we can do this question from the constraints that ,
- the wheel and the axle have the same angular speed or velocity
- the speed of the plank is equal to the speed of the axle at the topmost point .
thus ,
<em>since the wheel is pure rolling or not slipping,</em>
<em>⇒</em>
where
<em> - speed of the wheel</em>
<em> - angular speed of the wheel</em>
<em> - radius of the wheel</em>
<em>since the wheel traverses 1 m let's say in time '' ,</em>
<em></em>
∴
⇒
the speed at the topmost point of the axle is :
⇒
this is the speed of the plank too.
thus the distance covered by plank in time '' is ,
⇒