Recent content by merricksdad

It is in km/m because it is defined that way. There are three types of density. Most people know about volumetric mass density, because that is what we use in basic chemistry. But there is also surface mass density σ, which is for use with sheets of known thickness with a density already...

OK, so if anybody is still interested in this, I'm told that the key point of data in the question is to notice a mistake. That mistake is that density ρ is actually linear, not volumetric. Note that density is in kg/m not kg/m3. So part of this experiment was to test if we remembered chapter 1...

This teacher has a problem with the difference between 2D and 3D. He never defines which axis we're talking about. He seems to think of everything in 2D when he teaches the class, but then says words like "cone". Obviously, if you don't define which is x, y, and z, all you have is guessing left...

Nevermind. Yes, the teacher just confirmed that there is no z axis, and that y is the "cone" axis, and it isn't a cone, it's just a triangle. Makes sense now.

Does their formula simply calculate only the z portion, meaning that no matter what you plug in for ρ, the z-axis offset is always going to be 3/4h? And if so, don't I just calculate the x,y parts without considering z at all? It seems like the axis of the center of mass is going to be curve...

Balancing two integrals is far above my pay grade, so I'm not sure why we're even given this question. We've never seen anything similar in class. I've been examining http://physics.stackexchange.com/questions/38624/determining-the-center-of-mass-of-a-cone, which comes up with two balanced...

Homework Statement Find the center of mass of an inverted cone of height 1.5 m, if the cone's density at the point (x, y) is ρ(y)=y2 kg/m. Homework Equations The formula given for this problem is rcm=1/M * ∫rdm, where M is total mass, r is position, and m is mass. The Attempt at a Solution...

This is what we were shown a week later in the class which goes along with the lab. The class professor said he's not in control of the lab layout or requirements. Now I understand, and see why nobody got to that part of the lab. Thank you guys for helping out. It sounds like the professor was...

If I make these assumptions, yes I can make general vector component assumptions if I define one of the poles' strings as equaling one axis on the x-y plane, correct? But I'm making assumptions, right? And then would I not even need the Pythagorean theorem for anything except final calculation?

No trig functions on the calculator. No component calculation using same trig. Yup, just the Pythagorean theorem.

The poles are about the same heights. The position of the spring scale on the pole is about the same, but they are not at the top of the pole. We could have measured anything we wanted to with the ruler.

Sorry, I don't have a picture. It appears to be a simple physics project, with three poles, to which are attached a spring scale (reading Newtons), and from each scale is a cord which is fastened to the same weight ball, in the center, suspended above the table by the cords. The weight is not...

My name is Charlie. I'm studying geophysics at WMU in Kalamazoo Michigan. I'm struggling with understanding some of the physics lab content, primarily because some of it is poorly written, but also because my schedule doesn't match up with tutoring hours with department TA's and walk-in tutoring...

Homework Statement Given three spring scale readings, positioned at unknown angles, find the mass of the weight hanging from all three scales without using trig, and without measuring the angles. You have only a yard stick. This is a static equilibrium problem. Homework Equations Not allowed...