Recent content by the.flea

thats the thing i know how to integrate ln(x) where x is any function wih power 1. ln(x^3-1) let t = x^3-1 --> 3x^2dx=dt when i sub that in i don't know how to proceed =ln(t)dt/3x^2 --> maybe another substitution, ie let u=3x^2, du=6xdt, any help guys?

Homework Statement Integral of: 2x^3/(x^3-1) with respect to x. The Attempt at a Solution you can divide them to get sum of integrals. =int(2,x) + 2*int(1/(x^-1),x) =2x + 2*int(1/(x^-1),x) Im having trouble proceeding with the last part, i just need help getting started.

it all makes sense now. thanks for the help.

ok so we have the change in potentila enrgy = Wg but what are the calculations or formula for calculating Wa. that is what i need to understand. thnks

since it gets pushed sideways, the height changes vertically but the length of the rope remains the same. sqrt(15^2-5^2)=14.14

mg(df-di)=(9.8)(170)(14.14-15)=-1432.76J ? same as above i already calculated it?

but since it is a varying force, how would i use the scalar product, it isn't a constant force. should i only apply it to the final position where. what is the formula for Wa? since its the negative of Wg= mghcos180=-(9.8)(170)(0.86)=-1429 J. Wa should be +1429J but how can i calculate it...

Wa is the work done by the applied force. since at any point on the rope it is right angles to displacement cos90=0 so then tcos90=0. the answer in the book is 0 too. thats what i think.

so for a, are the calculations correct? tension will have its corresponding components, yes. and for c work due to tension is always 0 then (cos90) "why isn't the component of Force along x = M*g*Sinθ" how did you come to that conclusion? EDIT: how would one calculate the Wa without setting it...

for (a) gravity does not have any components. i think you misread the question. I've attached a diagram. for (b) since the change in kinetic energy is zero, mgh is equal to the negative of work for (c) i don't quite understand

given/known data A 170 kg crate hangs from end of rope length L = 15 m. You push horizontally on crate with varying force F to move distance d = 5 m sideways. (a) What magnitude be of F when crate is in final position? During crate's displacement, what are (b) the total work done on it, and (c)...

m=1.2g TlCl-->Tl + Cl mm=239.83g 1 : 1 : 1 n=0.005mol c=n/v c=0.0125mol/L solubility = c Ksp=[0.0125][0.0125] =1.56E-4 is this correct?

oh in part b it is a typo. so 6.5g (750ml) is irrelevant?

[SOLVED] Solubility Thallous Chloride Homework Statement 6.5g of thallous chloride is added to 750ml of water at room temperature to determine Ksp. The solution is stirred from time to time over a 4h period. Some thallous chloride does not dissolve and the solution is declared saturated. A...

Homework Statement Find all values of t so that the three vectors a=(1,2,3) b=(4,5,6) and c=(7,8,t) are coplanar.Homework Equations (axb) \bullet c=0The Attempt at a Solution (1, 2, 3) x (4, 5, 6) (-3,6,-3) (-3,6,-3)\bullet(7,8,t) 27=3t t=9 it says find ALL values of t. i keep getting 9...