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2 years ago

As the distance from a charged particle, "q", increases, what happens to the electric potential?

Physics

1 answer:

mr Goodwill [35]2 years ago

6 0

As the distance from a charged particle, "q", increases, the electric potential decreases.

<h3>Electric potential between particles</h3>

The electric potential between particles is the work done in moving a unit charge from infinity to a certain point against the electrical resistance of the field.

V = Kq/r

where;

K is Coulomb's constant
q is the magnitude of the charge
r is the distance between the charges

Thus, from the formula above, as the distance from a charged particle, "q", increases, the electric potential decreases.

Learn more about electric potential here: brainly.com/question/14306881

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4 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

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The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

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So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

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3 years ago

Which method do acoustic use to limit reverberations in music halls

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Answer:

reverberation time appropriate to the use and size of the room, adequate balance between direct and reverberant sound, intimacy and good sound diffusion in the room to obtain a uniform sound.

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3 years ago

A battery-operated car utilizes a 12.0 V system. Initially, the car is at rest at the base of a 195 m high hill. Some time later

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Answer:

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Explanation:

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q = 1683189.5/12 = 140265.8 C

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