Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy 
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
The frequency of oscillation is 2.153 Hz
What is the frequency of spring?
Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.
For the mass-spring system in this problem,
The Frequency of spring is calculated with the equation:
Where,
f = frequency of spring
k = spring constant = 64 N/m
m = mass attached to spring = 350g = 0.350 kg
a = maximum acceleration = 5.3 m/s^2
Substituting the values in the equation,
Hence,
The frequency of oscillation is 2.153 Hz
Learn more about frequency here:
<u>brainly.com/question/13978015</u>
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Explanation:
Particle moving in a circular path with a constant speed.
Answer:
625 W
Explanation:
Applying
P = W/t.................... Equation 1
Where p = power, W = Work, t = time
But,
W = Force (F) × distance (d)
W = Fd........................ Equation 2
Substitute equation 2 into equation 1
P = Fd/t.................... Equation 3
From the question,
Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds
Substitute these values into equation 3
P = (5000×30)/240
P = 625 Watt
