Recent content by wwshr87

I am sorry, I made a mistake. The value c=10^(10^9), which in MatLab is inf. I will look into Brent's method and see what I can do. Thank you.

I intend to solve the following problem n*log(n)=c, where log is base 10. n*log(n)=c, expressing this as n=10^(c/n), then this becomes n^n=10^c. Now this can be solved using a numerical method such as Lambert W function. The Lambert W function solves the problem x*e^(x)=c, where c is a...

Well that makes perfect sense. I knew about the mathematical explanation but was looking for the other explanation you provided. Thank you very much sir.

Why the area under the curve of the function y=1/x form 1 to inf is infinite. But if we take this area and revolve it around the y-axis we obtain a volume of pi ?

Ok got it, thanks for the explanation.

Ok, I understand that now. But I still don't get where the 50/L is coming from.

I have attached the solution. I cannot follow what they are doing; if anybody does please help.

When I draw the FBD of the 1kg weight; I addthe forces in the y-direction 50*sin(x)+50*sin(x)=1

Homework Statement Mass hanging from the pulley in 50kg, the mass on the rope is 1kg. Please see attachment for details. The answer will be in terms on L. Homework Equations The sum of the forces in the y-direction and x-direction are equal to 0. The Attempt at a Solution Adding...

Homework Statement A bobbin of M = 3 kg consists of a central cylinder of radius r = 5 cm and two end plates of radius R = 6 cm. It is placed on a slotted incline on which it will roll but not slip, and a mass m = 4.5 kg is suspended from a cord wound around the bobbin. It is observed that...

Homework Statement Find the acceleration of the block on the left in terms of M, m and g. See attachment Homework Equations F=ma The Attempt at a Solution From the free body diagram, -Mg+Mg+mg=(m+M)*a a=m*g/(m+M) I think this is correct, however the solution states the...

You know the 1 ohm resistor is in parallel with the 10V source, therefore the voltage across it is 10V. Now we can subtract from the 10V the drop across the other 1 ohm resistor which is 1*Iq, and subtract Vq. 10 - 1*Iq - Vq=0, we can plug in for Iq which is Vq^2+2*Vq. Solve for Vq, you'll...

I never heard of this before. But I can't give you an answer since I don't know much about heat transfer.

The force exerted by friction is Ff=(µk)(mg-Fsin(theta)). Friction is doing negative work because it opposes motion. Work done by friction is -(µk)*Ff*x. Work done by the force is F*cos(angle)*x. Does this help?

But, is this an incline plane problem. Where is F acting, can you sketch it?