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lesantik [10]
3 years ago
14

2. A solenoid. Suppose the south end of a bar magnet was introduced to the right end of this solenoid at a constant velocity. Wh

at direction would you expect the induced magnetic field to be in?4. To decrease the magnitude of current induced in the inductor using a bar magnet, you could (select all that apply):a. Decrease the strength of the magnet.b. Decrease the velocity of the magnet going into the solenoid.c. Decrease the number of coils in the solenoid.d. Increase the cross sectional area of the solenoid.6. List the TWO values you need to record to measure a voltage on the oscilloscope
Physics
1 answer:
Studentka2010 [4]3 years ago
5 0

Answer:

2) deflection must be towards the negative side of the voltage.

4) the correct statements are: b and c

Explanation:

2) This question is based on Faraday's law of induction, when we introduce a magnet in a solenoid an induced current is produced that generates a voltage that is given by

           E = - N d \phi_{B} / dt

where \phi_{B} = B. A

The bold are vectors

Therefore, when applying this formula to our case, the induction lines of the magnetic field increase as we approach the solenoid, as the South pole approaches the lines are in the direction of the magnet, therefore the normal to the solenoid that has an outgoing direction and the magnetic field has 180º between them and the cos 180 = -1; consequently the deflection must be towards the negative side of the voltage.

4) From the Faraday equation we can see that the inductive electromotive force depends

* The magnitude of B that changes over time

* The area of ​​the loop that changes over time

* The angle between B and the area that changes over time

* A combination of the above

With this analysis we will review the different alternatives given

a) False. It takes a temporary change and an absolute value of B

b) True. As the speed decreases, the change in B decreases, that is, dB / dt decreases

c) True. The current is induced in each turn, if there is a smaller number the total current will be smaller

d) False. A temporary change of area is needed, in addition to increasing the area the current increases

We can see that the correct statements are: b and c

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A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.
irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system (\tau), measured in Newton-meters, is:

\tau = I\cdot \alpha (1)

Where:

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\alpha - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

\alpha = \frac{\omega - \omega_{o}}{t} (2)

Where:

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\omega - Final angular speed, measured in radians per second.

t - Time, measured in seconds.

If we know that \tau = 3\,N\cdot m, \omega_{o} = 0\,\frac{rad}{s }, \omega = 145.875\,\frac{rad}{s} and t = 4\,s, then the moment of inertia of the motor is:

\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}

\alpha = 36.469\,\frac{rad}{s^{2}}

I = \frac{\tau}{\alpha}

I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }

I = 0.0823\,N\cdot m\cdot s^{2}

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

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If we put negative charge between two similar positive charges then what is it's equilibrium? And how?​
Gnesinka [82]

Your question has been heard loud and clear.

Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.

Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.

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Answer:

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Explanation:

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