/sci/ - Science & Math

>3 amps
RIP

>>11621194
Ah, okay. Sketch looks good.
>a
[math] u_\text{mid}=u(y=c/2)=(4U/2)(1-1/2)=U [/math]
>b
[math] \tau=\mu\cdot\partial u/\partial y=\mu\cdot(4U/c-8Uy/c^2), [/math] evaluate at 0, 15 and 30 mm.
>>11621277
Op-amps and MOSFETS aren't passive devices. Op-amps have an external power source usually not depicted in circuit diagrams. MOSFET have large, not infinite impedance (unlike op-amps).

>>11551076
You know the direction and magnitude of tension in AB and AD. There is only one other force acting on A. Because it is in equilibrium, you know the forces sum to zero. Graphically, this means you can arrange all the force vectors tail-to-tip to form a closed triangle. Use geometry and the law of cosines to find the angle and length of the third side. Parallelogram law is the same thing, if you look closely.
>>11551008
https://en.wikipedia.org/wiki/Canine_penis
>>11550265
Say you have a number less than 1 written in standard decimal form [math] q=0.a_1a_2a_3... [/math] where a1, a2, etc are integers [math]\in[0,9][/math]. We can see that q has non-terminating decimals. By definition this means [math]q=(a_1/10)+(a_2/100)+...=\sum_{n=1}^\infty a_n\cdot10^{-n} [/math]. If [math] a_1=a_2=...=a_n [/math] for all n then its easy to see this is a geometric series and using the classic formula for a geometric series you can get [math] q [/math] in the form of a fraction. It's easy to see how this works if you have a set of r numbers that repeat in the sum like [math] a_1=a_r, a_2=a_{r+1},... [/math] for fixed r.
>>11549761
I meant use superposition just to get current, but it is overkill. You can just sum voltage drops over the closed loop with all sources in place to get current. Then you have the complex power over each element. No need to combine powers after you compute them.

>>11551076
You know the direction and magnitude of tension in AB and AD. There is only one other force acting on A. Because it is in equilibrium, you know you the forces sum to zero. Graphically, however, this means you can arrange all the force vectors tail-to-tip to form a closed triangle. Use geometry and the law of cosines to find the angle and length of the third side. Parallelogram law is the same thing, if you look closely.
>>11551008
https://en.wikipedia.org/wiki/Canine_penis
>>11550265
Say you have a number less than 1 written in standard decimal form [math] q=0.a_1a_2a_3... [/math] where d1, d2, etc are integers [math]\in[0,9][/math]. We can see that q has non-terminating decimals. By definition this means [math]q=(a_1/10)+(a_2/100)+...=\sum_{n=1}^\infty a_n\cdot10^{-n} [/math]. If [math] a_1=a_2=...=a_n [/math] for all n then its easy to see this is a geometric series and using the classic formula for a geometric series you can get [math] q [/math] in the form of a fraction. It's also easy to see how this works if you have a set of m numbers that repeat in the sum like [math] a_1=a_r, a_2=a_{r+1},...,a_m=a_{r+m} [/math] for fixed r.
>>11549761
I think you might misunderstand. I meant use superposition just to get current, but that is really overkill. You can just sum voltage drops over the closed loop to get current. Then you have the complex power over each element. Don't use superposition, srry srry.