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Contact [7]
4 years ago
7

calculate the time (in years) it would take to send a message to the Galactic Capitol located at the center of our Milky Way gal

axy and then receive the reply. We live about 2.5x10^20 meters away from the Galaxy center. Assume the message leaves the Earth and travels at the speed of light and that they take 1 year to decide what to answer..
Physics
1 answer:
Alinara [238K]4 years ago
7 0

Answer:

The total time is \bf{52850~Yr}.

Explanation:

Given:

The distance between Galactic Capitol and Earth is, d = 2.5 \times 10^{20}~m

The speed of light is, c = 3 \times 10^{8}~m/s

The time taken by the message to reach and come back to Earth is given by

t &=& \dfrac{2d}{c}\\~~&=& \dfrac{2(2.5 \times 10^{20}~m)}{3 \times 10^{8}~m/s}\\~~&=& 1.6 \times 10^{12}~s\\~~&=& \dfrac{1.6 \times 10^{12}~s}{(86400 \times 365)~s}~Yr\\~~&\approx& 52849~Yr

In addition they take 1 year to decide to reply.

Thus, the total time is 25850~Yr.

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3 years ago
How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example
MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the (velocity)^2 and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

\text { Kinetic Energy }=\frac{1}{2} m v^{2}

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}

\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

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