@tatan I would start with the mass/rod stationary at the top, then calculate the change in potential energy when it has rotated all the way to the bottom. This change will then be equal to the rotational kinetic energy $\tfrac{1}{2}I\omega^2$ so you can calculate $\omega$.
@Scáthach in a transformer you want a core with a very high $\mu$ because that gives large magnetic fields. A solid iron core is good, but the changing magnetic fields will generate eddy currents in the core and those eddy currents dissipate energy as heat. OK so far?
Wait, what do they mean by "full cycle"? Does that mean that the rod becomes vertical again but this time above the suspension point? (And so that v is minimum, the velocity In that state is null?)
@Scáthach it shouldn't be that long. The resistance of the galvanometer and 5 ohm resistor is 4.75 ohms. So the total resistance is 20K + 4.75 which is approximately 20K ohms. Yes?
To work out the current we ignored the 4.75 ohms because that was a good approximation.
To work out the voltage across the galvanometer/shunt it is not a good approximation to ignore the 4.75 ohms. The voltage will be 0.1mA times 4.75 ohms = 0.475 mV.
So the current through the galvanometer will be 0.475mV/95 ohms = 0.005 mA
@JohnRennie Can you explain Newton's Law of Cooling please? I searched it on the internet but I am not able to properly understand it. In several places I have seen approzimations being made in problems related to this law but I can't make out where to apply approximation and where not to...
@tatan Newton's law of cooling isn't a fundamental law. In fact it's only an approximation. Cooling is a hugely complicated business because it depends on the convection currents in the air flowing around the body. When we do the experiment we find that Newton's law of cooling approximately describes the rate of heat loss.
A body cools from 67◦C to 37◦C. If this takes time t when the surrounding temperature is 27◦C, what will be the time taken if the surrounding temperature is 7◦C?
Yes... wait a sec.. I am posting
A body cools from 97◦C to 87◦C. If this takes time t when the surrounding temperature is 27◦C, what will be the time taken if the surrounding temperature is 7◦C?
@tatan Yes, you use the info you're given to calculate the constant $A$, then you assume $A$ doesn't change when you change the environment temperature.
A body cools from 67◦C to 37◦C. If this takes time t when the surrounding temperature is 27◦C, what will be the time taken if the surrounding temperature is 7◦C?
@JohnRennie from the second equation we get 30=60e^-{t'/A} or t'=A ln2 . From the first equation we got t=A ln(4)=2Aln(2)=2t' or t'=t/2 right? Please answer afer you have helped @Scáthach
@JohnRennie from the second equation we get 30=60e^-{t'/A} or t'=A ln2 . From the first equation we got t=A ln(4)=2Aln(2)=2t' or t'=t/2 right? Please answer afer you have helped @Scáthach
@Scáthach You are told that the wave in the column is the fundamental, and you know the relationship between the wavelength and the column length for the fundamental. Yes?
What you're doing is introducing an extra path length difference of $d(n-1)$ where $d$ is the thickness of the paper and $n$ is the refractive index of the paper.
You've worked out that the central fringe has been moved to the angle given by $d\sin\theta = t(n-1)$. The question is how many fringes is that away from the centre. Yes?
And to find that just substitute in your original equation $n\lambda = d\sin\theta$. We've worked out that $d\sin\theta = t(n-1)$ so we simply get $n\lambda = t(n-1)$.
when you calculate the gravitational potential energy of an object, then can you just apply the formula of PE = Mgh with the height taken as the height of the center of mass?
@Scáthach , if you have not figured out that yet, The phase differece between Source $A$ and Source $B$, and the phase difference between Source $A$ and Source $C$ would be$$ \delta = \frac{2\pi}{3}$$ And assuming the amplitude to be $a$, when they are combined by phasor method, the sum of horizontal component turns out to be zero. But for Vertical part, it is $$ a_r = 2(a \sin(\frac{2\pi}{3}) $$
which would return $$ a_r = a\sqrt3 $$
@MartianCactus Assuming( If the assumption is practical), that the change is gravitation potential is insignificant over the length of the body. Like, For a body like Pencil, when calculating its GPE with Earth, it can be taken as Mgh
@Scáthach For phase difference between $A$ and $C$,The path difference $$ x = \sqrt{(2d)^2+D^2} - D $$, $$= D\sqrt{1 + {\frac{2d}{D}}^2} - D $$ Then for small values of $d$ , $ {D\sqrt{1+ {\frac{2d}{D}}^2}} $ is approxiamately equals to $D(1 + \frac{(2d)^2}{2(D)^2}) $, which would make $$ x = \frac{2d^2}{D}$$ and from the part 1, $ \frac{d^2}{D} = \frac{2\lambda}{3} $, whic would make $x = \frac{4\lambda}{3} $ or $ \delta = \frac{2\pi}{3} $
@AjayMishra what about if 2 completely circular planets are separated by a distance? Cant you apply the formula for gravitational force by taking the distance between them as the distance between there center of masses? In that case, h is not smaller than r
Now, for a ring at a point on the axis you are the same distance from every part of the ring. So if $r$ is the distance to the ring then the potential is just $kQ/r$, where $Q$ is the charge on the ring. I can draw a diagram if that isn't clear.
But with the part cut out each one of those rings is missing one sixth of its length and therefore one sixth of its charge. So the potential for the remaining part of the ring reduces to $V = \frac{5}{6}\frac{kQ}{r}$.
When you set up your integral for the cone it will look just like it did before but with a factor of 5/6. That factor can be taken out of the integral, so after you've finished tearing your hair out doing the integral you'll get the same result as before multiplied by 5/6.
@AdvilSell I think they are just unrolling the cut cone and flattening it. You get a sector of a circle with radius equal to the cone edge length $r$. The perimeter of the sector (the curved bit) is the 5/6 of 2\pi R you're left with after the wedge is cut out.
I guess you can integrate that, but I'm not sure that's any easier that just multiplying the potential for the whole cone by 5/6.
I can confirm your result. I can also suggest you a neater way to derive it inspired by David Morin - Introduction to Classical Mechanics, check it out in a library if you have access.
The main idea is to use the symmetry of the equilateral triangle and split it into 4 smaller equilateral trian...
I can confirm your result. I can also suggest you a neater way to derive it inspired by David Morin - Introduction to Classical Mechanics, check it out in a library if you have access.
The main idea is to use the symmetry of the equilateral triangle and split it into 4 smaller equilateral trian...
