Recent content by 312213

Just noticed that the negative came from switching the signs of (5-x) into (x-5) and that I didn't notice the answer had that the other way around too. Problem solved

Pulling the 3 out only seems to make B=1/5 and everything else is the same. Can't see how that makes B negative either

Homework Statement Find the partial fractions of 1/((3x)(5-x)) Homework Equations The Attempt at a Solution 1/((3x)(5-x))=A/(3x)+B/(5-x)=(A(5-x)+B(3x))/((3x)(5-x)) 1=A(5-x)+B(3x) If x=0 1=A(5-0)+B(3*0)=5A => A=1/5 If x=5 1=A(5-5)+B(3*5)=B15 => B=1/15 So...

I see and understand now. Thanks for the help.

So then an acceptable answer would be (tan(x),Si(tan(x))?

I know that I can find the approximate number but I don't know where to go next to find the exact value of x for the first inflection point. I don't know what steps to take next to isolate x in one side and to have one x.

Homework Statement Find the coordinates of the first inflection point to the right of the origin. Homework Equations The Attempt at a Solution I know that the inflection point would equal zero for the second derivative of a function and that the second derivative of this function...

The work seems to have worked correctly up until the part of W=(Ffric)rcos37 0.3255=u((2)(9.8)cos37)(0.204)cos37 One of those cos37 shouldn't be there. Which is it? Or are either acceptable to be there, so long as only one is?

Ui=Kf+Usf+Ef mgh=0+1/2kx²+E (2)(9.8)(0.204sin37)=(1/2)(100)(0.204)²+E 0.3255J=E The nonconserved energy is the work of friction so W=(Ffric)rcos37 0.3255=u((2)(9.8)cos37)(0.204)cos37 u=0.128 (0.12763323293335829568080338705843) Does the nonconserved energy convert to mu like that?

Homework Statement A 2.00 kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m. The block is released from rest when the spring is unstretched, and the pulley is frictionless. The block moves 0.204 m down the incline before coming...

Homework Statement A spring of length 0.80 m rests along a frictionless 30° incline (a). A 2.6 kg mass, at rest against the end of the spring, compresses the spring by 0.10 m. (b) http://img64.imageshack.us/img64/5889/physics.jpg The mass is pushed down, compressing the spring an...

v1=-22.6 v2=17.7 (final momentum - initial momentum)/time (v2×m - v1×m)/time (17.7×0.54 - (-22.6)×0.54)/0.0018 (9.56 - (-12.2))/0.0018 21.76/0.0018 12088.9 Is this right?

v1=22.6 v2=17.7 (final momentum - initial momentum)/time (v2×m - v1×m)/time (17.7×0.54 - 22.6×0.54)/0.0018 (9.56 - 12.2)/0.0018 -2.63/0.0018 -1459.67 Is this correct?

Homework Statement A 0.54kg ball falls from rest at a height of 26m and rebounds up 16m. If the contact between the ball and ground lasted 1.8ms, what average force was exerted on the ball? Homework Equations F=ma v²=v0²+2a(x-x0) v=v0+at 1000ms=1s (1.8ms is 0.0018s) The Attempt at...

Homework Statement I have a vector word search that works by using vectors to find each letter on a coordinate plane. Each letter is obtained by using the current letter location and following the given vector to arrive at the new letter. Vectors are given as magnitude in a direction like "5...