You are designing a diving bell to withstand the pressure of seawater at a depth of 250 m. (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.) (b) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 cm in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (Ignore the small variation of pressure over the surface of the window.)

Video transcript

Hey everyone in this problem. We have a deep sea water vessel that operates 2.9 km below the surface of sea water and were asked to determine the pressure gauge would record at the operating death, we're going to assume that the density of water remains constant as depth increases. So we want to find the pressure gauge would record. Okay, so let's call that P. G pressure, the gauge. We call that the pressure gauge is going to be the difference between the pressure at our depth age minus the pressure at the surface. K peanuts. So we have p minus peanut and recall, we know that P is equal to p not plus ro G H. Okay, So this tells us that the difference P minus peanut is equal to rho G. H where rho is the density GS acceleration due to gravity and h is the depth that we're at. Alright, so we're looking at density. Okay, we're talking about seawater here so we can look at the density of seawater paying a table in our textbook or that our professor has provided And we see that we have kilograms per meter cubed. The acceleration due to gravity. G is 9.8 m per second squared. Ok. And our depth is 2.9 kilometers. So we have 2.9 kilometers. Everything else here that involves distances looking at meters. So let's convert this into meters. So we're gonna multiply by 1000 m per kilometer. Okay, The unit of kilometer will divide out and then we're gonna be left with just meters. Okay, So from kilometers to go two m, we multiply by 1000. And if we work all of this out, we are going to find we get 29,272, pascal's. Okay? We have kilogram per meter cubed times meter per second squared times meter. So we end up with kilogram per meter second squared. So we have pascal and if we approximate this. Okay. Right. In scientific notation with three significant digits, we have 2.93 times 10 to the exponents seven pascal's. Alright. So if we look at our answer traces, we see that we have answer choice. D. Okay. The pressure that a gauge would read at the operating depth is 2. times 10 to the seven pascal's. Thanks everyone for watching. I hope this video helped see you in the next one.