A cylindrical rod of mass M. length L and radius R has two cords wound around it whose ends are a
1 answer:
Answer:
T = mg/6
Explanation:
Draw a free body diagram (see attached). There are two tension forces acting upward at the edge of the cylinder, and weight at the center acting downwards.
The center rotates about the point where the cords touch the edge. Sum the torques about that point:
∑τ = Iα
mgr = (1/2 mr² + mr²) α
mgr = 3/2 mr² α
g = 3/2 r α
α = 2g / (3r)
(Notice that you have to use parallel axis theorem to find the moment of inertia of the cylinder about the point on its edge rather than its center.)
Now, sum of the forces in the y direction:
∑F = ma
2T − mg = m (-a)
2T − mg = -ma
Since a = αr:
2T − mg = -mαr
Substituting expression for α:
2T − mg = -m (2g / (3r)) r
2T − mg = -2/3 mg
2T = 1/3 mg
T = 1/6 mg
The tension in each cord is mg/6.
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Explanation:
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The mass of the copper cube is
The final temperature of both substance is
Generally form the law of thermal energy conservation,
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Generally the heat lost by the copper cube is mathematically represented as
The specific heat of copper is
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So
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Answer:
Current, I = 1000 A
Explanation:
It is given that,
Length of the copper wire, l = 7300 m
Resistance of copper line, R = 10 ohms
Magnetic field, B = 0.1 T
Resistivity,
We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :
r = 0.00199 m
or
The magnetic field on a current carrying wire is given by :
I = 1000 A
So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.