Answer:
Explanation:
I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .
As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.
If I am sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.
An inclined plane decreases the amount of force needed to move an object but increases the distance the onject needs to be moved. Since work = distance x force, whe amount of work stays the same.
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 
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Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding,
= 50 turns
number of turns in the secondary winding,
= 10 turns
the secondary load resistance,
= 250 Ω
Determine the turns ratio;

Now, determine the reflected resistance in the primary winding;

Therefore, the reflected resistance in the primary winding is 6250 Ω