Based on its location on the periodic table, which element is the best
2 answers:
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Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
Answer:
<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>
Explanation:
<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.
Here it is given that the object oscillates 20 times in 10 seconds.
So f = = 2Hz
The <em>time period</em> is defined as time taken by the object to complete one full oscillation.
T =
T= =0.5 s
<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>