@JohnRennie p/q=n means p=qn. That is right. But p=qn mean p/q=n only when we rule out q=0. How? Can you elaborate? I can say itβs true for p, q other than 0.
It's possible the oscilloscope is adjusting its display so that the waves start at zero at the left side of the screen. That would result in the two waves always being displayed as if they were in phase even when they aren't.
However you can calculate the phase angle from the two voltages so you can still measure it, though it's an indirect measurement.
Suppose the resistor and capacitor voltages are Vr and Vc, then the phase angle is given by: tan(ΞΈ) = Vc/β(Vr² + Vc²)
This might seem a little mysterious, but we can explain how this formula works with a diagram if you're interested.
@Ajay on page 64 there is a section explaining how to control the trigger
If you set it to trigger off channel 1 it should adjust channel 1 so it starts at zero on the left side of the screen, then you should be able to see the phase offset of channel 2.
When the scope displays the wave it has to decide when to start the trace. Suppose you have a voltage V = Vβ sin(Οt) if we were graphing this we would normally start our graph at t = 0 so the voltage V = 0 at the origin.
We talked about tone controls yesterday, and you could change your experiment very slightly to measure the behaviour of the circuit as a filter instead of measuring the phase angle.
But this reaction hardly goes at all. If you started with high initial concentrations of NHββΊ and Clβ» you'd only get tiny amounts of NHβ and HCl. Yes?
@KavinIshwaran That's because NHββΊ is a very weak acid and Clβ» is very weak base i.e. NHββΊ does not donate protons very strongly and Clβ» does not accept protons very strongly.
I have no idea how to do that. From the symmetry the current in the vertical wire will be zero so you can remove it. Also the circuit is symmetrical about its mid point so that relates the currents in the left and right halves.
The triangle is equilateral and the radius of circle is R.
> An infinite long hollow metallic cylinder of radius R and surface charge density Sigma is placed symmetrically with an imaginary surface of the shape of a prism. The length of prism is R and its three sides are all equal to 3R. The flux through the prism is
@Wolgwang For an equilateral triangle of side π the height of the centroid is aβ3/6 so the height of the centroid here is Rβ³ββ = R cos30
That means the angles marked have to be 30° and therefore that the angle subtended by the part of the circle outside the triangle is 60°.
And that means the total angle outside the triangle is 180° i.e. half the circle is outside and half is inside. So half of the charge on the cylinder is inside the triangular prism.
The last bright ring would have a path difference of zero, but that can only happen if the light is emitted parallel to the screen, so the 10th bright ring appears at infinity on the screen. OK so far?
You have done what we call a proof by contradiction. To do this we start by assuming 0/0 is determinate. If this is true then 0/0 must have a single value. But you have shown that 0/0 does not have a single value because you have shown it could be both 5 and 10.
And that means our initial assuming is wrong i.e. 0/0 is not determinate.
If we are looking at a point charge we use a sphere with the point its centre, or if we are looking at a line charge we use a cylinder with the axis along the line.
Problem Solving Strategies
Problem Solving Strategies