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Soloha48 [4]
4 years ago
5

Dos cargas q_1=-8μC y q_2=13μC se encuentran a una distancia r=0.12 m. ¿Cuál es la fuerza resultante sobre una tercera carga q_3

=-4μC, situada en el centro de la distancia que separa las cargas q_1 y q_2?
Physics
1 answer:
KiRa [710]4 years ago
8 0

Answer:

209.53 N

Explanation:

To find the force on the third charge you use the Coulomb formula:

F=k\frac{q_1q_2}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

The force on the third charge is the contribution of the force between the third charge and the other ones:

F=F_1+F_2

By taking into account that the third charge is at the middle of the distance between charge 1 and charge 2 you have r = 0.12m/2 = 0.06m

Furthermore, you take into account that the first charge repels the third charge and the second charge attracts the third charge.

By replacing you have:

F=k\frac{q_1q_3}{r^2}+k\frac{q_2q_3}{r^2}\\\\F=\frac{k}{r^2}q_3[q_1+q_2]\\\\F=\frac{(8.98*10^9Nm^2/C^2)(4*10^{-6}C)}{(0.06m)^2}[8*10^{-6}C+13*10^{-6}C]\\\\F=209.53\ N

Hence, the force between on the third charge is 209.53 N

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Explanation:

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The y-position of a damped oscillator as a function of time is shown in the figure.
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(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

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-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

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ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

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ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

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2(-0.857/cosω) = -0.57/cos2ω

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3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

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6cos²ω  - cosω - 6 = 0

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6y² - y - 6 = 0

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y = 1.1 or -0.92

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ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

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2 years ago
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Answer:

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We can obtain the distance travelled by using the following formula:

s = (u + v) t /2

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s = 116 × 11 /2

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