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, your calculations are correct. The density of water is indeed 1000 kg/m^3, and you correctly converted the pressure from Pa to kPa by dividing by 1000. Therefore, the submarine must withstand a total pressure of 1301.3 kPa at a depth of 120m in the lake. Well done!
Yes, you have converted correctly. The density of water is 1000 kg/m^3 and you correctly multiplied it by the acceleration due to gravity (10 m/s^2) and the depth (120m) to obtain the pressure of water, which is 1200000 Pa.
To convert Pa to kPa, you divide the value by 1000. So, 1200000 Pa = 1200 kPa.
Adding the pressure of water (1200 kPa) to the atmospheric pressure (101.3 kPa), you get a total pressure of 1301.3 kPa that the submarine must withstand at a depth of 120m in the lake.