So we have this pepper shaker made of glass with a print on it.
One fine dinner, it ran out of pepper, so I opened the lid to fill it in and noticed a peculiar thing – small particles of pepper dust were stuck to the inside glass walls but only in places, where the print wasn’t present on...
Okay, I finally got it right. The first mistake was that 4 in and 1 in are diameters, not radii...
Here's how one can look at it:
Now, obviously the "additional" oil that must be poured is the whole blue one, so both on the left and right.
V_{oil} = V_1 + V_2
We know that, because the...
Do you mean the additional volume of the 1-in layer in the cylinder? Yes. It's weight is:
W_1 = π \cdot (4 in)^2 \cdot 1 in \cdot ρ \cdot g = 6.87 N
Then we have:
\frac{V_{oil o}}{A_o} = \frac{W_1}{A_p}
And solving it I got V_{oil o} = 3.14 in^3
Homework Statement
The piston shown weighs 11 lbf. In its initial position, the piston is restrained from moving to the bottom of the cylinder by means of the metal stop. Assuming there is neither friction nor leakage between piston and cylinder, what volume of oil (S = 0.85) would have to...
Is it already known in physics what is happening in the electric field between the electrons? And I mean at the subatomic particles level.
Can someone share some materials from where I could learn to use and read Feynman diagrams in a more or less easy way?
I have a request for some explanation in Feynman diagram concerning repelling/attraction of electrical charges. I have been told that when two charges (say, e1(-) and the e2(-)) repel each other, a photon is being emitted from one of them, and the other charge is later being hit by this photon...
You mean the ball was in a free fall from this height?
The simplest equation you can get is h = \frac{gt^2}{2}. Your equation h = \frac{Ft^2}{m} is exactly this (after simplification), and the acceleration of gravity is hidden in the force:
h = \frac{Ft^2}{m} = \frac{mgt^2}{m} = gt^2,
though...
Yeah, it's not quite intuitive. Here happens to be the critical condition, any more or less load on either side would rotate it. Remember that less force is required on a longer arm and more on the shorter, to produce the same moment :)
Whether it is in equilibrium depends on the type of pivot. If it's a pin support, then there is no bending moment in the pivot but the beam could still be in equilibrium if the moment of the resultant force acting on the left side = moment of the resultant force acting on the right.
If it's a...