Recent content by IBY

Sorry. This is not from a class, so it is not top priority. It is just something I found while reading. Which is why I don't have Mathematica. The book tells me the answer, and kind of gives a vague description on what the mathematician did. I just want to know how he got that way. Anyways, I...

I don't have Mathematica. But I did follow one of the suggestion. So at least I know that: \int_x^\infty \int_x^\infty \int_0^\infty e^{it^2}e^{-iu^2}+e^{-it^2}e^{iu^2}dxdtdu=\sqrt{\frac{\pi}{2}} Now, another advice was to massage the limits of integration. Not exactly sure how I go about...

Wait, wouldn't cos(t^2-u^2)=\frac{e^{i(t^2-u^2)}+e^{-i(t^2-u^2)}}{2}?

Homework Statement This is an integral I came across while reading a book. It is: \int_0^\infty \int_x^\infty \int_x^\infty cos(t^2-u^2)dt du dx I know the solution is: \frac{1}{2}\sqrt{\frac{\pi}{2}} I want to know how it was solved. The Attempt at a Solution I don't know where to start...

Firstly, what is the total force on the 8.9kg object? That is the first thing you have to ask because you want to find the tension on that object. Also, because the 8.9 kg object is moving down, that means that total force is down. So, what is the total force? The total force is its weight minus...

Seems right to me.

The kinetic energy term lies the problem. You don't want to end up with neither w nor r, so don't replace v!

Oh, because in those cases, angular frequency would turn out negative, and there is no such thing. It would turn out negative because R^2 is being added by 4L/C, and since without the 4L/C, the square root term would be R, then that square root term must be greater than the R outside the square...

Firstly, how did you set up the conservation of energy? Secondly, did you use the fourth and fifth relevant equation you listed in order to substitute?

It would be nice if you actually attempted a solution. But you know what? I will give you a clue. Start with the x component.

No, on the y component, theta is on the right up, not right down. You got the 90 degrees right. Think about it, the shape of the area of vector x and y is a rectangle. All angles of a rectangle is 90 degrees, which means that if you split the rectangle with a diagonal, you see that the two...

I see that if I subtract omega 1 with 2, that gives me the correct solution, but why do I toss out omega 3 and 4, though? Why wasn't the negative R tossed out? It seems completely arbitrary.

Just to help you visualize the four solutions: \omega_1=\frac{R+\sqrt{R^2+4 \frac{L}{C}}}{2L} \omega_2=\frac{-R+\sqrt{R^2+4 \frac{L}{C}}}{2L} \omega_3=\frac{R-\sqrt{R^2+4 \frac{L}{C}}}{2L} \omega_4=\frac{-R-\sqrt{R^2+4 \frac{L}{C}}}{2L}

Oh right, thanks. One more problem, though, that leaves me with 4 solutions. The two of them is the one mentioned in my last post above, and two of them would be with the Rs in front shifted into negative.

Sin is opp/hyp, and in this case, opposite is the axis in which the box slides. Hypotenuse is g. If you draw a line between the arrows of g*sin(theta) and g, it forms a right triangle. Theta, in that triangle, is in the bottom left corner because the upper left forms 90-theta, and the right...