Recent content by ppkjref

i'm trying to dissolve an unknown acid salt and it is not soluble in water at stp. it does dissolve, however, when it undergoes heat and some stirring. is this property representative of most weak acids, that they are not easily miscible in water?

ln(2+(e^x) + 2/(2+(e^x)) Just needed a break I guess. Wasn't thinking.

the a/c fraction is u/(u^2), equivalent to 1/u which when integrated is 0?

Homework Statement S = integral symbol S (e^(2x))/((2+(e^x))^2) dx Homework Equations u = 2+(e^x) (e^x) = u-2 du = (e^x) dx The Attempt at a Solution S [(e^x)(e^x)]/((2+(e^x))^2) dx S [du(u-2)]/(u^2) For the first (e^x) dx I substituted du. And since there was only (e^x)...

nevermind i figured it out the answer is [(5x+7)^29]/145. stupid mistakes

Homework Statement S=integral symbol S(5x+7)^28 Homework Equations Substitution The Attempt at a Solution let u = 5x+7 du = 5 S u^28 du = (u^29)/29 du = [5(5x+7)^29]/29 is this correct? I'm not sure...

Ok I got 0.019 J as my final answer. Thanks for the help

nevermind I'm an idiot v1 = 0

dy=V1yt+ 0.5ay(t)2 and rearranged it in terms of t. I got 0.4403 s. how do you come to 0.4403 s. if there are two unknown variables (v1 and t)?

I'm not sure what you mean. Initially wouldn't there only be elastic potential stored? KE is 1/2mv^2 and the horizontal velocity is d/t? I don't know what time is

Elastic Potential Energy of Elastic Band* Homework Statement A stretched elastic band of mass 0.55 g is released so that its initial velocity is horizontal, and its initial position is 95 cm above the floor. What was the elastic potential energy stored in the stretched band if, when it lands...

wow that's embarrassing. I'm an idiot hah

i noted that it is 0/0 form so therefore I am likely to get a real number. [(2/(3+h))^2-(4/9)]/h = [(2/(3+h))(2/(3+h))-(4/9)]/h =[(4/(9+6h+h^2))-(4/9)]/h =[(36-(36+24h+4h^2))/(9(9+6h+h^2))]/h =[(-4h(h+6))/((3h+9)(3h+9))]/h =[(-4h^2(h+6))/((3h+9)(3h+9))] So the denominator is 81 and the...

Homework Statement Evaluate the limit, if it exists. the limit of ((2/(3+h))^2-(4/9))/h as h approaches 0.The Attempt at a Solution So far I got, [(2/(3+h))^2-(4/9)]/h = [(2/(3+h))(2/(3+h))-(4/9)]/h It seems nothing can cancel out when I find a common denominator.

Okay thanks for your help.