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Oduvanchick [21]
4 years ago
10

Consider two metallic rods mounted on insulated supports. One is neutral, the other positively charged. You bring the two rods c

lose to each, but without contact, and briefly ground the the neutral rod by touching it with your hand.
Physics
2 answers:
katen-ka-za [31]4 years ago
4 0

Answer:The rod touching your hand will have a Negative charge.

Explanation: This phenomenon is described as charging by induction,the other iron rod became charged as a result the repulsion force existing between the positively charged iron rod and the electronic forces around the election sphere. This make it different from the charging done by contact where both iron rods have close contact and electron transfer is achieved through contact. The shows that an electric charge can be persuaded and caused to transfer to another metallic particle without physical contact with the charged metallic particle.

anastassius [24]4 years ago
4 0

Answer:

The initial neutral rod would be induced with negative charges.

Explanation:

This process is called charging by induction. When you bring the charged rod close to that of the neutral, the charges in it would be redistributed. The positively charged rod attracts the negative charges within the neutral rod and repel like charges. Touching the neutral with hand causes the positive charge to be neutralised due to earthing. Thus the neutral rod has now been charged negatively.

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A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

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3 years ago
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3 years ago
The total negative charge on the electrons in 1mol of helium (atomic number 2, molar mass 4) is ________?
DaniilM [7]

Answer:

1.92\times 10^5 C

Explanation:

We are given that

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We have to find the total negative charge on the electrons in one mole of Helium.

We know that atomic number=Proton number

Proton number=Number of electrons=2

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1 mole of Helium=6.02\times 10^{23} atoms

We know that q=ne

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1 e=1.6\times 10^{-19}C

Using the formula

q=2\times 1.6\times 10^{-19}=3.2\times 10^{-19}C

Total negative charge in 1 mole=3.2\times 10^{-19}\times 6.02\times 10^{23}=1.92\times 10^5C

Hence, the total negative charge on the electrons in 1 mole of Helium=1.92\times 10^5 C

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