I can't imagine that this is going to do you much good, but
I'm sure going to enjoy solving it.
-------------------------
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.
-------------------------
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.)
================================================
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:
1: An electronic signal is the use of an electric current to encode information
2: Electronics: A message encoded by changing the voltage of an electric current.
3: I don't know :( Sorry
Explanation:
Answer:
It is not correct.
The correct conversion is:
1030 KJ = 246.18 KCal.
Explanation:
To know if 1030 KJ is equivalent to 245 KCal, we shall convert 1030 kJ to kcal. This can be obtained as follow:
4.184 KJ = 1 KCal
Therefore,
1030 KJ = 1030 KJ × 1 KCal / 4.184 KJ
1030 KJ = 246.18 KCal.
Thus, 1030 KJ is equivalent to 246.18 KCal.
From the above calculation, we can see that:
1030 KJ ≠ 245 KCal
Rather,
1030 KJ = 246.18 KCal.
Answer:
-36 m
Explanation:
In the diagram, the front of the black car is used as reference point which is point 0. The distance from the black car to the front of the red car is 15 m while the distance from the black car to the white van is - 21 m. The negative sign means the white van is behind the black car.
If the red car is used as the reference (point 0), the distance from the black car to the red car would be -15 m and the distance from the black car to the white van would be -21 m. Hence:
Distance from red car to white van = distance from the black car to the red car + distance from the black car to the white van = -15 + (-21)
Distance from red car to white van = -15 - 21 = -36 m
Answer:
Explanation:
Thus, we find 
(input rate of brine)
Since the solution is pumped out at a slower rate, thus it is accumulating
at the rate of 
Thus, after t minutes there will be 
gallons in tank
(concentration of salt in outflow) \cdot (output rate of brine)
Now, we substitute by these results in the 
to get
