Recent content by phoebz

The figure shows an aluminum wire of length L 1 = 60.0 cm, cross-sectional area 1.00 x 10-2 cm2, and density 2600 kg/m3, joined to a steel wire of density 7.80 g/cm3 and the same cross-sectional area. The compound wire, loaded with a block of mass 10.0 kg, is arranged so that the distance L...

I tried putting in my answer for Vair/Vsolid as 0.55 and it was wrong, so I guess I've messed up somewhere

Okay, for the weight in the air I have: W =Vρsφg The volume of the water displaced would be the volume of the geode...which we aren't given :s. But I know the geodes weight under water needs to be multiplied by two to equal the weight of the geode in the air so: Vρsφg = 2(Vρwg) solve for...

A geode is a hollow rock with a solid shell and an air-filled interior. Suppose a particular geode weighs twice as much in air as it does when completely submerged in water. If the density of the solid part of the geode is 3100 km/m^3 , what fraction of the geode's volume is hollow? The...

Hello, I am taking a biochemistry course right now, and I am so confused by this 'hydrophobic effect' and how it relates to entropy. Hydrophobic effect: THe exclusion of hydrophobic groups or molecules by water. (I get this part!) This appears to depend of the increase in entropy of solvent...

Awesome! I finally think I have it! Thank you everyone :D

The the length of the string? I am really not sure :s, I thought I had it figured

They add together to make the adjacent side of the angle? I tried going cos(theta) = [L cos(theta) +h] /L but the Lcos(theta)s both cancel out :s

I found the horizontal displacement so that I could find the angle theta, I found 'x' using the formula h= L- sqrt(L^2-x^2). Then I went sin(theta) =x/L to find the angle for each marble, does this work? And thank you TSny, I think I have the right value for the small marble now, 0.028m!

Awesome! I finally got values for the velocities. V'= 2.972m/s, v'=-0.743m/s. So this means the large marble moves forwards with ~3m/s and the small marble goes back the way it came with ~1m/s? Then I can find the heights of each marble after the collision: small: 1/2mv^2 =mgh +1/2mv'^2...

Do I have the right steps though? Or are my steps and math wrong? (thank you for all your help so far!)

Ok, I have changed that to: 10(3.715)^2 = 10v'^2 +15V'^2 sqrt(138) = sqrt(10) v'^2 + sqrt(15) V'^2 And when I solved for V' I got -0.000159? I think I messed up again, I apologize for my rusty algebra skills...

Okay, thanks! So this is what I came up with: v= 3.715m/s (From my previous post) m=20g V=0 M=30g momentum: mv=MV' + mv' 20g(3.715m/s) = 30gV' + 20gv' 74.3 = 30V' +20v' KE: 1/2mv^2 = 1/2mv'^2 +1/2MV'^2 10(3.715)^2...

To Doc Al: I have figured out the velocity of the 20g marble is 3.715m/s just as it is about to hit the 30g marble using h= L- sqrt(L^2-x^2) and PE=KE. I know that I have to use the law of conservation of KE and momentum to figure out the rest, but, I do not know when I should be breaking...

But how will I be able to find angles from this?