### Physics Homework#### 1. Calculate the pressure at the bottom of the Narragansett Bay, which is 56 m below sea level at its deepest point.A) $ 5.63094 \times 10^4 $ Pa B) $ 5.63094 \times 10^5 $ Pa C) $ 5.63094 \times 10^6 $ Pa D) $ 5.63094 \times 10^7 $ Pa E) $ 5.63094 \times 10^8 $ Pa F) $ 5.63094 \times 10^9 $ Pa #### 2. The pressure at the bottom of an average spot in the Narragansett Bay is only about $ 7.800 \times 10^4 $ Pa because most of the Bay is not as deep at its deepest spot. Calculate the average depth below sea level. *The answer is a little surprising - feel free to look this up if you want to check your result, but understand I rounded the pressure quoted above so the correct answer doesn’t agree exactly with what you’ll find.*A) 7.765 m B) 10.61 m C) 7.951 m D) 7.757 m E) 12.10 m F) 7.959 m #### 3. The point on Earth farthest below sea level (that we know about) is under the Pacific Ocean in the Mariana Trench at a depth of 11.034 km. Calculate the pressure there.A) $ 1.0813 \times 10^8 $ Pa B) $ 1.0824 \times 10^8 $ Pa C) $ 5.6309 \times 10^7 $ Pa D) $ 1.0662 \times 10^8 $ Pa E) $ 1.1084 \times 10^8 $ Pa F) $ 1.1095 \times 10^8 $ Pa ### The Physics of Deep-Sea Fish and Pressure#### Concept: Extreme Pressure and Deep-Sea Fish**4) Some fish that evolved to live at extreme depths essentially explode if they are brought to the surface. This happens for the same reason that you would essentially implode if you tried to go down to visit them without a very special submarine.**Let's approximate the cross-sectional area of a fish (as viewed from above) that might live in the Mariana Trench to be a rectangle of $2 \, \text{cm} \times 10 \, \text{cm} = 20 \, \text{cm}^2 = 2 \times 10^{-3} \, \text{m}^2$. Calculate the force this fish would feel pushing down on it due to the ocean water above it.### Multiple-Choice Question:A) $8 \times 10^9 \, \text{Pa}$B) $8 \times 10^{11} \, \text{Pa}$C) $6 \times 10^{12} \, \text{Pa}$D) $8 \times 10^{14} \, \text{Pa}$E) $9 \times 10^{11} \, \text{Pa}$F) $1 \times 10^{11} \, \text{Pa}$### Explanation:As an aside, even though this result is mind-boggling, the truth is even greater than this (for two reasons)! To perform this calculation, we assume the gravitational acceleration is constantly $9.81 \, \text{m/s}^2$.While this is a fine assumption most of the time, as we have seen in the course, the value of $g$ actually changes as you go farther from the surface of the Earth. At this depth, its magnitude would be greater than $9.81$, so the pressure calculation underestimates the actual pressure. For the same reason, and the fact that fluids aren’t technically incompressible, the density increases a bit as you go deeper into the ocean. Both these small errors in our approximations mean the pressure and thus the force here is larger than what this calculation shows.

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Description: ### Physics Homework#### 1. Calculate the pressure at the bottom of the Narragansett Bay, which is 56 m below sea level at its deepest point.A) $ 5.63094 \times 10^4 $ Pa B) $ 5.63094 \times 10^5 $ Pa C) $ 5.63094 \times 10^6 $ Pa D) \( 5.6

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1) Calculate the pressure at the bottom of the Narragansett Bay, which is 56 m below sea level at its deepest point. A) 5.63094 × 10ª Pa B) 5.63094 × 105 Pa C) 5.63094 × 106 Pa D) 5.63094 × 10' Pa E) 5.63094 × 10® Pa F) 5.63094 x 10° Pa

1) Calculate the pressure at the bottom of the Narragansett Bay, which is 56 m below sea level at its deepest point. A) 5.63094 × 10ª Pa B) 5.63094 × 105 Pa C) 5.63094 × 106 Pa D) 5.63094 × 10' Pa E) 5.63094 × 10® Pa F) 5.63094 x 10° Pa

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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