Recent content by Johnny Blade

Hello, I want to relearn and continue learning physics and math. I went to university studying physics for a few years but I never graduated. It's been three years and now I want to continue learning physics as a hobby. Right now I'm reading the Feynman lecture on physics but after that I'd...

Yes that would give you the velocity.

You're searching the time t when the velocity is 0. x(t)= bt^2 -ct^3, this finds the displacement as a function of time. You'll need another equation (related to this one) to find the velocity as a function of time.

Homework Statement What is the area of the portion of the sphere x^{2}+y^{2}+(z-a)^{2}=a^{2} that is inside the paraboloid z=x^{2}+y^{2} Homework Equations \int\int_{S} dS The Attempt at a Solution I used this \int\int_{S} dS=\int\int_{R}\sqrt{f^{2}_{x}+f^{2}_{y}+1}dx dy...

Bump.

Homework Statement Show that the wave equation becomes \left(1-\frac{V^{2}}{c^{2}}\right)\frac{\partial^{2}\psi'}{\partial x'^{2}}-\frac{1}{c^{2}}\frac{\partial^{2}\psi'}{\partial t'^{2}}+\frac{2V}{c^{2}}\frac{\partial^{2}\psi'}{\partial t' \partial x'} = 0 under a Galilean transform if the...

Homework Statement There's an object with mass m in movement in the horizontal axes. There's a force \textbf{P} of constant power acting on the object. Another force is the air drag which has the magnitude of \beta m v^{2}. I need to find the position x as a function of the speed v...

I'd like to know what are the general differences between a computer science program in university and programing course in a community college and the pros and cons for both.

It seems right, I would've done it a slightly different way. With 64 < c < 81 I changed it to sqrt(64) < sqrt(c) < sqrt(81), 1/8 > sqrt(c) > 1/9, then with the MVT you have sqrt(66)-8 = 1/sqrt(c), then you substitute 1/8 > sqrt(66) - 8 > 1/9. But that's just me.

That's another properties of the inequalities, if 0 < x < y, then 0 < 1/y < 1/x.

How did you get this 1/2sqrtx ≤ (sqrt(66)-8)/2 ≤ 1/2sqrtx? I would use the MVT to have sqrt(66)-8 = 1/sqrt(c), then transform the inequality 66 < c < 81 to put the sqrt(66)-8 in it.

Not really plugging, because you work the MVT with the inequality to prove the initial statement.

c in the interval ]64, 66[, so we can write 64 < c < 66, but we want 81 in there, 64 < c < 66 < 81. How can you use the property to only have 64, c and 81 in the inequality?

Do you remember the properties of the inequalities? The one you need to remember here is if x < y and y < z, therefore x < z.

You must remember the interval where c is defined. c is in ]a,b[, where a = 64, and b = 66, now we can write 64 < c < 66. What can you do with the c in this inequality and the one in the MVT?