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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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I think the temperature increases
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3 years ago
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An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr
notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

       Charge = Current \times Time

As two different current is passing at two different times, the net charge will be the different in current.  So,

        \text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

        V=\frac{k q}{R}

Here k = 9 * 10^9, q is the charge and R is the radius. As q=2 \times 10^{-6} \times t and R =17 cm = 0.17 m, then the voltage will be

        V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

The time is required to find to reach the voltage of 1500 V, so

1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}

So, 14 ms is required to reach the potential of 1500 V.

3 0
3 years ago
You drive a car 640 m to the east, then 340 m to the north what is the magnitude of your displacement
mixer [17]

Magnitude of displacement = \sqrt{640^2 + 340^2}

Adding the squares gives displacement = \sqrt{525,200}

Displacement =  \sqrt{525,200} ≈ 724.7m

6 0
3 years ago
a 100 g cart initially moving at 0.5 m/s collides elastically from a stationary 180 g cart. a) using the equation in the theory,
cupoosta [38]

A 100 g cart is moving at 0.5 m/s that collides elastically from a stationary 180 g cart. Final velocity is calculated to be 0.25m/s.

Collision in which there is no net loss in kinetic energy in the system as a result of the collision is known as elastic collision . Momentum and kinetic energy both are conserved quantities in elastic collisions.

Collision in which part of the kinetic energy is changed to some other form of energy is inelastic collision.

For an elastic collision, we use the formula,

m₁V₁i+ m₂V₂i = m₁V1f + m₂V₂f

For a perfectly elastic collision, the final velocity of the 100g cart will each be 1/2 the velocity of the initial velocity of the moving cart.

Final velocity = 0.5/2

=0.25 m/s.

To know more about elastic collision, refer

brainly.com/question/7694106

#SPJ4

7 0
1 year ago
If five joules were required to move a crate in 3.7 seconds, what power was applied?
AleksAgata [21]

Answer:

The answer to your question is 1.35 Watts

Explanation:

Data

Work = W = 5 J

time = t = 3.7 s

Power = P = ?

Formula

Power is a rate in which work is done or energy is transferred over time

P = \frac{W}{t}

Substitution

P = \frac{5}{3.7}

Result

P = 1.35 W

8 0
3 years ago
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