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Tom [10]
3 years ago
13

A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime

nter then briefly touches the electroscope with a finger. The finger is removed, followed by the removal of the positively charged object. What happens to the leaves of the electroscope when a negative charge is now brought near but not in contact with the top of the electroscope?
Physics
1 answer:
Olin [163]3 years ago
7 0

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

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A 2000 g of C 14 is left to decay radioactively the half-life of Corbin 14 is approximately 5700 years what fraction of that sam
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Answer:

  1/8

Explanation:

17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...

  (1/2)(1/2)(1/2) = 1/8

1/8 of the sample remains after 17,100 years.

8 0
3 years ago
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Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that
lawyer [7]

As we know by the first law of thermodynamics

Q = \Delta U + W

here we know that

Q = heat given to the system

\Delta U = U_f - U_i

W = work done by the system

now here we can say

\Delta U = 180 - 78 = 102 J

W = 64 J

now we can say that heat will be given as

Q = 64 + 102 = 166 J

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

8 0
3 years ago
The hydrogen and helium under the clouds of Jupiter are in liquid form due to _____.
asambeis [7]
<span>high pressure produced by the clouds because its the most likely!!!!!!!!!!</span>
8 0
3 years ago
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In the diagram, the crest of the wave is show by: <br> A <br> B<br> C <br> D
Ivenika [448]

Answer:

D.

Explanation:

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8 0
3 years ago
An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 4.95 cm2 . When the current is 12.5 A
Dafna1 [17]

I will try to define the net problem, without the intermediate message between the message as:

<em>An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 4.95 cm2 . When the current is 12.5 A , the energy stored is 0.390 J . </em>

<em>Part A: How many turns does the winding have?</em>

To solve the problem it is necessary to apply the concepts related to the storage of energy in an inductor and how it is possible to calculate from the inductance the number of turns of the system.

By definition we know that the energy stored in an inductor is given by,

E = \frac{1}{2} LI^2

Where,

L = Inductance

I = Current

In this way, clearing the Inductance in the previously given equation we have to

L = \frac{2E}{I}

L = \frac{2(0.39J)}{12.5A}

L = 0.0624H

In a system the inductance is given by

L =\frac{\mu_0 N^2A}{l}

Where l represents the length, however as we deal with the perimeter of a circle we have,

L =\frac{\mu_0 N^2A}{2\pi R}

Replacing our values we have

(0.0624)=\frac{(4\pi*10^{-7})(N)^2(12.5)}{2\pi (15.5/2*10^{-3})^2}

Re-arrange to find N,

N^2 = \frac{(0.0624)2\pi (15.5/2*10^{-3})}{(4\pi*10^{-7})(4.95*10^{-4})}

N^2 = 4884848.48

N = 2210.16\approx 2211 turns

Therefore the winding have 2211turns

8 0
3 years ago
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