Recent content by jahaan

Draw a circle... Think about how you can indicate cosine on the circle. You already know the definition of cosine, so you are on the right track...

Okay, r is the distance of the point you're evaluating to the axis of revolution. So \rho is generally a function of r too. It is a definite integral (I should have written \int_V instead of \int. You integrate over the volume of the body. That's why you only evaluate in one dimension for a...

Well, you just calculate it from the definition: I = \int r^{2} \rho \mathrm{d}V The difference here is that the integral has to be evaluated in two dimensions. For a thin rod you can eliminate two coordinates (it's a one-dimensional body); this is a two-dimensional body.

I think the problem should be interpreted as: "he throws the box, so it travels away from him at 5 m/s". Then use galileo's principle of relativity (to find the speed of the box with respect to the ground), and calculate the momentum.

Okay. It's true there's still a gravitational force, but the change in momentum is equal to the force times the time interval: \frac{\mathrm{d}p}{\mathrm{d}t} = F \Delta p = \int F\;\mathrm{d}t = F \Delta t = 0 So, even when there is a force (which I have never denied), momentum is conserved...

Mind the wording. They write "at the instant..." or "just after". That means that the time interval is extremely small, infinitesimal. The time just before impact and just after impact are extremely close. In such a short instance of time gravity has done no work on the system, thus momentum is...

Okay, but you still mention copper/iron, which wouldn't react fast enough to an applied voltage to achieve these frequencies. :rolleyes: But, if you could construct some antenna which can achieve frequencies high enough, you would see light.

What you are basically constructing is an antenna. I don't believe there are any antennas capable of generating/receiving frequencies as high as visible light. Your setup would probably have VERY high impedance for frequencies that high, so it wouldn't work. The 'fastest' antenas we have are...

By the way, the Euler relation is a very quick and handy way to derive any trigonometric identities you might forget during an exam. Try it ;-)

As long as it doesn't interact (like in outer space), its energy is constant. Only when it gets absorbed it can generate heat. By the way it's not "kind of like" the shining of a distant star. The shining of a distant star are photons (just like all other light). I don't know about...

I don't know what you mean by "cold interaction". The point is, a photon is a particle and it's emitted with a certain energy (this energy corresponds to a wavelength). It will keep this energy when it doesn't interact (gets absorbed) by stuff in it's way. UV-light (photons at the UV energy...

Ok, I'm sorry:redface: I'm new to this forum (it was just my 6th post), but i'll stick to hints next time!

If you know the identity \cos x = \frac{e^{i x}+ e^{-ix}}{2}, you get: \cos -i = \frac{e^{-i^2}+ e^{+i^2}}{2}=\frac{e^{1}+ e^{-1}}{2}=\cosh 1 So it's a purely real number ;-)

Bacle is right... The point is, the x's DON'T have to be the same throughout the whole argument. The x's only have to be the same inside the 'scope' of the quantifier, and the scope is the scentence between cases right after the quantifier. So, like Bacle said, if you have a set S= {a,b} as...

The reason for the 'mountain' instead of a peak is a process called 'wondowing'. The Fourier transform of a sine wave is indeed a spike at it's base frequency. BUT that is the Fourier transform of a sine wave that goes on forever (if you compute it by hand you take the integral from minus...