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mafiozo [28]
3 years ago
7

How do expectations relate to future schooling?

Physics
1 answer:
Snezhnost [94]3 years ago
8 0

Expectations relate to future schooling by choosing classes that are at your skill level.

Answer: Option B

<u>Explanation: </u>

Each individual has a different set of skills blessed by the Mighty and each should enhance them according to his/her will and interest. For the future prospects of schooling and teaching, there is one noticeable point to be acted is that students should not be forced to study the subject beyond their skills and capability.

We should remember that a fish cannot fly and a bird cannot swim. They both cannot be trained under the same study programmes. Although they both can know how to swim and fly but it’s their capability and interest over the subjects that can lead them further to rule over either the sky or the oceans.

So, the future school programming should be done in such a way that individuals have an authority to choose their subjects accordingly and learn more about their skills.

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Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o
Irina18 [472]

Answer:

a. q_C experiences the greatest net force and q_B experiences the smallest net force

b. Ratio of the greatest to the smallest net force= 9

Explanation:

<u>Electrostatic Forces </u>

Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude

\displaystyle F=K\frac{q_1q_2}{d^2}

If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.

Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.

a.

The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is

F_{TA}=-F_B-F_C+F_D

Computing the individual forces we have

\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}

\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}

\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}

The total force on qA is:

\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})

\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}

Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus

\displaystyle F_{TB}=F_A-F_C+F_D

\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}

\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}

Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus

\displaystyle F_{TC}=F_A+F_B+F_D

\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}

\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}

Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus

\displaystyle F_{TD}=-F_A-F_B-F_C

\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}

\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}

Comparing the relative values of all the forces

\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|

This means that qc experiences the greatest net force and qB experiences the smallest net force

b.

The ratio of the greatest to the smallest forces is

\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9

5 0
3 years ago
How does the mass of an object affect its acceleration?
solong [7]

Answer:

as the mass of an object is increased, the acceleration of the object is decreased & vice-versa.

Explanation:

hope this helps!

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2 years ago
A 1.00- and a 2.00- resistor are in parallel. What is the equivalent single resistance? ​
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Answer: R=2/3 ohms

Explanation:

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A 10kg box putting pushed with a force of 24 Newtons and a fraction of 12N.
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Answer:

The two dogs sitting here are already poor and ignorant

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How to increase the current in a circuit
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Answer:

So to increase current of the circuit what you can do is :

1. Use conductor of low resistivity, ¶.

2. Use conductor of small length.

3. Use thick wire.

4. Decrease the temperature of the circuit.

5. If operating temprature is high than use semiconductor, because it have negative temprature coefficient.

6. Minimise the circuit losses.

3 0
3 years ago
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