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Ainat [17]
3 years ago
9

Calculate the angular velocity of the earth in its orbit around the sun and about its axis.

Physics
1 answer:
Tomtit [17]3 years ago
8 0

A delightful problem !

I'm pretty sure that what we need here is the speeds, not the velocities,
and that's the way I'm going to do it.

Regular speed is  (distance covered) divided by (time to cover the distance) .

Angular speed is very much the same.
It's
                 (angle turned) divided by (time to turn the angle) .

<u>Earth's orbit around the sun</u>:

..... Once per year.
..... Roughly 360° in 365 days ..... <em>almost exactly 1° per day</em>.
Let's see what it is more accurately:

   (360°) / (<span>365.25636<span> days) = 0.985609° per day.

============================================

<u>Earth's rotation on its axis</u>:

..... Once per "day".
..... Roughly 360° in 24 hours ..... <em>almost exactly 15° per hour</em>.

This one is slightly trickier to do more accurately, because a day is
not necessarily 24 hours. It depends on what you call 1 day. 

-- If you say the day is the period of time between when the sun is
highest in the sky, then that averages out to 24 hours in the course
of a year.

-- If you say that the day is the period of time it takes for a star
to reach the same point in the sky tomorrow night, then that's </span></span>

                   23 hours, 56 minutes, 4.09 seconds  .

Using this to calculate the angular speed of rotation, you get

                 (360°) / (23h 56m 4.09s) = 15.041° per hour


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Answer:

Explanation:

Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes  . We shall apply conservation of angular momentum , because no external torque is acting .

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Applying law of conservation of angular momentum

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Putting the values ,

3 x .75 = .27 x ω₂

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Which three factors are used to calculate gravitational potential energy?
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Two point charges 3q and −8q (with q &gt; 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so
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Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

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x(2√2 - √3 ) = √3 L

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Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q  8q / L²

Q = 8q  (x² / L²)

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and its distance from x on negative x side = √3 L / (2√2 - √3 )

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