Recent content by abeltyukov

Homework Statement Hi, I have the following circuit: http://i137.photobucket.com/albums/q208/infinitbelt/ProblemSet3Circuit1.png" Also, the following is known: X=60 V R1= 10 Ω R2= 20 Ω R3= 10 Ω R4=5 Ω R5=20 Ω R6 = 5 Ω Homework Equations V = IR y = mx + b The...

Hi, I have the following circuit: http://i137.photobucket.com/albums/q208/infinitbelt/ProblemSet3Circuit1.png" Also, the following is known: X=60 V R1= 10 Ω R2= 20 Ω R3= 10 Ω R4=5 Ω R5=20 Ω R6 = 5 Ω How do I go about finding a V-I characteristic of this circuit? I know that the...

Thanks! I got the right answer but am still not sure why. Here's what I did: y = ((lambda)*(L)) / d ((0.523 * 1.408E-6) / 1.2) - (1.076E-7 - 0.523E-7) = 558.5 nm. Thanks!

So should y2 be y2 + y1? I am still a bit confused. Thanks!

I had: a*(y1 / L) = m1*lambda a*(y2 / L) = m2*lambda (lambda*L) / y1 = (2*lambda*L) / y2 lambda = [(y2)(L)] / [(y1)(L)] Is that right? Thanks!

Homework Statement The interference pattern on a screen 1.2 m behind a 710 line/mm diffraction grating is shown in Figure P22.45 (http://i137.photobucket.com/albums/q208/infinitbelt/p22-45alt.gif ), in which y1 = 52.3 cm and y2 = 107.6 cm. What is the wavelength of the light? 2. The...

Yeah, lol. Thanks again for your help.

Homework Statement Jose, whose mass is 90 kg, has just completed his first bungee jump and is now bouncing up and down at the end of the cord. His oscillations have an initial amplitude of 9 m and a period of 4.0 s.2. The attempt at a solution a) The spring constant of the bungee cord is...

Yes, sorry about that. Thank you very much for your help. I got the right answer of 0.223.

I think I have the wrong spring constant. The spring constant is actually unknown. I tried to calculate the k using the following equation: k = (m1+m2)((2pi)(f))^2 I then used A = (u * g (m1+m2)) / k Do I use m1 + m2 in both of the cases or do I just use one mass to solve for k...

How did you get that? Thanks!

Did you mean 35/3.8? If so, I get friction to be 0.939, which is not the right answer. Thanks!

So am I using the wrong acceleration? Should the equation be: A = (a(mass of bottom block)) / k? If so, I get u = 0.939, which still is the wrong answer. I think I misunderstood you. I get the box in the pickup truck. So the bottom block is like the truck and the top block is like the...

The restoring force is maximum when the amplitude is the highest, right? Here's what I tried doing: A = (a(m1+m2)) / k I solved for acceleration to get 6.03 m/s^2 I then used F = ma I found F to be 35 N I then used F = m1(u * g) I got u to equal 0.892, which is still the wrong...

Thanks! I wonder how I got the first part right. :smile: