Question

a 2404 lb force is applied to a surface that measures 70 in long and 16 in wide. determine the pressure exerted on the surface by the force in both psi and pa. 1 pa = 1.45×10⁻⁴ psi
psi
pa
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1. - / 4 points

calculate how much pressure (due to the liquid) would be on a 57 kg person if they were 248 m deep in...
a.)...freshwater. the density of freshwater is 1000 kg/m³. express your answer in both pascals and atmospheres. 1 atm = 1.013×10⁵ pa
pa
atm
b.)...saltwater. the density of saltwater is 1030 kg/m³. express your answer in both pascals and atmospheres. 1 atm = 1.013×10⁵ pa
pa
atm

Explanation:

First Problem (Pressure from Force on Surface)

Step1: Calculate the area of the surface

The surface is a rectangle, so area \( A = \text{length} \times \text{width} \). Given length \( l = 70 \) in and width \( w = 16 \) in, so \( A = 70 \times 16 = 1120 \) \( \text{in}^2 \).

Step2: Calculate pressure in psi

Pressure \( P = \frac{F}{A} \), where \( F = 2404 \) lb. So \( P_{\text{psi}} = \frac{2404}{1120} \approx 2.146 \) psi (rounded to three decimal places, but let's keep more for accuracy). Wait, actually \( 2404 \div 1120 = 2.14642857\ldots \approx 2.15 \)? Wait, no: \( 2404 \div 1120 \). Let's do it properly: \( 2404 \div 1120 = \frac{2404}{1120} = \frac{601}{280} \approx 2.146 \) psi.

Step3: Convert psi to Pa

Given \( 1 \) Pa \( = 1.45 \times 10^{-4} \) psi, so to convert psi to Pa, we use \( P_{\text{Pa}} = \frac{P_{\text{psi}}}{1.45 \times 10^{-4}} \). So \( P_{\text{Pa}} = \frac{2.14642857}{1.45 \times 10^{-4}} \approx \frac{2.14642857}{0.000145} \approx 14802.9556 \approx 1.48 \times 10^4 \) Pa? Wait, no: \( 2.14642857 \div 0.000145 = 2.14642857 \times \frac{1}{0.000145} = 2.14642857 \times 6896.5517 \approx 14800 \) Pa? Wait, actually the conversion factor from psi to Pa is \( 1 \) psi \( = 6894.76 \) Pa. Wait, the problem says \( 1 \) Pa \( = 1.45 \times 10^{-4} \) psi, so \( 1 \) psi \( = \frac{1}{1.45 \times 10^{-4}} \) Pa \( = \frac{10^4}{1.45} \approx 6896.55 \) Pa. So using that, \( P_{\text{Pa}} = 2.14642857 \times 6896.55 \approx 2.14642857 \times 6896.55 \approx 14800 \) Pa (exact calculation: \( 2.14642857 \times 6896.551724 \approx 2.14642857 \times 6896.55 \approx 14800 \) Pa).

Wait, let's redo Step2: Area is \( 70 \times 16 = 1120 \) in². Force is 2404 lb. So pressure in psi is \( P = F/A = 2404 / 1120 = 2.14642857 \) psi. Then convert to Pa: \( P_{\text{Pa}} = 2.14642857 \) psi \( \times (1 \text{ Pa} / 1.45 \times 10^{-4} \text{ psi}) = 2.14642857 / (1.45 \times 10^{-4}) \approx 14802.95 \) Pa, which is approximately \( 1.48 \times 10^4 \) Pa.

Step1: Recall the hydrostatic pressure formula

Hydrostatic pressure \( P =
ho g h \), where \(
ho \) is density, \( g = 9.8 \) m/s², \( h \) is depth.
Given \(
ho = 1000 \) kg/m³, \( h = 248 \) m, \( g = 9.8 \) m/s².

Step2: Calculate pressure in Pa

\( P =
ho g h = 1000 \times 9.8 \times 248 \).
First, \( 1000 \times 9.8 = 9800 \). Then \( 9800 \times 248 = 9800 \times (200 + 40 + 8) = 9800 \times 200 + 9800 \times 40 + 9800 \times 8 = 1,960,000 + 392,000 + 78,400 = 2,430,400 \) Pa.

Step3: Convert Pa to atm

Given \( 1 \) atm \( = 1.013 \times 10^5 \) Pa, so \( P_{\text{atm}} = \frac{2,430,400}{1.013 \times 10^5} \approx \frac{2.4304 \times 10^6}{1.013 \times 10^5} \approx 23.99 \approx 24.0 \) atm.

Second Problem (Sub - part b: Saltwater Pressure)

Step1: Hydrostatic pressure formula

\( P =
ho g h \), where \(
ho = 1030 \) kg/m³, \( h = 248 \) m, \( g = 9.8 \) m/s².

Step2: Calculate pressure in Pa

\( P = 1030 \times 9.8 \times 248 \).
First, \( 1030 \times 9.8 = 1030 \times (10 - 0.2) = 10300 - 206 = 10094 \). Then \( 10094 \times 248 \). Let's calculate: \( 10094 \times 200 = 2,018,800 \); \( 10094 \times 40 = 403,760 \); \( 10094 \times 8 = 80,752 \). Sum: \( 2,018,800 + 403,760 = 2,422,560 + 80,752 = 2,503,312 \) Pa.

Step3: Convert Pa to atm

\( P_{\text{atm}} = \frac{2,503,312}{1.013 \times 10^5} \approx \frac{2.503312 \times 10^6}{1.013 \times 10^5} \approx 24.71 \) atm.

Final Answers:

First Problem:

• psi: \( \approx 2.15 \) (or more accurately \( 2.146 \))
• Pa: \( \approx 1.48 \times 10^4 \) (or 14800)

Second Problem (a):

• Pa: \( 2.43 \times 10^6 \) (exact: 2430400)
• atm: \( \approx 24.0 \)

Second Problem (b):

• Pa: \( 2.50 \times 10^6 \) (exact: 2503312)
• atm: \( \approx 24.7 \)

Wait, let's recheck the first problem's pressure calculation. Area is 70*16=1120 in². Force is 2404 lb. So pressure in psi is 2404 / 1120 = 2.14642857 psi. Then converting to Pa: since 1 Pa = 1.45e-4 psi, so Pa = psi / (1.45e-4) = 2.14642857 / 0.000145 ≈ 14802.95 Pa, which is approximately 1.48×10⁴ Pa.

For the hydrostatic pressure (second problem), the formula is \( P =
ho g h \), where \(
ho \) is density, \( g = 9.8 \) m/s², \( h \) is depth. That's correct.

So summarizing:

1. First problem:

• psi: \( \frac{2404}{70 \times 16} = \frac{2404}{1120} \approx 2.146 \) psi
• Pa: \( 2.146 \div (1.45 \times 10^{-4}) \approx 14800 \) Pa

1. Second problem (a):

• Pa: \( 1000 \times 9.8 \times 248 = 2430400 \) Pa
• atm: \( 2430400 \div (1.013 \times 10^5) \approx 23.99 \approx 24.0 \) atm

1. Second problem (b):

• Pa: \( 1030 \times 9.8 \times 248 = 2503312 \) Pa
• atm: \( 2503312 \div (1.013 \times 10^5) \approx 24.71 \) atm

So the answers are:

First problem:

• psi: \( \boxed{2.15} \) (or more accurately \( \boxed{2.146} \))
• Pa: \( \boxed{14800} \) (or \( \boxed{1.48 \times 10^4} \))

Second problem (a):

• Pa: \( \boxed{2430400} \)
• atm: \( \boxed{24.0} \)

Second problem (b):

• Pa: \( \boxed{2503312} \)
• atm: \( \boxed{24.7} \)

(Note: Depending on rounding, the values might vary slightly. For example, using \( g = 9.81 \) m/s² would change the hydrostatic pressure slightly, but the problem doesn't specify, so 9.8 is standard.)

Answer:

(First Problem):
psi: \( \approx 2.15 \) (or more accurately \( 2.146 \))
Pa: \( \approx 1.48 \times 10^4 \) (or 14800)

Second Problem (Sub - part a: Freshwater Pressure)