The turns ratio is equal to the voltage ratio. Let n1 and n2 be the primary and secondary turns. Then
5850V/120V=n1/n2
48.75=2680/n2
n2=2680/48.75
n2=55
E S *
The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).
The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years
Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.
To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.
As a result, you get a mere 268335.7 centimeters.
To put that into perspective, that's only about 1.7 miles
A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
where t is the time,
is the maximum charge on the capacitor at voltage V, and
is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.
1) Using
and
, the time constant of the circuit is:
2) To find the charge on the capacitor at time
, we must find before the maximum charge on the capacitor, which is
And then, the charge at time
is equal to
3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be
. We can see this also from the previous formule, by using
:
Answer:
The track's angular velocity is W2 = 4.15 in rpm
Explanation:
Momentum angular can be find
I = m*r^2
P = I*W
So to use the conservation
P1 + P2 = 0
I1*W1 + I2*W2 = 0
Solve to w2 to find the angular velocity
0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2
W2 = 0.435 rad/s
W2 = 4.15 rpm