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 submarine that operates at a depth of 450 m below the surface of seawater. The submarine has a square glass window of length centimeters. Were asked to determine the net force on the window from the ocean water and air inside the submarine were told to assume that the air pressure inside the submarine is equal to the pressure at the surface of the ocean and that pressure remains constant on the entire window surface, the density of water remains constant as depth increases. Alright, so we're trying to find a net force and let's recall that the force F. Is equal to the pressure P times area. A. So when we're talking about the force on the window were given the dimensions of the window and so we can find the area A. We just need to find the pressure. So let's look at the net pressure p net and the net pressure on the window is going to be the pressure from the ocean pushing in on the window minus the pressure from the inside. Air pushing out on the window. So from the outside we have the ocean pushing in on the window. From the inside we have that air pushing out and so the net pressure is going to be the difference between those two. Now this is going to be equal to p minus P. Not okay. Where p ocean is this value P that we're looking at and P not is going to be equal to P inside because we're told that the air pressure is equal to the pressure at the surface of the ocean. Okay. So the pressure inside is just gonna be this p not which is the value of the pressure at the surface. And we know we can recall that P minus P. Not. Well this is equivalent to row times G times H. So we have the density times the gravitational acceleration times the depth. No the density of seawater. Okay. We're in seawater, we're talking about the density of seawater that's going to be kilograms per meter. Cute. And this is a value that you can look up in a table in your textbook that your professor provided the density of different liquids here. We're talking seawater. The gravitational acceleration will be 9.8 m/s squared. And the depth we're at is 450 m. And we can use these values because of some of these assumptions. Okay, we're told that the density of water remains constant as the depth increases. So we can use that constant um density. And again we already used that information about the pressure inside being equal to the pressure at the surface. And if we work this out, we're going to have four million 542,300. Our unit we have kilogram per meter cubed times meter per second squared times meter. So we get kilogram per meter second squared. And this is going to be equal to a pascal. So we found our p nets or pressure. Let's find the area in order to calculate our force F. So the area a were told that this is a square window. So it's just going to be equal to the length squared. The length we're given is 20 cm. Okay, so 20 cm times one m per 100 centimeters. Okay, So we divide by 100 to convert this into meters. This is gonna give us 0.2 m squared. And so we get an area of 0. m squared for that window. Now we have our pressure, we have our area and we can calculate our net force. So the net force F net is going to be equal to our net pressure times our area. Which is equal to 4,542, time zero point and we missed the unit. There we have pascal's Times the area of the window. 0.04 m squared. And this will give us a net force of 181, pascal meters square. Okay, I recall that pascal meter squared is the is a unit of newton. And so if we take three significant digits we get that. This is 1.82 times 10 to the five newtons. That is the net force. Alright, so we go back up to our answer choices. We see that we have answer choice. C The net force on the window is going to be 1.82 times 10 to the five newtons. Thanks everyone for watching. I hope this video helped see you in the next one.