Recent content by antiderivativ

Thanks for the reply! My new angle is 53.13. Is this better? :)

It turns out my number one is incorrect. The amplitude is half the total distance, so I should use 0.01 m instead. http://authors.ck12.org/wiki/index.php/Simple_Harmonic_Motion

80 = 10 log(I/10-12) = 10-4 W/m2 120 = 10 log(I/10-12) = 1 W/m2 P = 4\pi*r2 P = 4\pi*22 = 50.27 r = \sqrt{\frac{P}{I*4*\pi}} r = \sqrt{\frac{50.27}{I*4*\pi}} = 200 m

Thanks! <3

A section of an advertising sign consists of a long tube filled with neon gas having electrodes inside at both ends. A uniform electric field of 20 kN/C is set up between the electrodes, and neon ions accelerate along the length of the tube. Given that the ions each have a mass of 3.35 X 10-26kg...

Four point-charges are fixed at the corners of a 3.0m X 4.0m rectangle. The coordinates of the corners and the values of the charges are listed below. q1 = 100 microC (0, 4m), q2 = 36 microC (4m, 3m), q3 = 125 microC (0, 3m) and q4 = 32 microC (0,0). Compute the net electrostatic force acting on...

I'm sorry I don't understand how what you have told me allows me to use the inverse square law, so I will try it using the normal equations. However, thanks for pointing out that I do not have to add the D1 and D2. You say 104=10000 will allow me to use my inverse square law. I understand that...

I'm here to check my work again. I hope you don't mind. I'm going to try to check as many as I can today. :) The sound-level 2.0 m from a pneumatic chipper is 120 dB. Assuming it radiates uniformly in all directions, how far from it must you be in order for the level to drop 40 dB down to...

How do I mark this thread as solved? I don't see the option under Thread Tools.

Thanks kuruman and cephid. Both of you were very helpful! E = 18.1672 E = 0.5(M+m)v'2 v' = \sqrt{\frac{2E}{M+m}} v' = \sqrt{\frac{2*18.1672}{0.505}} v' = 8.4822 m/s v = \frac{(m+M)*v'}{m} v = \frac{0.505*8.4822}{0.005} initial velocity of the bullet = 856.7022 m/s

First part The bullet travels at a speed v. It has KE. E = .5K(A^2) E = .5(1614.8646)(0.15^2) KE = E = 18.1672 J Second Part The bullet hits the bock and moves the spring. The energy is no longer in the bullet as KE, but as spring energy. That's why KE = E. This is because of the law of...

My 10th post shows I know I need to apply the conservation of momentum formula and I tried to do so in that post. Since no comment was made on the work I showed, I have to consider that everything I did may not be right. Looking at Kuruman's 3rd post, I think that maybe I should say that the...

Also, on another note, is anyone having problems seeing my pictures? I've had a picture in every post, so if you can't see it, let me know. If you don't see pictures in my posts please tell me, so I can stop doing that. I prefer doing it because I like using my equation editor, but if the...

I'm STILL on this problem. I didn't think I'd get stuck on the simple harmonic stuff. I have much more difficult problems to try after I get this one. I know I'll need the conservation of momentum: mv=(m+M)v' where v' is after the impact. I think this because the block and bullet become one...

For Problem #1 Hooray! Thanks kuruman for checking. I have 18 problems to work on and now I feel confident about at least 1! For Problem #2: Thanks for the tip cepheid. Here is my new k. I’m saying that v is the initial velocity of the bullet while V is the initial velocity of the block...