PLEASE HELP ASAP!!!!!!!!!!!!! I dont understand this stuff at all.........


We know that standard pressure is one atmosphere, or 760 millimeters of mercury. This pressure results from the weight of gas molecules in the atmosphere. As a diver enters the water, he is subject to both water pressure and air pressure. Because water is much denser than air, the pressure increases rapidly as the diver descends. At the depth of 34 feet in fresh water, the diver is experiencing 2 atmospheres of pressure (one from air pressure and one from the 34 feet of water). For every additional 34 feet the diver descends he will be under an additional atmosphere of pressure. Since water pressure is proportional to depth, how many atmospheres of pressure would a diver experience at 102 feet? Why wouldn't this pressure squash the diver? Answering this second question may be easier if you think of the reason a person on land is not squashed by one atmosphere of pressure. Explain your answer in detail.

Respuesta :

At the depth of 34 feet in fresh water, the diver is experiencing 2 atmospheres of pressure (one from air pressure and one from the 34 feet of water). For every additional 34 feet the diver descends he will be under an additional atmosphere of pressure. Since water pressure is proportional to depth, how many atmospheres of pressure would a diver experience at 102 feet?


depth             pressure

34 feet           1 atm + 1 atm = 2 atm

2*34 feet        1 atm + 2 atm = 3 atm

3*34 feet        1 atm + 3 atm = 4 atm

=> pressure = 1 + n/34, where n is the number of feet under water.

102 feet => pressure = 1 atm + 102/34 atm = 1 atm + 3 atm= 4 atm

About the second question, the lungs will be at the same pressure than the external pressure, and the blood will also be pumped at higher pressure to the internal organs and so the balance is kept.

Why wouldn't this pressure squash the diver?

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