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-Dominant- [34]
3 years ago
14

A point charge of +3 C is located at the origin of a coordinate system and a second point charge of -6 C is at x = 1.0 m. At w

hat point on the x-axis is the electrical potential zero?
Physics
1 answer:
Butoxors [25]3 years ago
8 0

Answer:

The point at which the electrical potential is zero is x = +0.33 m.

Explanation:

By definition the electrical potential is:

V_{E} = \frac{K*q}{r}

Where:

K: is Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

r: is the distance

The point at which the electrical potential is zero can be calculated as follows:

V_{1} + V_{2} = 0

K(\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}) = 0    (1)

q₁ is the first charge = +3 mC

r₁ is the distance from the point to the first charge  

q₂ is the first charge = -6 mC

r₂ is the distance from the point to the second charge    

By replacing r₁ = 1 - r₂ into equation (1) we have:

K(\frac{q_{1}}{1 - r_{2}} + \frac{q_{2}}{r_{2}}) = 0   (2)

By solving equation (2) for r₂:

r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m

                 

Therefore, the point at which the electrical potential is zero is x = +0.33 m.

I hope it helps you!  

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Answer:

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The value of radial acceleration \alpha_{r} = 80 \frac{m}{s^{2} }

Explanation:

Angular acceleration = 50 \frac{rad}{s^{2} }

Radius of the disk = 0.8 m

Angular velocity = 10 \frac{rad}{s}

We know that tangential acceleration is given by the formula \alpha_{t} = r \alpha

Where r =  radius of the disk

\alpha = angular acceleration

⇒ \alpha_{t} = 0.8 × 50

⇒ \alpha_{t} = 40 \frac{m}{s^{2} }

This is the value of tangential acceleration.

Radial acceleration is given by

\alpha_{r} = \frac{V^{2} }{r}

Where V = velocity of the disk = r \omega

⇒ V = 0.8 × 10

⇒ V = 8 \frac{m}{s}

Radial acceleration

\alpha_{r} = \frac{8^{2} }{0.8}

\alpha_{r} = 80 \frac{m}{s^{2} }

This is the value of radial acceleration.

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4 years ago
The forces F 1, F 2, ..., F n acting at the same point P are said to be in equilibrium if the resultant force is zero, that is,
allochka39001 [22]
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3 years ago
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Mountain pull. A large mountain can slightly affect the direction of ""down"" as determined by a plumb line. Assume that we can
Nastasia [14]

Answer:

6.18 um

Explanation:

The plumb line will be pulled down by a combination of the gravitationall pull of Earth and of the mountain. The Earth pulls down and the mountain to the side. Because of this it will fall not in a straight line down, but slightly to the side. Since the plumb line will follow the compound gravity we can imagine a rectangle triangle formed by the plumb line, a vertical line that ends at the same height as the plumb line, and the sideways displacement.

The total gravity will be proportional to the plumb line lenght, the vertical line will be proportional to Earth's gravity and the sideways displacement to the mountain pull.

The gravity of Earth is 9.81 m/s^2

The pull of the mountain will be defined by Newton's law of universal gravitation:

F = G \frac{m1 * m2}{r^2}

Where

F: pull force

G: universal gravitational constant (6.67e-11 m^3/(kg * s)

m1: mass of the mountain

m2: mass of the plumb

r: distance between mountain and plumb (3 km in this case)

If we divide both sides by m2 we obtain the acceleration towards the mountain of the plumb

a = G \frac{m1}{r^2}

Now we need the mass of the mountain. This will be its volume times it's density. The volume depends on the radius (since we consider it as a sphere)

m1 = \delta * \frac{4}{3} * \pi * r^3

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This is very small compared to the pull of Earth, so we can make an approximation that the length of the plumb line is equal to vertical line.

We can use the principle of similar triangles to say that:

\frac{\Delta x}{L} = \frac{a}{g}

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Answer:

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