## King Group Title A cork of density 0.15 g/cm^3 floats in a bucket of water with 10 cm^3 of its volume above the surface of water.Find the mass of the cork. 2 years ago 2 years ago

1. King Group Title

do u know how to do this?

2. Gab02282 Group Title

$d=\frac{m}{V}$ Therefore, $m=dV$ Hope this helps, somehow. I'm unsure.

3. King Group Title

how?can u solve it?

4. King Group Title

help tomas......

5. Tomas.A Group Title

i think you need to use Archimedus law or something

6. King Group Title

|dw:1326975917374:dw|

7. Tomas.A Group Title

|dw:1326976017906:dw|

8. King Group Title

ya so now?

9. Tomas.A Group Title

also $F_a>mg$ if that matters lol

10. King Group Title

i know but then i am stuck at hw to find the missin volume..............

11. sachin_vishaul Group Title

hi

12. sachin_vishaul Group Title

u hv the same problem?

13. King Group Title

i posted fr u

14. sachin_vishaul Group Title

nanbenda

15. King Group Title

jacob help man!

16. King Group Title

wait sachin

I think you have to use Archimedes Principle and Density- Mass relationship. Still I am not getting the equation.

18. King Group Title

i know but how????

19. King Group Title

how is d=m/g

Sorry, i was saying whether we can use density = mass/volume?

21. King Group Title

yeah we can use anythin

Then it is simple. Mass= volume X density, so Mass= 10 * 0.15= 1.5 gm (Still u check the answer)

23. King Group Title

24. King Group Title

Saw it, but confused about the equation

26. King Group Title

volume is 10+some x

I will consult and say u tomorrow

28. nikvist Group Title

$\rho=0.15\,g/cm^3\quad,\quad \rho_w=1\,g/cm^3\quad,\quad V\uparrow=10\,cm^3$$\rho Vg=\rho_w V\downarrow g\quad\Rightarrow\quad \frac{V}{V\downarrow}=\frac{V\uparrow+V\downarrow}{V\downarrow}=\frac{V\uparrow}{V\downarrow}+1=\frac{\rho_w}{\rho}$$V\downarrow=\frac{\rho}{\rho_w-\rho}V\uparrow$$m=\rho(V\uparrow+V\downarrow)=\rho\left(1+\frac{\rho}{\rho_w-\rho}\right)V\uparrow=\frac{\rho\rho_w}{\rho_w-\rho}V\uparrow$$m=\frac{0.15}{0.85}\cdot 10\,g=1.765\,g$

29. JamesJ Group Title

nikvist, this a good answer. I understand your notation. But perhaps for the others could you explain in words what you've done here? Thanks.

30. salini Group Title

|dw:1326987424332:dw|For practical purposes water is incompressible so the submerged part would displace an amount of water equal to its own volume. so same mass that is what nikivist has done ...equate the mass of cork as a whole to mass of water displaced then manipulations gives us everything written in terms of density which we know(data in the question)

31. King Group Title

nikvist could u explain ure answer in words or JamesJ could u do it fr him?

32. salini Group Title

hmmm..... that is wat i did.............

Please explain in simple words. Are we to use that Density formula or not?

34. salini Group Title

I WILL TELL EVERYTHING FROM FIRST by archimedes principle, the wt of cork=wt of water displaced(buoyant force) this is because the body is floating upward forces=downward then wt of cork=volume*density*g wt of water displaced=volume(V)*density oif water*g write volume of cork as volume of upper immersed portion+vol of downward immersed portion(V) REMEMBER IT IS V that we need to calculate do it in simple terms after this |dw:1327073074048:dw|

35. salini Group Title

knowingV it is easy to calculate the mass of that part of cork+mass of upper part of cork...........got it adarsh?

Yeah

37. salini Group Title

u need to know here that by archimedes principle,buoyant force=wt of water displaced in this case as the cork is floating the wt mg equals the buoyant force so to to keep the cork at rest

Yeah, that's right.

39. salini Group Title