Recent content by maulucci

Ok thank you

Ok V=q/C V=V1+V2 q/C=q/C1+q/C2 C^-1=C1^-1+C2^-1

To show that they are equal charges q1=q2 And q=Cs^-1V q=(C1^-1+C2^-1)V (C1^-1+C2^-1)V=Cs^-1V Cs^-1=C1^-1+C2^-1

I tried to work it out and i was wondering if this was the correct way to show wht the question is asking for q1=C1V and q2=C2V q=q1+q2=(C1+C2)V CsV=q=(C1+C2)V Cs=C1+C2

Homework Statement Derive a relationship for Cs for a two capacitor series circuit with a resistor. Start by showing that the same charge separation q is present across the capacitor and each of the capacitors in series and that the voltage across Cs is equal to the sum of the potential...

ok so would a proof be showing that every vector is v=(Distance measure in meters)/(number of dots) with the direction

Homework Statement Show that you can measure the magnitude of the velocity vectors which have units of distance/time with a ruler that measures distance, if a certain condition is met. What is this necessary condition.Use the Equation and Figure to prove below that this can be done...

-k1x1=-k2x2=F F=-keq*x solving for x1 x1=[(k2)/(k1+k2)] +x2 putting into F=-k2x2+k2x1 F=x2 (-k1k2-k2^2+k2^2)/(k1+k2) simplified F=-[(k1k2)/(k1+k2)]*x2 then the Keq = (k1k2)/(k1+k2) ?

Homework Statement The problem is in the attachment Homework Equations F=-kx k1x1=k2x1 The Attempt at a Solution I got as my answer Keq= (k1k2)/(k1+k2) i just wanted to see if i had the right idea or no.