Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.
The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:
<em>70 km per second per megaparsec</em>.
We'll also need to know that 1 parsec = about 3.262 light years.
So the speed of your receding galaxy is
(Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =
(150 million) x (1 / 3,262,000) x (70 km/sec) =
<em>3,219 km/sec </em>in the direction away from us (rounded)
The waste products of a nuclear fission power plant can best be described as radioactive waste.
These are the by-products from the processes carried out that produce nuclear energy. This type of waste is highly dangerous. A lot of attention has to be paid to the collection and disposal of this waste as it must not reach any near by water bodies for example. It can be deadly for life.
Speed = (distance traveled) / (time to travel the distance).
Strange as it may seem, 'velocity' is completely different.
Velocity doesn't involve the total distance traveled at all.
Instead, 'velocity' is based on 'displacement' ... the distance
between the start-point and end-point, regardless of the route
taken to get there. So the displacement in driving once around
any closed path is zero, because you end up where you started.
Velocity =
(displacement during some time)
divided by
(time for the displacement)
AND the direction from the start-point to the end-point.
For the guy who drove 15 km to his destination in 10 min, and then
back to his starting point in 5 min, (assuming he returned by way of
the same 15-km route):
Speed = (15km + 15km) / (10min + 5min) = (30/15) (km/min)
= 2 km/min.
Velocity = (end location - start position) / (15 min) = Zero .
Answer:
<em>20.08 Volts</em>
Explanation:
<u>Parallel Connection of Capacitors</u>
The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is
They are both the same after connecting them, thus
Or, equivalently
The total charge of both capacitors is
We can compute the total charge by using the initial conditions where both capacitors were disconnected:
Now we compute Q1 from the equation above
The final voltage of any of the capacitors is
