Recent content by Miguel Velasquez

I agree with you, I am tired to check over and over my result, it doesn't seems to be nothing wrong with it, nothing wrong with math or reasoning, its a problem about definition. How do books define "Force exerted by a liquid in open liquid filled tank" is the question to answer. If we define...

Thanks to both for your response, sorry for late reply, there was a blackout earlier. Ill try to read carefully both answers to see if i can clarify my doubt.

I got that feeling too, i have been checking this result for weeks, with few sleep and can't find any mistake in my reasoning. Can anyone else confirm if this result is correct/wrong please?

I must be doing something wrong because books point the F_liq=MassLiquid*g. See by example "Serway - Physics for Scientists, 7th Ed, page 409, exercise 11." "11. A swimming pool has dimensions 30.0 m 10.0 m and a flat bottom. When the pool is filled to a depth of 2.00 m with fresh water, what...

Sorry maybe i have should pointing from the beginning my attempt to solve the problem was in the pdf file, i repost the link http://docdro.id/SYbWm95

Homework Statement [/B] I am trying to understand why books always point as a fact that hydrostatic force on the bottom of a open liquid filled tank doesn't depend on the force due atmospheric pressure because they these forces cancels each other. Homework Equations [/B] P=[P][/o]+ρgh F=P*A...

Wow! you were right, i plugged in a wrong value, thank you RUber! Could you tell me why the force must be averaged? What i did is use F=M_tot*g=(2.4Kg/m)(500m)(9.8m/s2), but this seems to be wrong.

When i plug in the data into your eq. i get 0.016333333 m which is not the right answer. The link seems to works for me, can anyone else test the link i gave? Thank you Ruber for trying.

A steel cable 3.00 cm 2 in cross-sectional area has a mass of 2.40 kg per meter of length. If 500 m of the cable is hung over a vertical cliff, how much does the cable stretch under its own weight? Take Y steel ϭ 2.00 ϫ 10 11 N/m 2 .Y=([L][/o]F)/(A*delta_L) My attempt of solution...

Thanks for reply my question James, problem is solved. Taking again a look at the figure you can notice the value of d actually increases when the sheet is being heated which would reduce the value of theta.

Oh now i can see where i am mistaken, the value of d is infact increasing when the sheet expand and not being reduced as i thought. This explains why the correct answer is letter a) it moves toward the centerline. When the value of d is increased, the argument of arcsin will be reduced. For some...

i think theta is the angle shown in the figure source of figure: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/doubsli.gif

If a sheet containing two very thin slits is heated (without damaging it), what happens to the angular location of the first-order interference minimum? a) It moves toward the centerline. b) It moves away from the centerline. c) It doesn't change.Condition for m-order interference minimum...