Recent content by gdbb

...What kind of "tough math" do you think you'll be doing as an environmental engineer? Haha. As an Enviro Engineer, you would mostly be dealing with biology and chemistry.

@connorm333: Calc BC covers Calc I in the first semester and Calc II in the second. It's pretty much the equivalent of taking two college math courses during your senior year of high school.

I think this would be easier to visualize if z were the vertical component of the problem, rather than y. So redo the equations like so: \begin{align} x &= v_x t \\ y &= v_y t \\ z &= 2.0 + v_z t - \frac {g t^2}{2} \end{align} We want the ball to fall into a hoop of diameter 0.5 m, which...

Oh, almost! Try adding units into your calculation to see if you can figure out where you went wrong.

You need to know the equation for the surface area of a sphere: SA = 4 \pi r^2

It's not 0.2 N, it's 1.2 N :)

This uses right triangle trig. If you plot the velocity vector of the ostrich on a graph, starting at the origin, can you figure out where it will end? You know the hypotenuse and you're looking for the x- and y-components.

If the car must exert no force on the road at the crest, what does that say about the force the road must exert on the car at the crest? :)

I have never heard of pre-engineering. However, extrapolating from its name, pre-engineering is most likely a two-year program/Associate's degree that teaches you the fundamentals of engineering and will prepare you for a Bachelor's in engineering. OR: it could be like pre-med or pre-law...

I had a MUCH harder time grasping concepts in my trig class than I did in my calculus class. Calculus was, by far, the easiest math class for me in high school.

Totally agree with cjl here. I actually had very little work to do most nights. Having all homework due once a week = five hours of work on Wednesday and Thursday nights. But that excludes studying (which I barely did freshman year...).

I'm talking about surface tension in fluid dynamics. I was just trying to give an extreme example of how little "kilograms per second squared" means in a Circular Motion context, as \frac{kg}{s^2} is also the unit of k, the spring constant, which is much more likely to be the case in the...

Well, force divided by radius gives units of \frac{kg}{s^2} , which, technically, is the unit for surface tension. However, I don't think you wanted to find that surface tension is equal to 0.17 \frac{kg}{s^2}. You could just note that, as the radius increased from r = a to r = b meters, the...

How can F_c be kept constant if you got a slope of 0.17 for an F_c vs. r graph?

You're welcome! Practice definitely makes perfect in physics.