Recent content by rculley1970

I just did, recalculated, and it worked. Thank you very much andrew. I can't believe it was such a small detail that turned it into such a huge problem. Once again, thank you very much.

no, that didn't work. I think it's the answer. I am doing something wrong but can't figure it out. I have to submit the work in 30 minutes so i guess I will leave it blank and find out the hard way.

Is that the correct answer for the problem? I have been working it out but putting it into the online answer form keeps saying it is wrong. For the friction of .15 I am getting a stopping distance of 1438 m. The problem asks for the answer in 3 significant figures only. I can't figure this...

The problem does not show any mass. The equation I get out of the book is: (coefficient of k)(g)=a Therefore, from the only given data from the problem: (.15)(9.8) = a = 1.47m/s^2

I come up with: a=1.47m/s^2 deltaX = 1437 m

problem: a car is traveling at 65.0km/h on a flat highway. coefficient of friction between road and tires is 0.15, what is the minimum distance in which the car will stop. I am assuming that I start out with: .15(9.8)=a which = 1.47m/s^2 plug that into: V^2=V0^2+2a(deltaX) and solve for...

Is there a link someone can provide that has a good tutorial on friction coefficient formulas and when to apply them. I seem to be having problems with this topic. Everyone knows the old M1 and M2 connected by a string and pully question. That is on a level surface. But if the surface is...

oops, i messed up the easy algebra and put .5t^2 on the denominator instead of multiplying deltaX by two. Such an easy mistake can kill a problem. Thanks for everyones help.

Problem: An object of mass 2.0kg starts from rest and slides down an inclined plane 69cm long in 0.90s. What net force is acting on the object along the incline? Now, I used what I was given and plugged it in the equation: deltaX=1/2at^2 + V0t I come up with acceleration = .0552...

Crashing out for the night. Working on it tomorrow along with some friction problems. (ughhh) Too many equations. Mostly memorization of equations and when to use them from what I can tell. Will look into the current problem tomorrow. Thank you for your help.

sqr root( (.029)^2 + (.287)^2) = .288

I keep coming up with 84.2 degrees but the website keeps telling me it's incorrect.

F(x)=-480(sin11) + 400(sin26) = 83.8 F(y)=480(cos11) + 400(cos26) = 831 a(x)=(83.8)/(2900) = .029 a(y)=(831)/(2900) = .287 a = .288

Odd, I can input the answers and it tells me if they are correct or incorrect. On the mag and acceleration, i got 914.8 and .288 respectively. It gave me credit for both of them. I keep getting the angle wrong though. will keep working on it.

Sorry, mass was given in part (b) of question: 2900Kg. Part (a) just asked for mag and dir. (b) asks for acceleration