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swat32
2 years ago
10

The distance between earth and sun is 15000000km. Light takes 499 seconds to reach earth from sun. Calculate the speed of light

from the data provided. Mention the unit as well.
Physics
2 answers:
iVinArrow [24]2 years ago
7 0

To solve the problem we must know about the relationship between Speed, distance, and Time.

<h3>What is the relationship between Speed, distance, and Time?</h3>

We know that sped, distance, and time all are in a relationship to each other. this relationship can be given as,

\rm{Speed = \dfrac{Distance}{Time}

The speed of the light is 30,060.12 km/sec.

Given to us

  • The distance between the earth and the sun is 15000000km
  • Light takes 499 seconds to reach earth from the sun.

We know that speed can be described as,

\rm{Speed = \dfrac{Distance}{Time}

Therefore,

<h3>What is the speed of the light?</h3>

\text{Speed of light} = \dfrac{\text{Distance between the earth and the sun}}{\text{Time taken by the light to travel the distance}}

Substitute the value,

\text{Speed of light} = \dfrac{15,000,000\ km}{499\ seconds}

\text{Speed of light} = 30,060.12\ km/sec

Hence, the speed of the light is  30,060.12 km/sec.

Learn more about Speed, distance, and Time:

brainly.com/question/15100898

fredd [130]2 years ago
6 0

Answer:

The speed of light is 30,060,120.24048 m/s.

Explanation:

15000000 km into m

1 km = 1000m

15000000 km = 15000000 × 1000 m

= 15000000000m

speed = distance travelled / time taken

or, speed = 15000000000/499 m/s

so, speed = 30,060,120.24048m/s

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2 years ago
The smallness of the critical angle θc for diamond means that light is easily "trapped" within a diamond and eventually emerges
hoa [83]

Answer:

26.6°

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By the Snell's law

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How does the configuration of the electric field occur between a "parallel plate" setup in a lab?What is the effect of conductor
sleet_krkn [62]

Explanation:

The magnitude of the electric field between the plates is given by

E = -ΔV/d

minus sign indicates Potential decreases in the direction of electric field

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D is the distance between the plates.

The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface.  That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.

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If the earth shrank until its radius were only one-quarter its present size without changing its mass what would a 20 n object w
Dahasolnce [82]

Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2

20N/4 = 5N

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