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
, the density of whatever you're sinking in, and the depth at which you are. In formula,
, 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 -
, while in B is
, 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
: 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 ![4m*3m*2m = 24m^3](https://tex.z-dn.net/?f=4m%2A3m%2A2m%20%3D%2024m%5E3)
2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: ![800* 24 = 1,92 *10^4kg](https://tex.z-dn.net/?f=800%2A%2024%20%3D%201%2C92%20%2A10%5E4kg)
2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's ![1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N](https://tex.z-dn.net/?f=1.92%20%2A10%5E4%20kg%20%2A%2010%20ms%5E%7B-2%7D%20%3D%201.92%20%2A%2010%5E5%20N)
2d.
again, what a surprise! ![800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa](https://tex.z-dn.net/?f=800%20%7Bkg%20%5Cover%20m%5E3%7D%20%2A%2010%20%7BN%20%5Cover%20kg%7D%7D%20%2A%202%20m%20%3D%201%2C6%2A%2010%5E4%20%7BN%5Cover%20m%5E2%7D%20%3D1.6%2A10%5E4%20Pa)
3. Yet again,
. ![1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa](https://tex.z-dn.net/?f=1000%20%7Bkg%20%5Cover%20m%5E3%7D%20%2A%2010%20%7BN%20%5Cover%20kg%7D%7D%20%2A%202%20m%20%3D%202%2A%2010%5E4%20%7BN%5Cover%20m%5E2%7D%20%3D2%2A10%5E4%20Pa)