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2 years ago

What is the distance between two spheres, each with a charge of 2.5 x 10-6 c, when the force between them is 0.50 n?

1 answer:

7 0

In this case, Coulomb's Law applies:

F = 1/(4πε₀) · (Q₁Q₂/r²)

You can solve it for r:

r = √[1/(4πε₀) · (Q₁Q₂/F<span>)]

Plugging in numbers:

r = </span>√[1/(4π·8.85×10⁻¹²) · (2.5×10⁻⁶)²/0.50]
= 0.335m

The correct answer is: the two charges are 0.335m apart.

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When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are gen

stealth61 [152]

Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

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2 years ago

Water is stored in 5 places on earth.Name each place an dnote the phrase that water is in liquid solid,gas

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Atmosphere - gas
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2 years ago

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

$I_2 = 4W/m^2$

Therefore the intensity at five times this distance from the source is $4W/m^2$

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3 years ago

How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?

myrzilka [38]

(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)

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2 years ago

Gravity is greater when there is

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More mass and less difference

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3 years ago

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