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SVEN [57.7K]
2 years ago
9

The electrostatic force between two charges is 3. 8 N. If the distance between the two charges is tripled, what is the new elect

rostatic force acting between the charges?
Physics
1 answer:
Dmitrij [34]2 years ago
8 0

Answer:

F = K Q1 Q2 / R^2         force between 2 charges separated by R

F2 / F1 = (R1 / R2^2)    dividing equations

F2 / F1 = (R1 / 3 R1)^2 = 1/3^2 = 1/9

The force would be reduced by a factor of 9

3.8 / 9 N = .42 N

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Explain how heat (thermal energy) can change the motion (kinetic energy) of the particles in objects.
Vika [28.1K]

Answer:

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Explanation:

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3 years ago
The d subshell can hold up to 10 electrons in an atom.<br> true or false
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Explanation:

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3 years ago
Read 2 more answers
A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied​
never [62]

Answer:

6N

Explanation:

Given parameters:

Pressure applied by the woman  = 300N/m²

Area = 0.02m²

Unknown:

Force applied  = ?

Solution:

Pressure is the force per unit area on a body

        Pressure  = \frac{force}{area}

         Force  = Pressure x area

        Force  = 300 x 0.02  = 6N

8 0
3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
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