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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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A long wave with a period of about 15 minutes will travel across the oceans at a speed of approximately _______.
Temka [501]

The speed of the wave  is mathematically given as

v=2266.66m/s

A long wave with a period of about 15 minutes will travel across the oceans at a speed of approximately v=2266.66m/s

<h3>Speed of the wave</h3>

Question Parameters:

A long wave with a period of about 15 minutes

Generally the equation for the Wave velocity  is mathematically given as

v=\lambda * frequency

Where

f=1/t

Therefore

v=\lambda * frequency

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For more information on Speed

brainly.com/question/4931057

7 0
2 years ago
In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?
kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰  (angle with respect x-axis)

Explanation:

E(t)  Electric Field is a vector, so we need to determine module and direction

E(t)  =  E(q₁)  + E (q₂)  Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂  respectively.

E(q₁) = K * q₁/ (d₁)²         K = 9 *10⁹   [N*m²/C²]    d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225      [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) =  9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9        tanα =  2,395      α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0
3 years ago
Read 2 more answers
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