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NARA [144]
3 years ago
5

Fill in the blanks. The electrostatic force between two objects is proportional to the ____________________ of the distance ____

________________.
Physics
2 answers:
Shkiper50 [21]3 years ago
6 0
Write an equation to calculate the force between two objects if the product of their charges is 10.0 × 10-4 C. (Note: Use the variable R for the distance between the charges.)

F = 900 ÷_________
Luba_88 [7]3 years ago
6 0

Answer:

The electrostatic force between two objects is proportional to the inverse of the square of the distance between the centres of the two objects

Explanation:

According to Coulomb's law of electrostatic attraction, the electrostatic force F between two charges or objects is directly proportional to the product of the absolute values of the charges or (charges on the objects) and inversely proportional to the square of their distance of separation.

Given that q_1 and q_2 are the charges or the charges on the objects, we can break down Coulomb's statement mathematically as follows;

F\alpha q_1q_2.............(1)  F is proportional to the product of the charges.

F\alpha \frac{1}{r^2}.......................(2) F is proportional to the inverse of the square of their distance r of separation.

This is an inverse-square relationship.

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The electric forces on both beads are equal.
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Which of the following would NOT affect the level at which a canoe floats in a pond
nataly862011 [7]
Sry i was knowing the answer i forgot ;(
5 0
3 years ago
Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to
Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

<em>I will take two assumptions:</em>

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the <u>linear density of the stretched wire</u>) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new <u>charge must be 10 times the original charge</u>, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

6 0
3 years ago
In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})

{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

Solving the above eqn:

X_{C} = 4R

6 0
3 years ago
Which law states that for any closed loop, the sum of the length segments times the magnetic field in the direction of the segme
dedylja [7]
That's Ampere's Law.  ( C ).
The magnetic permeability is the proportionality constant.
3 0
3 years ago
Read 2 more answers
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