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Finger [1]
3 years ago
7

Which vector below goes from (0, 0) to (-2, 3)?

Physics
1 answer:
TEA [102]3 years ago
8 0
Here is the answer!!!!!!!

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Which two options would INCREASE the electric force between two charged particles?
atroni [7]

The electric force between two charged particles can be increased by decreasing the distance between the two particles.

<h3>How to increase electric force between two charged particles.</h3>

The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while

increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.

Read more on Electric Force:

brainly.com/question/17692887

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2 years ago
Maureen takes notes in class. Wave Interactions
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3 years ago
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The axis of the earth is<br> Tilted about 23.5 degrees<br> Vertical<br> Vertical
melisa1 [442]

Answer:

The axis of rotation of the Earth is tilted at an angle of 23.5 degrees away from vertical, perpendicular to the plane of our planet's orbit around the sun. The tilt of the Earth's axis is important, in that it governs the warming strength of the sun's energy.

Explanation:

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5 0
3 years ago
g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the
Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

T^2 = \frac{4\pi^2 d^3}{2GM}

The given data are given with respect to known constants, for example the mass of the sun is

m_s = 1.989*10^{30}

The radius between the earth and the sun is given by

r = 149.6*10^9m

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

m = 8.2*1.989*10^{30}

d = 6.2*149.6*10^6

Substituting in Kepler's third law:

T^2 = \frac{4\pi^2 d^3}{2}

T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}

T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}

T = 120290789.7s

T = 120290789.7s(\frac{1year}{31536000s})

T \approx 3.8143 years

Therefore the period of this star is 3.8years

7 0
3 years ago
Does the tide that the moon raises on the earth different?
alukav5142 [94]

Answer:

No the gravity of the moon pulls the water making high tide

Explanation:

7 0
3 years ago
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