Recent content by Anna55

Thanks for all the help I have gotten! :) I really appreciate it. :)

Sorry for not writing what the variables stand for in the beginning. Yes, Wm is the weight of the metal. Yes, Vm is the volume of the metal. The change in reading of the scale will be pwVmg.

For water the heat capacity is the greatest when it is a liquid. However is not water an exception? What is the overall rule?

Before the metal is submerged in the water. Ft=W Submerged metal Fb+Ft=W Fb+Ft-W= (They cancel out) Fb= pwVdg= pwVmg To summarize everything. Is this correct?

At what phase is the heat capacity the greatest? Is it when the substance is a solid, liquid or gas? Please also explain why. I believe that the heat capacity is the greatest when the subastance is a solid.

Fb=Wm pw+Vd+g=mg pm+Vd+g=mg pm+Vd+g

Yes, the symbols are correct. Fs+Fb+Ft=Ww Fs+Fb+Ft-Ww=0 Fs+Fb+Ft-Ww-(Fs-Ww)= Fs+Fb+Ft-Ww-Fs+Ww= Fb+Ft= pwVdg+Ft= Is this correct? How should I continue?

In the lab you place a beaker that is half full of water (density pw) on a scale. You now use a light string to suspend a piece of metal of volume V in the water. The metal is completely submerged, and none of the water spills out of the beaker. Give a symbolic expression for the change in...

Here are my results: http://imageshack.us/photo/my-images/853/pic2uoe.jpg/ Thank you for the advice of the webpage. Ok, thank you for informing me about the rules.

Thank you! How should you make the picture URL? It is a JEPG image.

I have taken a picture of the results, however how should I insert the URL of the picture?

Hello! By performing electrophoresis, I have created a filter paper which shows the protein types from different species of fish, wild pig and chicken. Unfortunately, problems have arisen regarding the analysis of the results since most protein samples vary between 15 and 25 kilodaltons. Can you...

Thank you very much ehild! I understand now.

If I have understood this correctly, the solution should be this: 2aΔx=vf2- vo2 a=-µg 2 -µg Δx=vf2- vo2 vf=0 2 -µg Δx=- vo2 Δx=- vo2/2 -µg Δx=-252/(2×-0.95×9.8) Δx=33.6 m

A 1000 kg car traveling rightward at 25 m/s slams on the breaks and skids to a stop (with locked wheels). Find the distance required to stop it. Coefficient of friction between the tires and the road is 0.95. Let g=9.8 m/s^2 Solution: 2aΔx=vf2- vo2 a=µg 2 µg Δx=vf2- vo2 vf=0 2 µg Δx=-...