Recent content by rabar789

Hang on, I think I'm on the right track: y-y1 = m(x-x1) --> y = 2x-4 And then make that equal to the derived equation? 2x-4 = 1-x^2 ? Which should equal x^2 + 2x - 5 = 0 ? I put that in the quadratic formula, and I'm getting an answer that's close. Am I doing the math wrong?

Hey everyone, I'm a new Calculus student, and this problem on derivatives is giving me lots of trouble. I can't get around it! Please help. Homework Statement Find the coordinates of all points on the graph of y=1-x^2 at which the tangent line passes through the point (2,0). Homework...

Nevermind, I figured out my answer; I might as well post the solution here for reference for others! 1/f = (n-1)(1/R1 - 1/R2) (In this case, both R's should be positive.) 0.038 = (0.58)(1/R1 - 1/2) 0.038 = (0.58/R1) - 0.29 (Solve for R1) R1 = 1.768 cm

[SOLVED] Contact Lenses Problem Homework Statement A contact lens is made of plastic with an index of refraction of 1.58. The lens has a focal length of +26.0 cm, and its inner surface has a radius of curvature of +20.0 mm. What is the radius of curvature of the outer surface? Homework...

This is the exact statement. This was the problem, no one in the class knew whether or not the 5 m/s occurred immediately and where. And yes, 10 meters does sound a bit short; unless it's one of those goofy circus cannons that just shoot the guy out a little bit...

This is an annoying problem for which I have a close answer to, not sure where I went wrong. Homework Statement Cannonball Man (mass = 75kg) is stuff into his circus cannon, compressing a giant spring by 1.5 meters. He is launched vertically upward at 5 m/s reaching a maximum height...

Ah! Thanks! I took that answer, plugged it into all my other equations and solved for the coefficient of static friction, getting 0.15976. It seems right, but I have no way of checking. I appreciate the help!

Wait, now I'm really confused. How do I convert 36 revolutions per minute to "X" seconds per revolution?

OK so this is what I tried: Ffriction = Fc (coefficient)(Fnormal mg) = (mass)(v^2/r) I cancel out the masses and solve for the coefficient and I get 1.226, which I believe is impossible because it's greater than one. I follow your logic, but maybe I did something wrong in the math...?

[SOLVED] Circular Motion Problem Homework Statement A coin is placed 0.11 meters from the center of a rotating turntable. The speed of the turntable is slowly increased; the coin remains fixed on the turntable until 36 rpm is reached and the coin slides off. What is the coefficient of the...

It is?! Thanks so much! I'm not so great at physics, but I'm trying. And thank God my teacher doesn't care about sig figs, one less thing to worry about.

0.866 is the cosine of 30 degress. Also, 0.5 is the sine of 30, not to be confused with the coefficient of static friction, 0.500.

See, I took my original FY equation, FY = Ften(sin30) + Fnormal - Fg = 0 and rearranged it to Fnormal = Fg - Ften(sin), or FN = 500 - 0.5Ften. Then I took my original FX equation, FX = Ften(cos30) - Ffric = 0 and rearranged it to Ffriction = Ften(cos30), or Ffric = 0.866Ften. Then, since I...

Hang on, I had a burst of realization; how about this: FY Fnormal = 500N - 0.5Ftension FX Ffriction = 0.866Ftension Ffriction = (coefficient)(Fnormal) 0.866Ftention = (0.500)(500N - 0.5Ftention) Ftention = 224.014N

I'm not sure I follow you on that; I don't know the Fnormal or the Ffriction, and I need one of them plus the static friction coefficient to find the other, right? Normally, Fnormal would just equal Fg, but since there's the Ftension(sin30) force, that also has to be taken into account, right?