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Mashcka [7]
2 years ago
8

We see a full moon by reflected sunlight. How much earlier did the light that enters our eye leave the sun? the earth-moon and e

arth-sun distances are 3. 8 x 105 km and 1. 5 x 108 km
Physics
1 answer:
Tju [1.3M]2 years ago
6 0

The time taken by the light reflected from sun to reach on earth will be 8.4 minutes.

To find the answer, we need to know about the distance travelled by light.

<h3>How to find the time taken by the light reflected from sun to reach on earth?</h3>
  • So, in order to solve this problem, we must first know how far the moon is from Earth and how far the Sun is from the moon.
  • These distances are given as 3.8×10^5 km (Earth-Moon) and 1.5×10^8 km (Sun- Earth).
  • Since the Moon and Sun are on opposite sides of Earth during a full moon, the light's distance traveled equals,

         d=(1.5*10^8km)+2(3.8*10^5km)=1.51*10^8km=1.51*10^{11}m

  • As we know that light travels at a speed of 300,000 km per second. then, the time taken by the light reflected from sun to reach on earth will be,

                      t=\frac{1.51*10^{11}}{3*10^8}=503.33 s\\t=\frac{503.33}{60}=8.4min

Thus, the time it takes for the light from the Sun to reach Earth and be recognized as 8.4 minutes.

Learn more about distance here:

brainly.com/question/11495758

#SPJ4

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Answer:

v_{2f} = \frac{2vm_1}{m_2 + m_1}

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

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After the collision

P_f = m_1v_{1f} + m_2v_{2f}

So using the law of momentum conservation

P_i = P_f

m_1v = m_1v_{1f} + m_2v_{2f}

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Their kinetic energy is also conserved before and after collision

m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2

m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2

From here we can plug in v_{1f} = v - v_{2f}\frac{m_2}{m_1}

m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2

m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2

m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2

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v_{2f}(1 + \frac{m_2}{m_1}) = 2v

v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}

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