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Kisachek [45]
3 years ago
8

A point charge of -2 µC is located at the origin. A second point charge of 6 µC is at x = 1 m, y = 0.5 m. Find the x and y coord

inates of the position at which an electron would be in equilibrium
Physics
1 answer:
Soloha48 [4]3 years ago
3 0

Answer:

x coordinate = -1.66 m

y coordinate is = -0.825m

Explanation:

Suppose z be the distance form the first charge and z + sqrt(1^2 +.5^2) be the distance from the second So z + sqrt(1+.25) = z + 1.12

We have k*2.0x10^-6/s^2 = k*6x10^-6/(s+1.12)^2

0.0356s^2 -0.019s-0.0897=0  

s=1.876m

The angle of the line between the two charges is arctan(.5/1) = 26.6o

x coordinate = -1.876*cos(26.6) = -1.66m

y coordinate is -1.876*sin(26.6) = -0.825m

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Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

              ½ a t² - c t + b = 0

we can see that this is a quadratic equation whose solution is

             t = [c ±√(c² - 4 (½ a) b)]  / 2

for the results to be real, the square root must be real, so the radicand must be greater than zero

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3 0
3 years ago
Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r
MakcuM [25]

Answer:

  • Current = 0.33 A

Explanation:

  • For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

\dashrightarrow \: \:  \sf V = 2 \times 5 = 10 V

Three resistor of 5\Omega, 10\Omega, 15\Omega are connected in Series, so the net resistance:

\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}

\dashrightarrow \: \:  \sf R = 5 + 10 + 15

{ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \:  \Omega}}}}

According to ohm's law:

\dashrightarrow  \sf\: \: V = IR

\dashrightarrow  \sf \: \: I = \dfrac{V}{R}

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 {\pmb{\sf{\Omega}}} we get:

\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}

{ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}

\thereforeThe electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

4 0
2 years ago
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Answer:

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

Explanation:

Given

Number of protons = 92

Radius of nucleus r_n = 7.4 * 10^{-15} m

Distance of the electrons r_1 = 1.0 * 10^ {-10} m

Part 1

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21}  N/C

Part 2

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13}  N/C

Part 3

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hence, the solution is

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

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7 0
3 years ago
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Grace [21]
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Serga [27]

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When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.

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