/sci/ - Science & Math

>>11381809
>PV=nRT
Yuck.
[eqn] \frac{P\nu}{RT}=1+\frac{B}{\nu}\implies\nu=\frac{RT}{2P}+\sqrt{\Bigg(\frac{RT}{2P}\Bigg)^2+\frac{RTB}{P}} [/eqn] The discriminant is going to be positive so we have [math]-B<RT/4P[/math]. You should know that the second virial coefficient is actually always negative so we are just saying that a positive number is less than a big positive number. As far as the plus/minus goes, only the plus will give you the proper specific volume.
>>11384046
I will assume that the specific heat of water is a constant 1 BTU/lbm-R. You want to heat water from 40 F (500 Rankine) to 90 F (550 Rankine). At 1 atm, water boils at 671 Rankine. By basic energy balance,
[math] c_pm_c(550-500)=-c_pm_h(550-671) [/math]. So [math] m_h/m_c\approx0.4. [/math] Therefore, for every gallon of cold water in your tub, you need about 0.4 gallons of boiling water to bring it from 40 to 90 F (and it needs to be well mixed).
>>11384051
>"khan academy young's double slit"
>first result
>section: kindergarten thru grade 12 physics
>video
*<3~Use Your Brain <3~*
>>11383916
>>11383919
embarrassing desu

>>11381809
>PV=nRT
Yuck.
[eqn] \frac{P\nu}{RT}=1+\frac{B}{\nu}\implies\nu=\frac{RT}{2P}+\sqrt{\Bigg(\frac{RT}{2P}\Bigg)^2+\frac{RTB}{P}} [/eqn] The discriminant is going to be positive so we have [math]-B<RT/4P[/math]. You should know that the second virial coefficient is actually always negative so we are just saying that a positive number is less than a big positive number. As far as the plus/minus goes, only the plus will give you the proper specific volume.
>>11384046
I will assume the specific heat of water is a constant 1 BTU/lbm-R. You want to heat water from 40 F (500 Rankine) to 90 F (550 Rankine). At 1 atm, water boils at 671 Rankine. By basic energy balance,
[math] c_pm_c(550-500)=-c_pm_h(550-671) [/math]. So [math] m_h/m_c\approx0.4. [/math] Therefore, for every gallon of cold water in your tub, you need about 0.4 gallons of boiling water to bring it from 40 to 90 F (and it needs to be well mixed).
>>11383916
>>11383919
embarrassing desu

>>11381313
>is there a simple formula for such a relation?
Amazingly, there is.

In general, the relationship between the deflection [math] y(x) [/math] in an elastic, prismatic beam with length [math] L [/math] under a distributed load [math] \omega(x) [/math] is governed by the following fourth order ODE.
[eqn] EI\frac{\text{d}^4y}{\text{d}x^4}=-\omega(x) [/eqn]
The term [math] EI [/math] is something called flexural rigidity and is the product of the modulus of elasticity (material property) and the second moment of area of the beam's cross section (geometric property), but nevermind that. Since your load is "evenly spread," it is constant. So [math] \omega(x)=\omega_0 [/math]. Solving the differential equation above is now easy. However, as you know, you get constant of integration whenever a separable diff'eq is solved. So we need four boundary conditions to eliminate those. Because you beam is simply supported on both ends, boundary conditions are [math] y(0)=y(L)=0 [/math] and [math] y''(0)=y''(L)=0 [/math] (this is because there is no bending moment at a simple support). Getting in bed with all that algebra gives us
[eqn] y(x)=\frac{-\omega_0}{24EI}(x^4-2Lx^3+L^3x) [/eqn]
The maximum deflection obvious occurs at the midpoint. So
[eqn] y_{min}=y(L/2)=\frac{-5\omega_0L^4}{384EI} [/eqn]
and presto you have an explicit formula for the maximum deflection. Notice that deflection is proportional to load and to the fourth power of its length.

Therefore, if you cut the length in half you can increase load by sixteen times for the same deflection. (Hint: all of these formulas can be found in the appendix of Mechanic of Materials by Beers.)

We are given [math] \mu,\ R,\ \omega_i,\ \omega_f=\omega_i/100,\ \rho [/math]. The height is not given, but will be represented by [math]
H [/math]. First of all, we need to determine the torque on the cylinder to get the equation of motion. Shear stress is given by
[eqn] \tau=\mu\frac{\Delta v}{\Delta r}=\mu\frac{R\omega}{R}=\mu\omega [/eqn]
Torque is (stress*area*moment arm). So torque is given by
[eqn] T=\tau A R=2\pi R^2H\mu\omega [/eqn]
But, the torque on a body is the rate of change of angular momentum. And angular momentum is given by
[eqn] L=I\omega=\frac{1}{2}mR^2\omega=\frac{1}{2}\rho\pi HR^4\omega\ \implies\ T=\dot{L}=\frac{1}{2}\rho\pi HR^4\dot{\omega} [/eqn]
So,
[eqn] \dot{\omega}=-\frac{4\mu}{\rho R^2}\omega [/eqn]
where the negative sign appear because torque opposes motion. Solve the ODE and set proper boundary conditions to get
[eqn] t=\frac{\rho R^2}{4\mu}\ln 100 [/eqn]
where ln is the natural logarithm. Now you can plug in values. I probably made a mistake. Kiss my ass.

>>11193036
Hint: for small displacements [math]\frac{\Delta L}{\Delta x}=2[/math] where L is the natural length of the springs.
From the free body diagram and applying Newton's second law,
[eqn]m\ddot{x}+2cx=0[/eqn]
therefore we have that x(t) is the sum of a sine and a cosine with
[eqn] \omega=\frac{\sqrt{8mc}}{2m} [/eqn]

Apply yourself!

>>11193049
?

>>11193036
Hint: for small displacements [math]\frac{\Delta L}{\Delta x}=2[/math] where L is the natural length of the springs.
From the free body diagram and applying Newton's second law,
[eqn]m\ddot{x}+2cx=0[/eqn]
therefore we have that x(t) is the sum of a sine and a cosine with
[eqn]\omega=\frac{\sqrt{8mc}}{2m}[/eqn]

Apply yourself!

don't really understand your drawing but probably superposition. Are those Vs in the circles sources or voltmeters? If sources, you got a short circuit/infinite current in the loop by V3. If meters, V3 is obviously zero.

>>11189232
don't really your drawing but probably superposition. Are those Vs in the circles sources or voltmeters? If sources, you got a short circuit/infinite current in the loop by V3. If meters, V3 is obviously zero.

>She is HOT.
>She is COOL.
>We communicate the most using SCIENCE.

>giving commies (You)s

But [math]\frac{dy}{dx}[/math] IS a (limit of a) ratio.

[eqn]\frac{dy}{dx}\equiv\lim_{\Delta x\to0}\frac{y(x+\Delta x)-y(x)}{\Delta x}[/eqn]

The limit of a ratio is equal to the ratio of the limits of the numerator to denominator (provided the denominator does not go to zero, but it doesn't even matter). You can treat dy/dx as a fraction without any consequence so long as you are not dealing with partials or a higher order. Math nerds get fucked.

>>11052829
based realist

But [math]\frac{\text{d}y}{\text{d}x}[/math] IS a (limit of a) ratio.
[eqn]\frac{\text{d}y}{\text{d}x}\equiv\lim_{\Delta x\to0}\frac{y(x+\Delta x)-y(x)}{\Delta x}[/eqn]
The limit of a ratio is equal to the ratio of the limits of the numerator to denominator (provided the denominator does not go to zero, but it doesn't even matter).
You can treat dy/dx as a fraction without any consequence so long as you are not dealing with partials or a higher order. Math nerds get fucked.
>>11052829
based realist

>>11049647
Uhh, we already have an incel general up.