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3 years ago

How much force is required to move a electron through an electric field with strength of 1.375 x 10^19 N/C?

Physics

1 answer:

Luda [366]3 years ago

4 0

Answer:

The magnitude of the force required to move the electron through the given field is 2.203 N

Explanation:

Given;

The field strength of the electron, E = 1.375 x 10¹⁹ N/C

charge of electron, q = 1.602 x 10⁻¹⁹ C

The magnitude of the force required to move the electron through the given field is calculated as follows;

F = Eq

F = (1.375 x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)

F = 2.203 N

Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N

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Answer:

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Explanation:

As we know that child is hanging by rope

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2) Tension in the rope which act vertically upwards

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2 years ago

A toy car pushed across the floor is observed to slow down and stop based on Newton’s first law of motion what is the best concl

Maksim231197 [3]

An object in motion eventually stops because of friction.

<h3>What is Newton’s first law of motion?</h3>

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So we can conclude that An object in motion eventually stops is the correct answer.

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2 years ago

If a sulfur atom has 16 protons, 16 electrons, and 16 neutrons, its atomic mass is:

Stolb23 [73]

Answer:

Explanation:

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3 years ago

Which option is a microscopic property that helps determine whether or not two substances will dissolve with each other? (1 poin

Bumek [7]

Answer:

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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

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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

3 years ago