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olya-2409 [2.1K]
2 years ago
11

a 4,000 kilogram rocket has accelerates at a rate of 35 m/s2. How much force is required to do this?​

Physics
1 answer:
WARRIOR [948]2 years ago
5 0

\Large{\underline{\sf Given:-}}

  • Mass=4000kg
  • Acceleration=35m/s^2

\Large{\underline{\sf To\:Find:-}}

  • Force

\Large{\underline{\sf Formula:-}}

\boxed{\sf Force=Mass\times Acceleration}

\Large{\underline{\sf Solution:-}}

\\ \sf\longmapsto Force=4000(35)

\\ \sf\longmapsto Force=14000N

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A brick with a mass of 20 kg and a weight of 196 N is released on a frictionless plane at an angle of 30°. What is the accelerat
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At which of the following angles will the sunlight received at a location on Earth spread out over the largest area?
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The angle at which the sunlight received at a location on Earth spread out over the largest area is 10°. Last option is correct.

<h3>What is sunlight?</h3>

The light coming from the Sun reaching the Earth's surface is called as Sunlight.

When the sun is overhead, the intensity is high because sun's rays are perpendicular to the earth's surface, so the energy spreads over a small area and the heat is too high in that region.

When, the angle is smaller, the sunlight will spread out over a larger area.

Thus, at 10° the sunlight received at a location on Earth spread out over the largest area. Last option is correct.

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brainly.com/question/23504828

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Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

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<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

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By using Kepler's third law, this can be written as,

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Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

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