How much absolute (total) pressure must a submarine withstand at a depth of 120m in the lake. Give answer in kPa. (Use 10m/s^2 for g).
h=120m
p_water = 1 g/cm^3 = 1000 kg/m^3
p_atmosphere = 101.3kPa
P_total = ?
P_water =density *g*h
=1000*10*120
=1200000Pa
=1200kPa
P_total =1200 kPa +101.3 kPa= 1301.3 kPa
Did I convert correctly for the density of water and from Pa to kPa?Thanks for your help.
Yes, you converted correctly. The total pressure that a submarine must withstand at a depth of 120m in the lake is 1301.3 kPa.
Well, it looks like you did your calculations correctly! However, I must say, that's a lot of pressure for a submarine to handle! They must really be under pressure to perform at that depth. Keep up the good work, and remember to always keep your answers as light as possible!
Yes, you have converted correctly. The density of water is commonly expressed in kg/m^3, which you used. And you correctly converted Pa (pascals) to kPa (kilopascals) by dividing by 1000. So your calculation of the absolute pressure the submarine must withstand at a depth of 120m in the lake is 1301.3 kPa. Well done!
Yes, you converted correctly for the density of water and from pascals (Pa) to kilopascals (kPa). The density of water is 1000 kg/m^3, and you correctly multiplied it by the acceleration due to gravity (g) and the depth (h) to find the pressure exerted by the water alone, which is 1200000 Pa or 1200 kPa.
Then, you added the atmospheric pressure (101.3 kPa) to the pressure exerted by the water to obtain the total absolute pressure that the submarine must withstand at a depth of 120 m in the lake, which is 1301.3 kPa.
So, your calculations and conversion were correct. Well done!