What a delightful little problem !
Here's how I see it:
When 'C' is touched to 'A', charge flows to 'C' until the two of them are equally charged. So now, 'A' has half of its original charge, and 'C' has the other half.
Then, when 'C' is touched to 'B', charge flows to it until the two of <u>them</u> are equally charged. How much is that ? Well, just before they touch, 'C' has half of an original charge, and 'B' has a full one, so 1/4 of an original charge flows from 'B' to 'C', and then each of them has 3/4 of an original charge.
To review what we have now: 'A' has 1/2 of its original charge, and 'B' has 3/4 of it.
The force between any two charges is:
F = (a constant) x (one charge) x (the other one) / (the distance between them)².
For 'A' and 'B', the distance doesn't change, so we can leave that out of our formula.
The original force between them was 3 = (some constant) x (1 charge) x (1 charge).
The new force between them is F = (the same constant) x (1/2) x (3/4) .
Divide the first equation by the second one, and you have a proportion:
3 / F = 1 / ( 1/2 x 3/4 )
Cross-multiply this proportion:
3 (1/2 x 3/4) = F
F = 3/2 x 3/4 = 9/8 = <em>1.125 newton</em>.
That's my story, and I'm sticking to it.
The distance covered by the plane is 600 km
Explanation:
The motion of the plane is a uniform motion, so at constant velocity, therefore we can use the following equation
![d=vt](https://tex.z-dn.net/?f=d%3Dvt)
where
d is the distance covered
v is the velocity
t is the time interval considered
For the plane in this problem,
v = 400 km/h
t = 1.5 h
Substituting, we find the distance covered:
![d=(400)(1.5)=600 km](https://tex.z-dn.net/?f=d%3D%28400%29%281.5%29%3D600%20km)
Learn more about distance:
brainly.com/question/3969582
#LearnwithBrainly
To solve this problem it will be necessary to apply the concepts related to the electric potential in terms of the variation of the current and inductance. From this definition, we will start to find the load, which is dependent on the current as a function of time.
![\Delta V = L \frac{dI}{dt}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20L%20%5Cfrac%7BdI%7D%7Bdt%7D)
Here,
L = Inductance
Rate of change of current
If we take the equation and put the variation of the current as a function of time, in terms of the voltage in terms of the inductance we would have
![\frac{dI}{dt} = \frac{\Delta V}{L}\\\frac{dI}{dt} = \frac{5.0mV}{3.6mH}\\\frac{dI}{dt} = 1.389A/s](https://tex.z-dn.net/?f=%5Cfrac%7BdI%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5CDelta%20V%7D%7BL%7D%5C%5C%5Cfrac%7BdI%7D%7Bdt%7D%20%3D%20%5Cfrac%7B5.0mV%7D%7B3.6mH%7D%5C%5C%5Cfrac%7BdI%7D%7Bdt%7D%20%3D%201.389A%2Fs)
The current as a function of time will be then,
![I(t) = I_0 + (\frac{dI}{dt})t\\I(t) = I_0 + (\frac{\Delta V}{L})t](https://tex.z-dn.net/?f=I%28t%29%20%3D%20I_0%20%2B%20%28%5Cfrac%7BdI%7D%7Bdt%7D%29t%5C%5CI%28t%29%20%3D%20I_0%20%2B%20%28%5Cfrac%7B%5CDelta%20V%7D%7BL%7D%29t)
The charge is the integral of the current in each variation of the time, then
![Q = \int (I(t)) dt](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%20%28I%28t%29%29%20dt)
Equation the terms we will have,
![Q = \int_0^t (I_0 + (\frac{\Delta V}{L})t) dt \\Q = \bigg[I_0t + \frac{1}{2}\frac{\Delta V}{L}t^2\bigg]^{5.0s}_{0s}\\Q = (1) (5.0s)+\frac{1}{2} (1.389A/s)(5.0s)^2 \\Q = 22C](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint_0%5Et%20%28I_0%20%2B%20%28%5Cfrac%7B%5CDelta%20V%7D%7BL%7D%29t%29%20dt%20%5C%5CQ%20%3D%20%5Cbigg%5BI_0t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20V%7D%7BL%7Dt%5E2%5Cbigg%5D%5E%7B5.0s%7D_%7B0s%7D%5C%5CQ%20%3D%20%281%29%20%285.0s%29%2B%5Cfrac%7B1%7D%7B2%7D%20%281.389A%2Fs%29%285.0s%29%5E2%20%5C%5CQ%20%3D%2022C)