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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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Which of the following elements most likely has the highest boiling point?

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Answer:

Explanationalluminiim

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3 years ago

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Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?

Xelga [282]

<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

30 cm = 0.3 m (distance)

=> Work done = Force × Distance

=> 10 × 0.3

=> 3 Joules

$\small \boxed{work \: done \: by \: Ryan \: = 3 \: Joules}$

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3 years ago

A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci

larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 = e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

#SPJ1

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2 years ago

A student walks 3 blocks east, 4 blocks north, and 3 blocks west. What is the displacement of the student?

12345 [234]

4 blocks north because he is it not asking for north east

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3 years ago

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r

dusya [7]

Answer:

28 degree C

Explanation:

We are given that

$T_1=4.00^{\circ}$

$R_1=217.7 \Ohm$

$R_2=215.1\Ohm$

$\alpha=-5.00\times 10^{-4}C^{-1}$

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

$\alpha(T_2-T_1)=\frac{R_2}{R_2}-1$

Using the formula

$-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1$

$-5\times 10^{-4}(T_2-4)=0.988-1=-0.012$

$T_2-4=\frac{0.012}{5\times 10^{-4}}=24$

$T_2=24+4=28^{\circ}C$

Hence, the temperature on a spring day 28 degree C.

7 0

3 years ago