Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
We are given the mass of an <span>aluminum sculpture which is 145 kg and a horizontal force equal to 668 Newtons. The coefficient of friction can be determined by dividing the horizontal force by the weight of the object. In this case, 668 N / 145 * 9.8 equal to coeff of friction of 0.47</span>
<span> </span>For any prism-shaped geometry, the volume
(V) is assumed by the product of cross-sectional area (A) and height (h).
<span> V = Ah </span>
<span>
Distinguishing with respect to time gives the
relationship between the rates.
dV/dt = A*dh/dt</span>
<span> in the meantime the area is not altering </span>
<span>
dV/dt = π*(1 ft)^2*(-0.5 ft/min) </span>
<span>
dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min
Water is draining from the tank at the rate of π/2
ft^3/min.</span>
