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Soloha48 [4]
3 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]3 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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Answer:

8.91 J

Explanation:

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radius, r = 0.22 m

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n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

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3 years ago
What is the value of x in the equation below 1+2e^x+1=9
GuDViN [60]
<h2>Answer: 1.252</h2>

Explanation:

We are given this equation and we need to find the value of x:

1+2e^x+1=9   (1)

Firstly, we have to clear x:

2e^x=9-1-1  

2e^x=7  

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Applying<u> Natural Logarithm</u> on both sides of the equation (2):

ln(e^x)=ln(\frac{7}{2})     (3)

xln(e)=ln(\frac{7}{2})     (4)

According to the Natural Logarithm rules xln(e)=x, so (4) can be written as:

x=ln(\frac{7}{2})     (5)

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x=1.252    

3 0
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3 0
3 years ago
A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m
belka [17]

Answer:

Energy stored in the capacitor is U=2.7\times 10^{-5}\ J        

Explanation:

It is given that,

Charge, q=1.5\ \mu C=1.5\times 10^{-6}\ C

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

U=\dfrac{1}{2}q\times V

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U=2.7\times 10^{-5}\ J

So, the potential energy is stored in the capacitor is U=2.7\times 10^{-5}\ J. Hence, this is the required solution.

4 0
3 years ago
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