The range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x 
90° 0
180° - 375 π x 
We have current carrying wire in a form of a circle placed in a uniform magnetic field.
We have to the range of potential energies of the wire-field system for different orientations of the circle.
<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>
The formula to calculate the magnetic potential energy is -
U = M.B = MB cos 
where -
M is the Dipole Moment.
B is the Magnetic Field Intensity.
According to the question, we have -
U = M.B = MB cos 
We can write M = IA (I is current and A is cross sectional Area)
U = IAB cos 
U = Iπ
B cos 
For
= 0° →
U(Max) = MB cos(0) = MB = Iπ
B = 5 × π ×
× 3 ×
=
375 π x
.
For
= 90° →
U = MB cos (90) = 0
For
= 180° →
U(Min) = MB cos(0) = - MB = - Iπ
B = - 5 × π ×
× 3 ×
=
- 375 π x
.
Hence, the range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x 
90° 0
180° - 375 π x 
To solve more questions on Magnetic potential energy, visit the link below-
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<span>The change in internal energy is only gravitional PE because the tube is being drug up at a constant speed. Since it is at a constant speed, the change in KE is 0.
Change in PE = m*g*h = 78 kg * 10 m/s^2 * 30 m = 23400 J
Work done on the system is from the force
Work = force * distance = 350 N * 120 m = 42000 J
So, work added 42000 J to the system, but the rider's energy only increased 23400 J. Therefore, friction took up the difference. Friction is where the thermal energy comes from
Q = 42000 J - 23400 J = 18600 J.
Therfore, friction generated 18600 J of heat to the surroundings.</span>
I think it is this because compression stroke it needs to be compressed then open up when started.
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:
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<h3>Hope this helps </h3>