For a given substance, its tells you the mass occupied by one unit of volume of that substance. In essence, is a measure of how well the molecules of...
Given the above, $"Volume"$ $=$ $"Mass"/rho$ $=$ $((489.1-241.3)*cancelg)/(1.00*cancelg*cm^-3)~=250*cm^3$. Note how the dimensions cancel to give me an answer in $cm^3$, as is required for a volume. Given a...
The density, $rho$, of mercury $=$ $13,593*kg*m^-3$; it is $13.6*g*cm^-3$ in more accessible units. Anyway, mercury metal is pretty dense stuff. If you ever found a pool of mercury, you...
By definition, $"density,"$ $rho="mass"/"volume"$... ..and thus to get the $"mass"$ we take the product.... $"mass"=rhoxx"volume"$...and here are quoted both $rho$ and $"volume"$. And so..... $"mass"=underbrace(13.6*g*cancel(cm^-3))_"density"xxunderbrace(16.8*cancel(cm^3))_"volume"=underbrace(??*g)_"mass"$. Often, in the lab you...
The pressure caused by the fluid is given by: $sf(P=rhogh)$ $sf(rho)$ is the $sf(g)$ is the acceleration due to gravity $sf(h)$ is the height of the column Since the...
We must measure the volume of the plot of the land, and then multiply by the : $"1 hectare"$ $=$ $100*mxx100*m=10000*m^2$ We multiply this value by the depth, to give...
Because we filled the vessel with $(125-50)*g=75*g$ water, which has a VOLUME of $75*mL$. When it was filled with an unknown liquid, the volume of the liquid is still $75*mL$,...
will support a column of mercury $760$ $"mm"$ high. See Because pressure is measured $"force per unit area"$, typically, the mercury column will be designed to be thin so...