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vladimir2022 [97]
3 years ago
12

When two oppositely charged electrodes are immersed in a solution, positively charged ions are attracted to the negative electro

de and gain electrons. The negatively charged ions are attracted to the positive electrode and release electrons. What would be the net effect of this on conduction of electricity through the solution?
Physics
1 answer:
e-lub [12.9K]3 years ago
5 0

Answer:

conductivity of solution is reduced.

Explanation:

When two oppositely charged electrodes are immersed in a solution, positively charged ions are attracted to the negative electrode and gain electrons. The negatively charged ions are attracted to the positive electrode and release electrons.

Due to the process mentioned above , the negatively charged ions are accumulated at the positive electrode and the positively charged ions are accumulated at the negative electrode . This accumulation prevents further attraction of ions at oppositely charged electrodes because the incoming ions face repulsion from already accumulated ions at electrodes.  Further , it creates an emf acting in opposite direction . It reduces the current through the solution. Hence conductivity of solution is reduced.

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Answer:

Q=977216.256\ J=977.216\ kJ

Explanation:

Given:

  • mass of  steam, m=384\ g
  • temperature of steam, T_{is}=100^{\circ}C
  • temperature of resultant water, T_{fw}=31^{\circ}C

We have,

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<em>When we cool the steam of 100°C then firstly it loses its latent heat to convert into water of 100°C and the further cools the water.</em>

<u>Now the heat removed from steam to achieve the final state of water:</u>

\rm Q=latent\ heat\ of\ vapourization+sensible\ heat\ of\ water

Q=m(L+c.\Delta T)

Q=384(2256+4.186\times (100-31))

Q=977216.256\ J=977.216\ kJ

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An electron in a mercury atom changes from energy level b to a higher energy level when the
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1eV = 1.6 * 10^-19 J

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Explain how a form of<br> energy is used in daily life
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Answer:

When we talk about residential uses of energy, these are the most basic uses of energy . They include watching television, washing clothes, heating and lighting at home , taking a shower, working from home on your laptop or computer, running appliances and cooking.

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3 years ago
A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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Answer: dakdadakdadakdadakda

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