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gtnhenbr [62]
2 years ago
9

An initially electrically neutral conducting sphere is placed on an insulating stand. A negatively-charged glass rod is brought

near, but does not touch the sphere. Without moving the rod, a wire is then attached to the sphere that connects it to earth ground. The rod and wire are then removed simultaneously. What is the final charge on the sphere?a) negative
b) positive
c) neutral
d) It has a fifty percent chance of having a positive charge and a fifty percent chance of having a negative charge.
Physics
1 answer:
balu736 [363]2 years ago
7 0

Answer: Option (c) is the correct answer

Explanation:

It is known that when a neutral object comes in contact with a charged object then an opposite charge develops on the neutral object. But this development of charge occurs only when both the objects come in contact and if they are not in contact with each other then there occurs no charge on the neutral object.

As in the given situation, negatively-charged glass rod is brought near, but does not touch the sphere. So, there will occur no charge on the sphere.

Thus, we can conclude that final charge on the sphere is neutral.

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In which situations can you conclude that the object is undergoing a net interaction with one or more other objects? (Select all
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Answer:

<em>4, A proton in a particle accelerator moves faster and faster.</em>

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When a 100-Ω resistor is connected across the terminals of a battery of emf ε and internal resistance r, the battery delivers 0.
gladu [14]

Answer:

The emf and internal resistance of the battery are 9 volt and 2.04 Ω

Explanation:

Given that,

First resistorR = 100\ \Omega

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Second resistorR=200\ \Omega

PowerP=0.401\ W

We need to calculate the emf and internal resistance

Using formula of emf

P=\dfrac{e^2}{R+r}

For first resistor

0.794=\dfrac{e^2}{100+r}....(I)

For second resistor

0.401=\dfrac{e^2}{200+r}.....(II)

From equation (I) and (II)

\dfrac{795}{401}=\dfrac{200+r}{100+r}

Using component of dividend rule

\dfrac{393}{401}=\dfrac{100}{100+r}

r =2.04\ \Omega

Now, Put the value of internal resistance in equation (I)

0.794=\dfrac{e^2}{100+2.04}

e^2=0.794(100+2.04)

e=\sqrt{0.794(100+2.04)}

e=9\ volt

Hence, The emf and internal resistance of the battery are 9 volt and 2.04 Ω

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