/sci/ - Science & Math

>>12739592
>im the only one that posts satoko, and i think ive done so maaaaybe 5-6 times, and the most recent was like a month ago
>im not even good at math
Irrelevant.

Each beaker contains volume [math] V [/math] of water at density [math] \rho [/math] and a ball with volume [math] V_B [/math] at a density [math] \rho_2 [/math] or [math] \rho_1 [/math]. We have [math] \rho_2>\rho>\rho_1 [/math] (where 2 represents the steel and 1 the ping pong ball). There are three forces on the steel ball: tension (upward), buoyancy (upward), and weight (downward). The ball is in equilibrium so the sum of these forces is zero. We get tension [math] T=(\rho_2-\rho)gV_B [/math]. Now consider the entire container on the right. The total force needed to support it is its total weight, minus the tension in the string. So [eqn] F_2=\rho gV+\rho_2 gV_B-T=\rho g V+\rho gV_B [/eqn]
The force required to support the left hand container is simply its total weight. [eqn] F_1=\rho gV+\rho_1gV_B [/eqn]
But we already established that [math] \rho_1<\rho [/math]. Therefore [math]
F_1<F_2 [/math] and the container tips right.

inb4 some retard gets confused about internal forces
inb4 some retard claims I am wrong (you can literally find a video of this exact setup proving I'm right)
inb4 we have to have this thread again

>>11435267
they are wrong
>>11435184
>>11435225
>>11435236
you are also wrong
>>11435188
you have the right conclusion but with dogshit reasoning, you might as well be wrong

Each beaker contains volume [math] V [/math] of water at density [math] \rho [/math] and a ball with volume [math] V_B [/math] at a density [math] \rho_2 [/math] or [math] \rho_1 [/math]. We have [math] \rho_2>\rho>\rho1 [/math] (where 2 represents the steel and 1 the ping pong ball). There are three forces on the steel ball: tension (upward), buoyancy (upward), and weight (downward). The ball is in equilibrium so the sum of these forces is zero. We get tension [math] T=(\rho_2-\rho)gV_B [/math]. Now consider the entire container on the right. The total force needed to support it is its total weight, minus the tension in the string. So [eqn] F_2=\rho gV+\rho_2 gV_B-T=\rho g V+\rho gV_B [/eqn]
The force required to support the left hand container is simply its total weight. [eqn] F_1=\rho gV+\rho_1gV_B [/eqn]
But we already established that [math] \rho_1<\rho [/math]. Therefore [math]
F_1<F_2 [/math] and the container tips right.

inb4 some retard gets confused about internal forces
inb4 some retard claims I am wrong (you can literally find a video of this exact setup proving I'm right)
inb4 we have to have this thread again

We already have 13 other threads dedicated to this. Use one of those, retard.

>>11343196
Exact DEs can be homogeneous and vice versa, they aren't mutually exclusive. I typed up another post a while back that explains why we bother studying exact DEs. >>/sci/thread/S11250238#p11260811
>>11343267
>using calculus to prove elementary geometry
wat
just use law of sines https://en.wikipedia.org/wiki/Angle_bisector_theorem#Proof_1
>>11343366
>the voltages are slightly different
The theoretical voltage predicted by Kirchhoff, Thevin, superposition, etc. should all be exactly the same unless you are doing something wrong. If the experimental value is differing from that unique theoretical value, it's likely related to what this anon >>11343379 said. Also, remember that the resistors have pretty wide tolerances on them, usually like 5-10%, sometimes even up to 15% error from the color code value printed on their side. Lastly, it's a pretty tiny effect for what you are doing, but the resistance of a single resistor will definitely change with temperature and mechanical stress and all that.

>>11231626
>Is separation of variables the right approach?
Yes. You mistake was in the 3rd line, not including the "constant" of integration, which turns out to be a function of a single variable y.
[eqn] \frac{\partial}{\partial x}u(x,y)-2u(x,y)=u_x-2u=0 [/eqn]
Separate the variables and integrate and you get
[eqn] \ln |u|+C(y)=2x \implies \exp\big(\ln |u|+C(y)\big)=e^{2x} [/eqn]
so
[eqn] e^{C(y)}e^{\ln|u|}=C'(y)u=e^{2x}\implies u(x,y)=f(y)e^{2x} [/eqn]

What have you tried?

>>11164438
[eqn]\langle f(t),g(t)\rangle=\int_1^2f(t)\cdot g(t)\text{ d}t[/eqn]
where 1 and 2 are the bounds of the interval

Anybody who says it doesn't take off is most likely trolling.
>>11141617
>The conveyor is set to match the speed of the wheels whatever they may be
That doesn't make any sense. If you like, we can eliminate the wheels altogether and just picture the plane as a frictionless block of ice on the conveyor. No matter how fast the conveyor speeds--it can be wound up to 8 times the take-off speed if you like--there is STILL a net force propelling the plane to the left of the pic. Since the plane has motion, there must be air over the wings and therefore lift.
Draw a FUCKING free body diagram.

>why am I falling for b8

Anybody who says it doesn't take off is most likely trolling.
>>11141617
>The conveyor is set to match the speed of the wheels whatever they may be
That doesn't make any sense. If you like, we can eliminate the wheels altogether and just picture the plane as a frictionless block of ice on the conveyor. No matter how fast the conveyor speeds--it can be wound up to 8 times the take-off speed if you like--there is STILL a net force propelling the plane to the left of the pic. Since the plane had motion, there must be air of the wings and therefore lift.
Draw a FUCKING free body diagram.

>why am I falling for b8