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Radda [10]
3 years ago
11

Which affects the electrostatic charge more, charge or distance?

Physics
2 answers:
denpristay [2]3 years ago
8 0

Answer:

Charge

Explanation:

makvit [3.9K]3 years ago
3 0

Answer:

The electric force decrease as the distance increase and vice versa

As it obeys inverse square law that states ththaforce is inversly proportional to square if distance

so if the distance increased to twice the force will be decreased to quarter.

And that is coloumb's law of electrostatic force:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

F=k(Q1.Q2)/r2

So as you increase the distance between charged objects, Force(attraction/repulsion) weakens by the square of their distance.

As you decrease the distance between charged objects, Force strengthens by the square of their distance.

Inverse relationships are common in nature. In electrostatics, the electrical force between two charged objects is inversely related to the distance of separation between the two objects. Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance. These observations are commonly made during demonstrations and lab experiments. Consider a charged plastic golf tube being brought near a collection of paper bits at rest upon a table. The electrical interaction is so small at large distances that the golf tube does not seem to exert an influence upon the paper bits. Yet if the tube is brought closer, an attractive interaction is observed and the strength is so significant that the paper bits are lifted off the table. In a similar manner, charged balloons are observed to exert their greatest influence upon other charged objects when the separation distance is reduced. Electrostatic force and distance are inversely related.

The pattern between electrostatic force and distance can be further characterized as an inverse square relationship. Careful observations show that the electrostatic force between two point charges varies inversely with the square of the distance of separation between the two charges. That is, the factor by which the electrostatic force is changed is the inverse of the square of the factor by which the separation distance is changed. So if the separation distance is doubled (increased by a factor of 2), then the electrostatic force is decreased by a factor of four (2 raised to the second power). And if the separation distance is tripled (increased by a factor of 3), then the electrostatic force is decreased by a factor of nine (3 raised to the second power). This square effect makes distance of double importance in its impact upon electrostatic

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If I exert a 203 N force on a 58 kg box, what will the acceleration of the box be?
Anit [1.1K]

Answer:

<h2>3.5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{203}{58} =  \frac{7}{2}   \\

We have the final answer as

<h3>3.5 m/s²</h3>

Hope this helps you

7 0
3 years ago
Most irregular galaxies are thought to have formed _____
Musya8 [376]
Most irregular galaxies were have thought to have been formed from collisions from other galaxies <span />
6 0
4 years ago
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Please help me on this position-time graph assignment!!!
saw5 [17]

\text{Hello there! :)}

\text{The velocity is essentially the slope of the graph in a position vs. time graph, so:}

\text{Given: From 0 - 8 sec, velocity of 10 m/s.}\\\\\text{From 8 - 18 sec, velocity of -5 m/s. (In opposite direction because of the negative sign)}\\\\\\\text{From 18 - 22 sec, velocity of 0 m/s. (Object at rest)}\\\\\text{From 22 - 24 sec, velocity of 5 m/s.}\\\\\text{Graph the position vs. time graph using the velocities as the slope of the line }

6 0
4 years ago
A motorboat traveling on a straight course slows
never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

  • 65 km/h → 18.0556 m/s
  • 35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

  • v² = v₀² + 2aΔx

Substitute the known values into the equation.

  • (9.72222)² = (18.0556)² + 2a(45)
  • 94.52156173 = 326.0046914 + 90a
  • -231.4831296 = 90a
  • a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0
3 years ago
A car which is traveling at a velocity of 1.6 m/s undergoes an acceleration of 9.2 m/s over a distance of 540 m. How fast is it
mixas84 [53]

Answer:

Vf = 99.7 m/s

Explanation:

In order to find the final velocity of the car, we will use the third equation of motion. The third equation of motion is given as follows:

2as = V_{f}^{2} - V_{i}^{2}

where,

a = acceleration of the car = 9.2 m/s²

s = distance traveled by the car = 540 m

Vf = Final Speed of the car = ?

Vi = Initial Speed of the car  1.6 m/s

2(9.2\ m/s^{2})(540\ m) = (V_{f})^{2}-(1.6\ m/s)^{2}\\V_{f}^{2} = 9936\ m^{2}/s^{2} + 2.56\ m^{2}/s^{2}\\\\V_{f} = \sqrt{9938.56\ m^{2}/s^{2}}

<u>Vf = 99.7 m/s</u>

8 0
3 years ago
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