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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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3 years ago

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2 years ago

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Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes attached to them. Face to face, th

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Answer:

0.78 m

Explanation:

By the conservation of energy, the energy that they gain from potential energy, must be equal to the kinetic energy. So, for Adolf:

Ep = Ek

ma*g*ha = ma*va²/2

Where ma is the mass of Adolf, g is the gravity acceleration (10 m/s²), ha is the height that he reached, and va is the velocity. So:

100*10*0.51 = 100*va²/2

50va² = 510

va² = 10.2

va = √10.2

va = 3.20 m/s

Before the push, both of them are in rest, so the momentum must be 0. The system is conservative, so the momentum after the push must be equal to the momentum before the push:

ma*va + me*ve = 0, where me and ve are the mass and velocity of Ed. So:

100*3.20 + 81ve = 0

81ve = 320

ve = 3.95 m/s

By the conservation of energy for Ed:

me*g*he = me*ve²/2

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2 years ago

Light travels 300 000 000 m/s and one year has approximately 32 000 000 second a light year is the distance light travels in one

Lelu [443]

Explanation:

It is given that,

Speed of light, $v=300 000 000\ m/s=3\times 10^8\ m/s$

Seconds in 1 year, $t=32 000 000=32\times 10^6\ s$

We need to find the distance traveled by light in one year. Speed of an object is given by :

$v=\dfrac{d}{t}$

So,

$d=v\times t\\\\d=3\times 10^8\times 32\times 10^6\\\\d=9.6\times 10^{15}\ m$

Since,

$1\ \text{light year}=9.46\times 10^{15}\ m\\\\1\ m=\dfrac{1}{9.46\times 10^{15}}\ \text{ly}\\\\9.6\times 10^{15}\ m=\dfrac{9.6\times 10^{15}}{9.46\times 10^{15}}\\\\d=1.01\ \text{ly}$

So, the distance covered by light is 1.01 light years.

8 0

2 years ago

What are the2 factors thatincrease theelectric forcebetweenobjects?​

What are the2 factors thatincrease theelectric forcebetweenobjects?