Physics question on density!urgent help?

Question

Hello I need help on the following question part c) I need to see if I am doing it correctly!

http://www.collegeboard.com/prod_downloads/ap/students/physics/physics_b_frq_02.pdf

For part c where it says to calculate the density of the liquid. To determine the density one needs to find the difference in vertical distance that the mass is not travelling due to the opposing buoyance force.

Does that mean the vertical distance from the object to the top of the surface?

Anyhow the following formula was shown to me :

k x delta h= mass of liquid x gravity

k x delta h = density of liquid x volume of mass x gravity

density of liquid = k x delta h / (volume off mass x gravity)

is this correct?

Thank you!!

Answer 1

Your question left out some info. There are 7 questions, all of which have a part (c). I guess you mean question 6?

The "difference in vertical distance" tip refers to the difference between the spring stretch distance with the beaker removed versus the spring stretch distance with the object submerged.

I could just say yes, your formulas are correct. But I'd rather try to explain why. I assume since you didn't ask about part a, you have found the spring constant.

With the object hanging into the fluid, the forces are in equilibrium: net force,
Fnet = 0.

So the forces have this relationship
Fnet = 0 = Weight - buoyancy - spring tension
Or
Weight = buoyancy + spring tension
W = b + st

So, what do you know about those forces?
W?
The weight is m*g but also is related to the spring constant by
W = k*d1
where d1 is the spring stretch distance with the beaker removed.
b?
b = weight of displaced fluid.
b = Vm*rho*g
where Vm is the volume of the object, rho is the density, and g is 9.8 m/s^2.
st?
Spring tension, in the mode with the object in the fluid, is
st = k*d2
where d2 is the spring stretch distance with the object in the fluid.

So plug into
W = b + st
what we just said.
k*d1 = Vm*rho*g + k*d2
Simplify
k*(d1-d2) = Vm*rho*g
You found k and d1 in working out (a), and you can measure d2. You know g, need rho, ... but Vm is still undetermined.

The beaker has gradations on it. You can get the volume of the object from how much the fluid surface rose when the object was lowered. Plug that in and solve for rho.

Answer 2

Just more physics jargon ... I am not even sure if this is correct anymore ... Photons and gluons are strictly massless. They are the gauge fields of unbroken symmetries (unlike the weak vector bosons). Due to the strong force and the self-interactions of the gluons (the strong force originates from a non-Abelian group -- so there is self interaction among the gauge fields -- the gauge fields themselves are charged), there may be gluon condensation and the condensation appears massive. As they are massless, they do not directly interact with Higgs boson at the tree-level. At higher-loop levels, there are interactions because you can have all sorts of virtual particles (i.e. particles that are not 'on-shell'). The same likely goes for the interactions with gravitons (the particle interpretation of the gravitational field). Photons curve around large bodies. Photons can not escape black holes once they are in the event horizon of the black holes. However, those two things are somewhat independent, unless the curvature of the photon trajectory is induced by the black hole.