b. the forces of attraction among them limit their motion.
Explanation:
It is given that,
Diameter of the circular loop, d = 1.5 cm
Radius of the circular loop, r = 0.0075 m
Magnetic field, 
(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :
I = 32.22 A
(b) The magnetic field on a current carrying wire is given by :
r = 0.00238 m
Hence, this is the required solution.
(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²
<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>
<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>
Option A overweight
HOPE IT HELPS!!
Answer:
a) Θ = ω₀*t + ½αt² To complete first revolution 2π rads = 0*t + ½αt² and to complete the first and second combined 4π rads = 0*t + ½α(t+0.810s)² Divide second by first: 2 = (t + 0.810s)² / t² This is quadratic in t and has roots at t = -0.336 s ← ignore and t = 1.96 s ◄ b) Use either equation from above: 2π rads = 0*t + ½α(1.96s)² α = 3.27 rad/s² ◄ Hope this helps!
Explanation:
