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2 years ago

What are the2 factors thatincrease theelectric forcebetweenobjects?

Physics

1 answer:

harkovskaia [24]2 years ago

4 0

The two factors that affect the electric force are the charges and the separation between the objects

Explanation:

The magnitude of the electrostatic force between two objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two objects

Looking at the equation, we see that the electric force between objects depends on two factors:

The charges of the objects: the force is directly proportional to the product of the two charges, therefore the greater the charges, the stronger the force
The distance between the two objects: the force is inversely proportional to the square of the distance, therefore the smaller the distance between the objects, the stronger the force

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

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38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

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2 years ago

Describe the ratio of kinetic energy and potential energy the skateboarder will have at each of the following points in her ride

amm1812

Answer:

no answer

Explanation:

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2 years ago

Raj is trying to make a diagram to show what he has learned about nuclear fusion.

KIM [24]

No, he should place the He atom and energy on the right, and the H atoms and the heat and energy on the left.

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2 years ago

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A parallel-plate capacitor has a voltage of 391 v applied across its plates, then the voltage source is removed. what is the vol

andrezito [222]

When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling $C_0$ the capacitance of the capacitor in air, the charge Q, the capacitance $C_0$ and the voltage ( $V_0=391 V$ ) are related by

$C_0 =\frac{Q}{V_0}$ (1)

when the source is disconnected the charge Q remains on the capacitor.

When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):

$C=k C_0 = 5.4 C_0$