Recent content by Friend of Kalina

This is why we need to go back to slide rules. Precision was always fixed by the physical dimensions of the rule. The internal precision was always maintained, but at the end, you couldn't read it any closer than the tolerance of the markings, so restraining yourself to the number of digits in...

Only thing I see is if you are trying to get the ball to go 1 m before it hits the ground, you need to consider where it hits the ground. The horizontal force required to get the ball of fixed mass up to fixed horizontal velocity is essentially constant (ignoring air resistance), but the...

Given the diagram, I saw the problem as the force for someone turning with their leg at an angle to the ground and the foot/ankle pinned. The force is applied in line with the tibia (legs are don't bend laterally), with the leg at a 60 degree angle to the ground. Force in line with the leg...

At a guess, it's from the dimensions of the force at the knee, where the angular force at the knee is broken into it's horizontal and vertical components of -200 and 346, which correspond to a 30/60/90 triangle having sides (2, 2##\sqrt (3)##, 4). It's unclear whether OP was given the angle and...

Now I do. Because I can see how a mass with a changing angular velocity and changing distance reduces to the constant angular momentum. Otherwise, I was having trouble seeing how a mass at a distance of, say, 100,000r, on a "nearly" radial approach to the axis and with "near zero" angular...

What I come up with is that the angular momentum of a mass ## m ## moving in a straight line at velocity ## \frac v 2 ## at an angle ##\Theta## between the direction of motion and an offset fixed axis, at a distance ## d ## from that axis, has (constant) angular momentum ## \omega = {mvd}...

Okay. And what is its angular momentum at distance 2r from the axle? Or its angular momentum at t = -infinity, r = -infinity when it's moving in an essentially radial direction? Your definition requires that for any fixed axis, the angular momentum of the clay is constantly varying, unless...

The clay moving vertically upwards has zero angular momentum. It is moving in a straight line. If you try to impose an arbitrary axis about which it "rotates," you need to explain away an angular momentum under acceleration in a system with no friction and no external forces. If the wheel is...

The linear velocity of the clay is opposed to the tangential velocity of the wheel at the point of impact. Consider the case where the relative mass of the clay is such that the wheel, on its fixed axis, stops... It is not a closed system. Neither linear nor angular momentum are conserved.

I thought your posts were perfectly clear, but did not resolve the OPs confusion between 1 amu of M, and 1 amu FOR EACH amu of O (or MO, for that matter). I do not know how to get "3 atoms" or "4 atoms," either. They certainly aren't contained in the 1:1 ratio between M and O in the original...

You know how many g-mols of each (1.725). And you (should) know Avagadro's number, which is the number of atoms per g-mol. With that information, it is left as an exercise for the reader... But I don't see that this gets you closer to the answer for your original question, which can be...

1. The ball on the string doesn't fall to the center because it starts with a velocity tangent to the circle. The string pulls it into the center of the circle, the velocity imparts an inertial component to keep going straight, the result is that the ball turns along the circumference of the...

You found that 100 grams of MO had 1.725 mol O. But by molecular formula, for each mol of O, you have one mol of M. Right? Formula of MO means one mol of each per molecule. Consider if you had started with 200 grams of MO instead of 100, and and found that you had 3.45 mol O instead of 1.725...

By that reasoning, molar mass of O is 27.6. Try again. Here's a hint. With MO, you have 1 mol of M for EACH MOL of O. For your initial 100g of MO, how many mols of O do you have, how many mols of M do you have, what is the mass of that number of mols, and, therefore, what is the mass of a...

Let's say you start with a mol of LN2 at one atmosphere at its boiling temperature, and you fix a cap on it (constant volume). You can look up the volume it displaces and the temperature on your own... Add heat until it's all evaporated. As it evaporates, pressure rises, and boiling...