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yarga [219]
3 years ago
12

Dos cargas puntuales están fijas en el eje x: q1 = 6.0µC está en el origen, O, con x1 = 0.0 cm, y q2 = –3.0 µC está situada en e

l punto A, con x2 = 8.0 cm. ¿Cuál es la intensidad del campo eléctrico en el punto medio entre las dos cargas?
Physics
1 answer:
erik [133]3 years ago
8 0

Answer:

E_total = 1.30 10¹⁰ C / m²

Explanation:

The intensity of the electric field is

     E = k q / r²

on a positive charge proof

The total electric field at the midpoint is

as q₁= 6 10⁻⁶ C the field is outgoing to the right

for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore

E_total = E₁ + E₂

E_total = k q₁ / r₁² + k q₂ / r₂²

r₁ = r₂ = r = 4 10⁻² m

E_total = k/r² (q₁ + q₂)

 we calculate

E_total = 9 10⁹ / (4 10⁻²)²   (6.0 10⁻⁶ +3.0 10⁻⁶)

E_total = 1.30 10¹⁰ C / m²

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When astronauts travel to the moon, their bodies experience a lower gravitational pull than on Earth. Which type of pull are the
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They are experiencing the pull of leaving the atmosphere

3 0
3 years ago
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A box, initially at rest, has 36.7 N of force exerted on it for 2.81 s. If the box has a mass of 7.41 kg, what was its velocity
ElenaW [278]

Answer:

13.91 m/s

Explanation:

First we need to find the acceleration:

Acceleration = Force/mass

Acceleration = 36.7N/7.41 kg

Acceleration = 4.95 m/s² (rounded to two decimal places)

Then we find the velocity:

Velocity = Acceleration * Time

Velocity = 4.95 m/s² * 2.81 s

Velocity = 13.91 m/s (rounded to two decimal places)

8 0
2 years ago
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During normal beating, a heart creates a maximum 3.95-mV potential across 0.305 m of a person’s chest, creating a 0.75-Hz electr
WINSTONCH [101]

Answer:

E = 0.0130 V/m.

Explanation:

The electric field is related to the potential difference as follows:

E = \frac{\Delta V}{d}

<u>Where:</u>

E: is electric field

ΔV: is the potential difference = 3.95 mV  

d: is the distance of a person's chest = 0.305 m

Then, the electric field is:

E = \frac{\Delta V}{d} = \frac{3.95 \cdot 10^{-3} V}{0.305 m} = 0.0130 V/m

Therefore, the maximum electric field created is 0.0130 V/m.

I hope it helps you!

7 0
3 years ago
uniform electric field exists between two parallel plates separated by 2.0 cm. The intensity of the field is 15 kN/C. What is th
adoni [48]

Answer:

potential difference V= 300 volts

Explanation:

Given:

d= 2.0 cm = 0.02m

E = 15 kN/C = 15 × 10³ N/C

For a uniform field between two plates, the Electric Filed Intensity (E) is proportional to the potential difference (V) and inversely  proportional to distance between the plates.

E= V/d

⇒ V= E×d = 15 × 10³ N/C × 0.02 m = 300 volts  (∴1 Nm/C = 1 J/C= 1 volts)

7 0
3 years ago
****PLEASE HELP**** WILL MARK BRAINLIEST
allochka39001 [22]

Answer:

The Current decreases

Explanation:

HOPE THIS HELPS!

4 0
2 years ago
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