+- Twitist Forums (http://twitist.com)
+-- Forum: Facebook forums (/forum-14.html)
+--- Forum: General facebook and life forums (/forum-25.html)
+--- Thread: Physics question? (/thread-146559.html)
Physics question? - Erraticsleet369 - 04-28-2014 03:11 AM
I know ask is probably filled with these, but I got some homework that I can't really solve by myself and I would really appreciate it if someone could help me out. So here goes:
"A bracelet that weights 0.49 N on air, drops to 0.44 N. How much Copper and Gold is in the bracelet, if the density of Copper is 8,9 g/cm^3 and the density of Gold is 19,3 g/cm^3.
I'm really not sure how to figure this out, if anyone can help me, I'd be grateful.
- RickB - 04-28-2014 03:20 AM
I think you left out part of the question. I think you meant to say: "A bracelet that weights 0.49 N in air, drops to 0.44 N when submerged in water."
This means the buoyant force on the bracelet is: Fb = 0.49N−0.44N = 0.05N
You also know (I hope) that the buoyant force equals the weight of the displaced water. That means the displaced water weighs 0.05N. And since water's density is 1000kg/m³, we can figure out the volume of the displaced water:
volume of displaced water = (mass of displaced water)/(density of water)
= (0.05N/g)/(1000kg/m³)
= 5.1×10^-6 cubic meters
= 5.1 cubic centimeters.
Of course the volume of the displaced water also equals the volume of the bracelet, so:
Volume of bracelet = 5.1 cm³
Also we know the bracelet weighs 0.49N, so its mass is:
mass of bracelet = 0.49N/g = 0.05 kg = 50 grams
The mass of copper plus the mass of gold must therefore add up to 50 grams:
m_copper + m_gold = 50 grams
Also, the volume of copper plus the volume of gold must add up to 5.1 cm³. Recall that volume = mass/density; then we can write:
m_copper/Ï_copper + m_gold/Ï_gold = 5.1 cm³
(where "Ï_copper" and "Ï_gold" are the given densities of copper and gold.)
So, now you have two equations in two unknowns (m_copper and m_gold). Use the algebra of simultaneous equations to solve.