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Complete Question
How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.
The output from the secondary coil is 12 V
Answer:
The value is
Explanation:
From the equation we are told that
The input voltage is
The number of turns of the primary coil is
The output from the secondary is
From the transformer equation
Here is the number of turns in the secondary coil
=>
=>
=>
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as
with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as
with ω denoting the angular velocity, which we are given. With that, the above becomes:
Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?
Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:
and so 45 rev/min = 4.71 rad/s.
A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.