Recent content by cunhasb

I think I finally got this... ∫(x+2)√(x-3)dx ∫(x-3+5)√(x-3)dx u=(x-3)^1/2 u^2=(x-3) x=u^2-3 dx=2u ∫(x+2)√(x-3)dx=(u^2+5)*(u)*(2u) du = 2∫(u^4+ 5u^2) = 2∫(u^4 du +5∫u^2) = 2*1/5*u^5 + 2*5*1/3*u^3 = 2/5*u^5 + 10/3*u^3 u*3(2/5u^2 +10/3) + k (x-3)^(3/2) * (2/5(x-3) + 10/3) + K...

I think the last one is wrong... here is the one I think might be right... Can you please tell me if it is, and if not I would really appreciate if you could show me the way... thanks... ∫(x+2)√(x-3)dx ∫(x-3+5)√(x-3)dx u=(x-3)^1/2 u^2=(x-3) x=u^2-3 dx=2u...

∫(x+2)√(x-3)dx ∫(x-3+5)√(x-3)dx u=(x-3)^1/2 u^2=(x-3) x=u^2-3 dx=1/du ∫(x+2)√(x-3)dx=(u^2+5)*(u^2) du = 5∫(u^4) du +∫(u^2) =5/5*u^5 + 1/3*u^3 =(x-3)^(5/2) - 1/3(x-3)^3/2 + K Is it right now? Thank you very much... Thank you so much guys...

Hoping anyone could give a hand on this... this is what I've gotten so far... ∫(x+2)√(x-3)dx u=(x-3)^1/2 u^2=(x-3) x=u^2-3 dx=2u du ∫(x+2)√(x-3)dx=(u^2-3+2)*(u)*(2u) du =2∫(u^2-1)*(u)*(u) = 2∫(u^2-1)*(u^2) =2∫(u^4-u^2) =(2*1/5*u^5) - ((2/5)*2*(1/3)*u^3) =2/5*(x-3)^(5/2) -...

So the answer will be i=1/5(ln(t+3)^2+ln(t+7)^4+k... now about the 1/L this is the 1/5 (I had 5-henry inductor)... Is it right?

I hope anyone could give me a hand on this one... The induced emf in a 5-henry inductor varies as V=(6t+26)/(t^2+10t+21). Recalling that current i=1/L∫Vdt, find the formula for the current i as a function of time t. This is what I've gotten so far...

I hope anyone could give me a hand on this one... The induced emf in a 5-henry inductor varies as V=(6t+26)/(t^2+10t+21). Recalling that current i=1/L∫Vdt, find the formula for the current i as a function of time t. This is what I've gotten so far...

thank you guys... Sorry not to thank you before... but I was away for a few days... Thank you so much...

Could anyone give a hand in solving this problem? In shallow ocean water, the frequency of the wave action (number of wave crests passing a given point per second) is given by f= [(g/(2PiL))tanh(2Pih/L)]^1/2, where g is the acceleration due to gravity, L is the distance between wave crests...

Thank you so much guys... well... I think I got it... correct me if I'm wrong... 1) Rc= 40 micro V(0) = 80 T=? V = 5V V=v(0) e^(-t/RC) 5=80*e^(-t/RC) 5/80=e^(-t/RC) RC*ln0.0625=-t t=110.903 microseconds... 2) Rc= 0.0001 V(0) = 100 T=? V = 20V V=v(0) e^(-t/RC) 20=100*e^(-t/RC)...

I would really appreciate if anyone could help me figure out these problems... An RC circuit has a time constant of 40 microseconds. If the capacitor is first charged to a voltage of 80 volts and the RC circuit is then closed upon itself, after what time will the capacitor voltage be equal to...

I would really appreciate if anyone could help me figure out these problems... An RC circuit has a time constant of 40 microseconds. If the capacitor is first charged to a voltage of 80 volts and the RC circuit is then closed upon itself, after what time will the capacitor voltage be equal to...