Answer:
Check body of the explanation
Explanation:
Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about [tex]10 ms^-^2[/tex], the density of whatever you're sinking in, and the depth at which you are. In formula, [tex]p(h) = p_0 + \rho g h[/tex], and the pressure is the same for every point of the tank at the same depth.
At this point, we can start answering!
1a. The pressure at A is - not counting atmosferic pressure - [tex]1000 * 10 * 1 = 10^4 Pa[/tex], while in B is [tex]1000*10*2 = 2*10^4 Pa[/tex], so it's half of it.
1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!
1c. Ditto. Same depth? same pressure!
1d. Usual equation, this time density is 800. Pressure is [tex]800*10*2 = 1,6*10^4 Pa[/tex]: Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water
2a. The volume is simply the product, so [tex]4m*3m*2m = 24m^3[/tex]
2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: [tex]800* 24 = 1,92 *10^4kg[/tex]
2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's [tex]1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N[/tex]
2d. [tex]\rho gh[/tex] again, what a surprise! [tex]800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa[/tex]
3. Yet again, [tex]\rho gh[/tex]. [tex]1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa[/tex]
