Acceleration = change in velocity/time
By F = ma,
6 = 33 x change in velocity / 9
change in velocity = +1.636 m/s
Electric motors require electricity to move the motor parts and do work (like an electric fan).
Elecgric generators actually burn diesel fuel to spin a motor around and around and around to GENERATE, or MAKE, electricity that you can then use to power your fans and lights in your house.
I can't imagine that this is going to do you much good, but
I'm sure going to enjoy solving it.
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Skip this whole first section.
It was an attempt to master a bunch of trees, while
the forest was right there in front of me all the time.
Drop down below the double line.
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Kepler's 3rd law says:
(square of the orbital period) / (cube of the orbital radius) = constant
T₀² = K R₀³
I put the zero subscripts in there, because you doubled 'R'
and I need to know how that affected 'T'.
new-T² = K(2 R₀)³
new-T² = 8 K (R₀)³ = 8 old-T₀²
<u> new-T = √8 old-T</u> <=== that's what I was after
I just teased out the Moon's new orbital period if it's distance were doubled.
Instead of 1 month, it's now √8 months.
To put a somewhat sharper point on it, the moon's period of revolution
changes from 27.322 days to 27.322√8 = 77.278 days (rounded) .
Using 385,000 km for the moon's current average distance, the current orbital speed is
(2π x 385,000 km) / (27.322 days) = 1,024.7 m/s
(One online source says 1.023 km, so we're not doing too badly so far.)
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I'm such a dummy. I don't need to go through all of that.
If the moon were twice as far from Earth as it really is, then it would
average 770,000 km instead of the present 385,000 km.
That's 120.86 times the Earth's radius of 6,371 km.
So the acceleration of gravity out there would be
(1 / 120.86)² of the (9.807 m/s²) that it is here on the surface.
new-G = 0.000671 m/s²
Distance a dropped object falls = 1/2 g t²
In the first second, that's 1/2 g (1)² = 1/2 g
For an orbiting object, every second is the "first"second, because ...
as we often explain orbital motion qualitatively ... the Earth "falls away"
just as fast as the curved orbit falls.
Distance an object falls in the 1st second =
1/2 G = 0.000336 m/s = <em>0.336 millimeter per second</em>
I estimate the probability of a mistake somewhere during this process
at approx 99.99% . But I don't have anything better right now, and I've
wasted too much time on it already, so I'll stick with it.
Answer:
Step-by-step explanation:
<u>1. Find acceleration:</u>
<u>2. Find distance traveled:</u>
3. Find work done by friction:
(Work formula when angle between Force and Displacement vectors are 0°) 
Answer: Within any frame of reference that is accelerating
Special relativity was proposed on 1905 by Einstein, who developed his theory based on the following two postulates:
1. <em>The laws of physics are the same in all inertial systems. There is no preferential system.
</em>
2. <em>The speed of light in vacuum has the same value for all inertial systems.
</em>
Focusing on the first postulate, it can be affirmed that <u>any measurement on a body is made with reference to the system in which it is being measured</u>.
Now, taking into account that an inertial reference system is the one that complies with the principle of inertia:
<em>"For a body to have acceleration, an external force must act on it."</em>
The correct answer is
Within any frame of reference that is accelerating
(Newton's second law)
(Kinematic equation)