Physics Help Forum Mass of water acting on submarine.

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 Feb 29th 2016, 01:23 PM #1 Junior Member   Join Date: Feb 2016 Posts: 3 Mass of water acting on submarine. A submarine pressure hull in the form of a circular cylinder is of external Ø10m and length 200m. It dives to the bottom of the Mariana Trench which is 11.52km deep. What will be the mass of water acting on the submarine's surface in terms of London buses. p=1020kg/m3 g=9.81m/s2 London bus=7t I would be thankful if anyone could explain it to me. Thank you!
 Feb 29th 2016, 10:33 PM #2 Junior Member   Join Date: Feb 2016 Posts: 16 According to Archimedes, the force of buoyancy on a body equals to the weight of the fluid being replaced by the body Volume = Pi*(r^2)*L mass=Volume*1020 Number of bus= mass/7000
 Mar 1st 2016, 04:32 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,289 I wonder if this is actually how the problem is worded: "What will be the mass of water acting on the submarine's surface ..." To me this means calculating the mass of water that lies above the submarine - i.e. the length times width of the submarine times the depth under the surface of the water times the density of water. "
 Mar 1st 2016, 12:06 PM #4 Junior Member   Join Date: Feb 2016 Posts: 3 Solved That is how to do it: Area of submarine: 2π5*200=2000π=6283.19 m2 Pressure: P=1020*9.81*11520=115271424 Pa [1 N/m2] I would like to change it to kg instead of N so: 115271424:9.81=11750400 [kg/m2] Then: 6283.19*11750400=7.38*10^10 kg Instead od kg they asked me for London Buses so: 7.38*10^10:7000=10547142.25 London Buses
 Mar 1st 2016, 01:47 PM #5 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,289 Two comments: 1. You haven't included the area of the two ends of the submarine cylinder. 2. Be careful taking calculations to so many digits of accuracy. You use values for g of 9.81 (3 digits of accuracy), you're given the weight of a bus to only 1 digit, and you rounded off the surface area of the sub to 6283.19 m^2 (6 digits). Hence your answer is accurate to 3 digits at most, not ten digits. So 1.05 x 10^7 London busses is a better answer than 10547142.25 London Buses. Last edited by ChipB; Mar 1st 2016 at 02:25 PM.
Mar 1st 2016, 04:47 PM   #6
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 Originally Posted by ChipB Two comments: 1. You haven't included the area of the two ends of the submarine cylinder. 2. Be careful taking calculations to so many digits of accuracy. You use values for g of 9.81 (3 digits of accuracy), you're given the weight of a bus to only 1 digit, and you rounded off the surface area of the sub to 6283.19 m^2 (6 digits). Hence your answer is accurate to 3 digits at most, not ten digits. So 1.05 x 10^7 London busses is a better answer than 10547142.25 London Buses.
Yes I do know I haven't included the area of two ends but that is how my teacher explained it.

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