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Damm [24]
3 years ago
14

A laser pulse takes 2.56 seconds to travel from Earth to the Moon and return. Use this to calculate how far away the Moon is. Ho

w might this time delay affect conversations between an astronaut on the Moon and someone back on Earth
Physics
1 answer:
Deffense [45]3 years ago
7 0

Answer:

x = 3.84\times 10^{5}\,km

This delay affect conversations in the sense that recipient will be receive the message just 2.56 seconds after the message was sent.

Explanation:

The laser pulse has an ondulatory nature as electromagnetic wave, which can travel in the void. Speed of light is constant and distance between Earth and the Moon is:

x = 0.5\cdot c\cdot \Delta t

x = 0.5\cdot (3\times 10^{5}\,\frac{km}{s})\cdot (2.56\,s)

x = 3.84\times 10^{5}\,km

This delay affect conversations in the sense that recipient will be receive the message just 2.56 seconds after the message was sent.

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Which of the following statements is FALSE regarding the law of conservation of energy?
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svlad2 [7]

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Explanation:

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A student makes a model of the sun-Earth system by swinging a ball around her head. Using this model, the student is trying to e
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Read 2 more answers
Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. What is the veloc
Nuetrik [128]

The velocity of B after elastic collision is 3.45m/s

This type of collision is an elastic collision and we can use a formula to solve this problem.

<h3>Elastic Collision</h3>

v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1 - m_2}{m_1 + m_2}u_2

The data given are;

  • m1 = 281kg
  • u1 = 2.82m/s
  • m2 = 209kg
  • u2 = -1.72m/s
  • v1 = ?

Let's substitute the values into the equation.

v_1 = \frac{2*281*2.82}{281+209} -\frac{281-209}{281+209}(-1.72)\\v_1 = 3.45m/s

From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.

Learn more about elastic collision here;

brainly.com/question/7694106

4 0
2 years ago
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