Water pressure vs. area of an object

[ebeck1]

I've been looking at this question for about 1 hour now, and cannot figure out the relationship between the distance an object is at the bottom of the ocean vs. its area... The question is:
A solid copper ball with a diameter of 3.20 m at sea level is placed at the bottom of the ocean, at a depth of 7.0 km. If the density of the seawater is 1030 kg/m3, how much does the diameter of the ball decrease when it reaches bottom?

I used the equation P=Po+pgh to solve for the pressure at the bottom of the ocean but cannot find a relation of initial pressure and initial are to final pressure and final area...

 

[Pyrrhus]

This problem is obviously about volume elasticity, don't they give you the bulk modulus? look for tables in your textbook about bulk modulus of copper and its alloys.

 

[kyle8921]

I can provide some insight into this question. The relationship between water pressure and the area of an object is a fundamental concept in fluid mechanics. When an object is submerged in water, it experiences a force due to the weight of the water above it. This force is directly proportional to the area of the object. This means that the larger the area of the object, the greater the force it experiences from the water pressure.

In your question, the solid copper ball has a larger area compared to a smaller object, such as a sphere with a diameter of 1 meter. This means that the copper ball will experience a greater force from the water pressure at the bottom of the ocean compared to the smaller object. This can be seen in the equation you used, where the pressure (P) is directly proportional to the density of the seawater (p), the acceleration due to gravity (g), and the height (h) of the water column.

Now, as the copper ball descends to the bottom of the ocean, it will experience an increase in pressure due to the increasing height of the water column. This increase in pressure will cause the ball to compress, resulting in a decrease in its diameter. The amount of decrease in diameter can be calculated using the equation for compressibility, which takes into account the initial diameter, the change in pressure, and the bulk modulus of the material (in this case, copper).

In summary, the relationship between water pressure and the area of an object is directly proportional, and this can be seen in the equation you used. As the object descends to the bottom of the ocean, the increase in water pressure will cause it to compress, resulting in a decrease in its diameter. This decrease can be calculated using the equation for compressibility. I hope this helps to clarify the relationship between water pressure and the area of an object in this scenario.