Recent content by fundoo

ln xn = t x = et dx = et dt so initial eqn becomes \int t^n e^t dt and now integrate by parts

You have made wrong partial fractions. that's not the way. first separate the terms in numerator so Ur = 2/(r+1)(r+2) - 1/r(r+1)(r+2) and now splitting them gives Ur = 2(1/(r+1) - 1/(R+2)) - 1/2 (1/r(r+1) - 1/(r+1)(r+2) so the final Answer will be = 2(1/2 - 1/(r+2)) - 1/2 (1/2 - 1/(r+1)(r+2))

dont we consider the perpendicular length from the centre of mass, so that will be c/2. now even if we take the distance from centre to left corner, it will be c2/4 + c2/16 = 5c2/16. so torque will be mg x \sqrt{5}c/4 and equating it to Ia = (5mc2/8 + 5mc2/4)a gives a =...

Ans is Kinetic Energy. http://unfairstudy.uni.cc/3/20/416/1492/img/image008.gif KE is always positive, because both m and v2 are both positive. Read more Here :: http://unfairstudy.uni.cc/3/20/416/1492/index.html

if you are asking for P(AnB) = P(A)xP(B) then read Here :: http://unfairstudy.uni.cc/4/41/591/2373/index.html Under the Section "Independent Probabilities".

Homework Statement A uniform plate of mass m is suspended in the way shown. determine immediately after the connection at B has been released ;: A) angular aceleration of plate B) acceleration of its centre of mass Homework Equations before b is detached t1 + t2 = mg (Linear equillibrium) t1...

for second question :: probability that A miss :p(A): 30/100 probability that B miss :p(B): 60/100 Probability that both Miss :: p(AnB) = p(A) x p(B) = 30/100 * 60/100 = 18/100 (As both Events are independent) so the Probability that the Target is shot = 1 - Both Miss 1 - 18/100 =...