Question

As an object's depth below the surface of a body of salt water increases, so does the pressure acting on the object due to atmospheric and water conditions. the rate at which pressure increases is approximately 111111 pounds per square inch (\text{psi})(psi)left parenthesis, start text, p, s, i, end text, right parenthesis for every increase in depth of 252525 feet (\text{ft})(ft)left parenthesis, start text, f, t, end text, right parenthesis. the pressure at the surface of the water is 15\ \text{psi}15 psi15, space, start text, p, s, i, end text. rounded to the nearest foot, at what depth will the pressure acting on the object be 50\ \text{psi}50 psi50, space, start text, p, s, i, end text?

Answer 1

There appears to be an error in the rate of pressure increase provided in the question. Assuming a corrected rate of 1 psi per 25 ft, the required depth to reach 50 psi is 875 feet.

Step-by-step explanation:

The student is asking about the relationship between an object's depth in salt water and the pressure exerted on it. We know the pressure at the surface (15 psi) and the rate at which the pressure increases with depth (approximately 111111 psi for every 25 feet). To find the depth at which the pressure acting on the object is 50 psi, we first determine the pressure difference between the target and initial pressures, which is 50 psi - 15 psi = 35 psi.

Next, we calculate how many 25-foot increments are needed to account for the 35 psi increase by dividing the pressure difference by the rate. So we divide 35 psi by the rate of 111111 psi/25 ft, which gives us:

35 psi / (111111 psi/25 ft) = 0.007875 ft ≈ 0.7875 ft (rounded to three decimal places).

Since 0.7875 is much less than one increment of 25 ft, there seems to be a typo or error in the rate given. Assuming a correct and reasonable rate, such as an increase of 1 psi for every 25 ft, we would do the following:

35 psi / (1 psi/25 ft) = 875 ft

Therefore, the depth at which the pressure is 50 psi would be 875 feet below the surface, rounded to the nearest foot.

Answer 2

To calculate the required depth for a pressure of 50 psi, we increment the starting pressure of 15 psi by the rate given until we reach 50 psi. Due to a very high rate of increase (111,111 psi per 25 feet), any increase in depth results in an immediate exceeding of 50 psi, suggesting a potential error in the question's data.

Step-by-step explanation:

To calculate the depth at which the pressure acting on the object in salt water will be 50 psi, we need to account for the initial pressure at the surface and the incremental increase in pressure with depth. The pressure at the surface is given as 15 psi, and we are told that for every 25 feet of depth, the pressure increases by 111,111 psi.

Finding the depth is a matter of determining how many increments of 25 feet are needed for the pressure to go from 15 psi to 50 psi. The difference in pressure is 50 psi - 15 psi = 35 psi. We'll need to divide this difference by the rate of increase in pressure per 25 feet, which is 111,111 psi, to get the number of increments:

Increments = 35 psi / 111,111 psi = 0.000315 increments (since we're only looking for a single step increase, we're expecting a fractional result). Now, we multiply the number of increments by 25 feet to get the depth:

Depth = 0.000315 increments × 25 ft/increment = 0.007875 ft ≈ 0.01 feet

Since we want to provide the depth rounded to the nearest foot, we round 0.01 feet to 0 feet. Therefore, at the surface of the water (0 feet depth), the pressure is already 15 psi, and since the increase rate is far too high in the given condition (111,111 psi per 25 feet), the smallest increase in depth would immediately exceed our target pressure of 50 psi. It appears there may be a typo or misunderstanding in the rate of pressure increase as mentioned in the question since it isn't practical or physically plausible.

Conditions such as the water's density and the presence of atmospheric pressure influence pressure increases with depth, and these must be accounted for when considering real-life applications.