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2 years ago

Which two characteristics increase the strength of the gravitational force

Physics

1 answer:

snow_tiger [21]2 years ago

4 0

Answer:

The answer is A and C.

Explanation:

Only two factors are relevant when dealing with the gravitational force between two objects - their mass and their distance apart from one another. Gravity's force is proportional to the product of the masses of the two objects and is inversely proportional to the square of the distance between them.

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An electric charge, A, is placed carefully between two other charges, B and C, and experiences no net electric force. Do B

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Answer:

I do not have enough information to tell

Explanation:

This is deduced due to the fact that if the net force due to B and C on A is zero, the charges on B and C could either be positive or negative depending on the charge on A.

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3 years ago

Breaks in the Earths crust called faults from where plates meet true or false

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False theyre called plate boundries

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3 years ago

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What does the index of refraction directly measure? the angle between the incident ray and the normal line the angle between the

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<span>The bending of light in a medium</span>

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4 years ago

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You are walking in the forest and see a bear. According to the Cannon-Bard theory, what happens next?

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3 years ago

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We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

7 0

3 years ago