Consider a circuit with a main wire that branches into two other wires. If the current is 10 A in the main wire and 4 A in one o
2 answers:
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The center of mass of the two objects is the average position of the parts of the two object system
The center of mass of block <em>A</em>, and block <em>B</em> after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>
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Reason:
The given parameters are;
The position of block A and block B at t = 0 is x = 0
The mass of block A, m₁ = 8 kg
Mass of block B, m₂ = 16 kg
Speed of block <em>A</em> = 0 m/s
Speed of block <em>B</em>, v₂ = 10 m/s
Location of the center of mass of the two object at t = 3 s; Required
Solution;
The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)
The location of block <em>B</em>, = v₂ × t
The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m
The center of mass of two masses are given as follows;
The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)
Learn more about the center of mass here:
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d= * tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip