Recent content by Gmaximus

You can't do that! Use l'hospital's rule. \lim_{x \rightarrow 0} \frac {tanx}{4x} =\lim_{x \rightarrow 0} \frac {secxtanx}{4}

Two plates, potential difference V =1500 which are .012m apart. What speed particle can pass through them without being deflected?

never mind guys, i did it right. The book was wrong. The problem was solve for X. Thanks anyway!

F_1=x F_2=4000lbs A_1=1.5 in A_2=21 in So, i did: F_1=1.5/21 * 4000 Gave me 280 pounds. My book, however, disagrees.

Does the outside pressure matter?

I just forgot to add what i put for I_c It was: I_c=1/2*M_c*R^2[/tex] The example in the textbook derives it in a general form, with mass M, and Radius R, and a constant density. I Tried it with M=2 and R=2 And got the above number.

I don't believe that's right, because i kind of ad libbed a step where i used the relation i mentioned before. i'd show my work but its very long. Here's how it worked: M_s=Massofsqaure I_s=1/6*M_S(2R)^2 M_s=M+M_c M_c=(4-pi)R^2 That last part I am unsure of.

With random numbers, (R=2 and M=2), i got: I=7.57

But, the way i see it, the square has an area of 4r^2, and the circle has an area of pir^2

Also, if the bottle has two openings facing in the same direction, how does that work? Do you add the area of the two holes and use that for area, or is it two separate cases?

How can you determine the net force of a bottle who's contents are kept under 2 atm of pressure, and it has a 1 cm opening for example? a1v1=a2v2 doesn't seem to apply, and the flowrate=dav doesn't seem to either. I suspect it is bernoulli's equation, but I am unsure. I know P=F/A, but how...

Oh, Duh. I shouldn't try to help people so late at night. Obviously, g is constant for everything. Funny how stupid i can be. Sorry Mr.Jeffy, I sincerely apologize for confounding you further.

a_1v_1=a_2v_2 Flow Rate=Dav Where D is density. And look up "Bernoulli's Equation"

But the magnitude of the acceleration due to gravity is directly related to the mass... Well then, I am confused. Oh well!

I don't think this can be done without a mass of the box. Do you have the actual text of the problem?