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Fudgin [204]
3 years ago
6

Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 7.0 nC. Using insulatin

g handles, the blocks are moved so they touch each other. After touching for a few seconds, the blocks are separated (again using insulating handles). What is the final charge of block A?
Physics
2 answers:
maxonik [38]3 years ago
6 0

Answer:

Block A will have a final charge of 3.5nC.

Explanation:

This is because at the point of contact with Block B, which is electrically positive, the electrons in Block A will be attracted to the excess 'unpaired' protons in block B. Hence, the electrons will flow into Block B causing unpaired protons to remain in Block A.

This process is called Charging by Conduction.

This charging process will continue until the charges are evenly distributed between both objects.

In case you're wondering, "<em>how's all this possible within a few seconds</em>?", remember that electrons travel very fast and so, this process is a rather rapid one.

Liula [17]3 years ago
5 0

Answer:

3.5 nC

Explanation:

When both metal blocks get in touch each other, as they are assumed perfect conductors, excess charge is evenly spread out on the surfaces on both blocks.

Due to the total charge must be conserved, this means that if one block had a charge Q, and the other block was neutral, after separated, each of them must have exactly the half of the total excess charge.

So, if the total charge was 7 nC, each block must keep 3.5 nC, i.e., the final charge of block A is 3.5 nC.

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Olenka [21]

Answer:

12.8 Watts

Explanation:

P = VI

P = (2 V) (6.4 A)

P = 12.8 W

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4 years ago
How much heat is required to heat 2 kg of water from 25°C to 40°C?
Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

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3 years ago
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The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

4 0
3 years ago
Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?
liberstina [14]

Answer : The maximum concentration of silver ion is 5\times 10^{-12}m

Solution : Given,

K_{sp} for AgBr = 5\times 10^{-13}

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

NaBr(aq)\rightleftharpoons Na^++Br^-

The concentration of NaBr solution is 0.1 m that means,

[Na^+]=[Br^-]=0.1m

The equilibrium reaction for AgBr is,

                          AgBr\rightleftharpoons Ag^++Br^-

At equilibrium                     s       s

The expression for solubility product constant for AgBr is,

K_{sp}=[Ag^+][Br^-]

The concentration of Ag^+ = s

The concentration of Br^- = 0.1 + s

Now put all the given values in K_{sp} expression, we get

5\times 10^{-13}=(s)(0.1+s)

By rearranging the terms, we get the value of 's'

s=5\times 10^{-12}m

Therefore, the maximum concentration of silver ion is 5\times 10^{-12}m.

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