D. 128.25 because force=mass x acceleration
The field strength needed to produce a 24.0 V peak emf is 0.73T.
To find the answer, we need to know about the expression of emf.
What's the expression of peak emf produced in a rotating rectangular loops?
- The peak emf produced in a rotating loops= N×B×A×w
- N= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequency
- So, B = emf/(N×A×w)
<h3>What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?</h3>
- N= 300, A= 5cm × 5.22cm = 0.05m × 0.0522m = 0.00261 m²
- Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/s
- Now, B= 24/(300×0.00261×42)
B= 24/(300×0.00261×42) = 0.73T
Thus, we can conclude that the magnetic field is 0.73T.
Learn more about the electromagnetic force here:
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Answer:0.669
Explanation:
Given
mass of clock 93 kg
Initial force required to move it 610 N
After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity
Initially static friction is acting which is more than kinetic friction
thus 613 force is required to overcome static friction


The answer to this question is D or the last one