Answer:
6Ω
Explanation:
The 6Ω resistor and the 3Ω resistor are in parallel.
R = 1 / (1/6 + 1/3)
R = 1 / (1/6 + 2/6)
R = 1 / (3/6)
R = 6/3
R = 2
The 8Ω resistor and the other 8Ω resistor are also in parallel.
R = 1 / (1/8 + 1/8)
R = 1 / (2/8)
R = 8/2
R = 4
This 4Ω resistance is in series with the 4Ω resistor.
R = 4 + 4
R = 8
The 2Ω resistor and the 6Ω resistor are also in series.
R = 2 + 6
R = 8
These two 8Ω resistances are in parallel.
R = 1 / (1/8 + 1/8)
R = 1 / (2/8)
R = 8/2
R = 4
Finally, this 4Ω resistance is in series with the 2Ω resistance we found earlier.
R = 4 + 2
R = 6
The total equivalent resistance is 6Ω.
Answer:
10
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cords unwind the rod rotates. Find the tension in the cords as they unwind.
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.