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
Answer:
F-F(gr) = ma
a= {F-F(gr)}/m =
=(15-10)/15=0.33 m/s² (upward)
Answer:
a)- 1.799 rad/sec²
b)- 17.6 x 10ˉ³Nm
Explanation:
ω₀ = 720 rev/min x (1 min/60 sec) x (2π rad / 1 rev) = 24π rad/s
a) Assuming a constant angular acceleration, the formula will be
α = (ωf -ω₀) / t
As final state of the grindstone is at rest, so ωf =0
⇒ α = (0-24π) / 41.9 = - 1.799 rad/sec²
b)Moment of inertia I for a disk about its central axis
I = ½mr²
where m=2kg and radius 'r'= 0.099m
I = ½(2)(0.099²)
I = 9.8 x 10ˉ³ kgm²
Next is to determine the frictional torque exerted on the grindstone, that caused it to stop, applying the rotational equivalent of the Newton's 2nd law:
τ = I α =>(9.8 x 10ˉ³)(- 1.799)
τ = - 17.6 x 10ˉ³Nm
(The negative sign indicates that the frictional torque opposes to the rotation of the grindstone).
Answer:
1) 50°
Explanation:
We need to find the angle of incidence first before finding the angle of reflection.
Angle of incidence = 90° - 40°
= 50°
Since the angle of incidence is the same as the angle of reflection, the angle of reflection here would be 50°.
Answer:
False
Explanation:
The depending on what type of pressure is enclosed in a object, the smaller the object the more pressure because that pressure has limited room to escape.
