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PilotLPTM [1.2K]
1 year ago
9

when charges mutually repel and distribute themselves on the surface of conductors, what becomes of the electric field inside th

e conductor?
Physics
1 answer:
AleksAgata [21]1 year ago
3 0

The charges align themselves so that the conductor's internal field is zero.

<h3>What occurs if a charged surface is in close proximity to a conducting surface?</h3>

Induced charges are created on the conductor when a charge is brought close to it. The internal free charges of the conductor, however, are gathered throughout its surface because the electric field inside the conductor must be zero in order to defeat the electric field of induced charges.

<h3>What takes place within a conductor?</h3>

A substance that has a lot of free electrons accessible for the flow of current is said to be a conductor. Since there are numerous electrons, a powerful force of repulsion exists between them as well. As a result, the electrons move to lessen their attraction for one another.

To know more about conductor's visit:-

brainly.com/question/18084972

#SPJ4

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The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.
Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0
3 years ago
A large spool in an electrician's workshop has 65 m of insulation-coated wire coiled around it. When the electrician connects a
Art [367]

Answer:

40.34\ \text{m}

Explanation:

L_1 = Length of wire = 65 m

I_1 = Initial current = 1.8 A

I_2 = Final current = 2.9 A

We know

R\propto \dfrac{1}{I}

and

R\propto L

\dfrac{V}{I}\propto L\\\Rightarrow L\propto \dfrac{1}{I}

so

\dfrac{L_2}{L_1}=\dfrac{I_1}{I_2}\\\Rightarrow L_2=\dfrac{I_1}{I_2}L_1\\\Rightarrow L_2=\dfrac{1.8}{2.9}\times 65\\\Rightarrow L_2=40.34\ \text{m}

The length of the wire remaining on the spool is 40.34\ \text{m}.

8 0
3 years ago
a 5.00 × 105 kg rocket is accelerating straight up. Its engines produce 1.50 × 107 of thrust, and air resistance is 4.50 × 106 N
Angelina_Jolie [31]
This one is simple :)

use the equation F = ma and then re-arrange for a.

a = F / m

But there is a trick, so be careful. the question gives you wind resistance. Simply subtract the wind resistance from the thrust of the rocket to get the net force upward.

1.50 * 10^7N - 4.5 8 10^6N = 1.05 * 10^7N

So,

a =  \frac{F}{m} =  \frac{1.05*10^7N}{5.00*10^5kg}
3 0
3 years ago
If the object is placed 6 cm from a concave mirror the image distance 12 cm what is the image discige of the objects move from t
Alika [10]

Answer:

The image distance is -9.3cm.

6 0
2 years ago
Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta
Sergeu [11.5K]

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

Heat capacity at constant volume = 4.9 cal/mol.K

Heat capacity at constant pressure = 6.9 cal/mol.K

Initial temperature = 300 K

Final temperature = 320 K

We need to calculate the heat flow in to gas at constant pressure

Using formula of heat

\Delta H_{1}=nC_{p}\times\Delta T

Put the value into the formula

\Delta H_{1}=3\times6.9\times(320-300)

\Delta H_{1}=414\ cal

We need to calculate the heat flow in to gas at constant volume

Using formula of heat

\Delta H_{1}=nC_{v}\times\Delta T

Put the value into the formula

\Delta H_{1}=3\times4.9\times(300-320)

\Delta H_{1}=-294\ cal

We need to calculate the heat flows into the gas during two steps

Using formula of total heat

\Delta H_{T}=\Delta H_{1}+\Delta H_{2}

\Delta H_{T}=414-294

\Delta H_{T}=120\ cal

Hence, The heat flows into the gas during this two-step process is 120 cal.

7 0
4 years ago
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