Recent content by estel

This is the problem really. Originally, the regular period of pulsars led to them being called "little green men" in the belief that the radio waves were sent by sentient beings, and the original radio sources discovered still carry the LGM label.

Heya I'm in a not-too different situation to yourself. I'm about to start my final year of my Astrophysics degree at Adelaide, Australia. My first 2 years of uni were in a double degree in Maths and Applied Physics, and that gave me a good solid grounding for astro. I guess it depends if you...

Of course, at the centre of the sun, gravitation itself is very low in comparison to that at the surface. However, as has been mentioned, due to the pressure of the sun, the core has a very high particle density for such a high temperature. Pretty sure Chris nailed the rest - and can I...

OK - to get potential, in this case, since it's got spherical symmetry, you can exploit Gauss' law Take the total charge inside the sphere by integrating the charge density over the whole volume. Then, by Gauss' Law, the potential outside the sphere is the same as that for a point charge of...

put the constant on the side integrated with respect to t, so when you exponentiate, you get a constant multiplier on that side Qexp(kAt)

1- you can do it the eqbm way. You've got to balance the following forces: Tension, weight, pull. weight is down, the pull is back, so the tension must act at an angle to counter both of these. The vertical component of the tension is 160N, the horizontal component is 100N

use the universal flux rule. emf = d(phi)/dt = B da/dt = Bl dx/dt = B v

You're missing a constant of integration, and theta has disappered. Also, jumping back to question 1 - if the powe has been turned ON, the bit won't settle to the ambient temperature, it will stop at the temperature such that the electrical energy being converted to thermal energy = energy lost...

It's cylindrical, so take the current enclosed in a loop of radius "s" around the cylinder, which lies so that the area of the loop is perpendicular to the current. The magnetic field will then be parallel to the Amperian loop.