Module 1.1 Exercises

From 'Fundamentals of Physics (12th ed.) book

Introduction

This is my attempt to understand Physics from the very basics. The first chapter was Measurement and this journal entry will cover the exercises from this chapter.

My math isn’t mathing and I admit that I have a rather poor math skill. As such, I asked Google’s Gemini to review and correct my solutions.

Problem 1

Earth is approximately a sphere and the radius is approximately $ 6.37 \times 10^6 \space m $. What is the (a) circumference in kilometer, (b) surface area in kilometer squared, and (c) volume in cubic kilometer?

My attempt

Circumference of the Earth:

$ C = 2 \pi R $

$ C = (2) \times (3.14) \times (6.37 \times 10^6 \space m) $

$ C = (6.28) \times (6.37 \times 10^6 \space m) $

$ C = 40 \times 10^6 \space m $

$C = 4.0 \times 10^7 \space m$

$C = 4.0 \times 10^7 \space m \left(\frac{1 \space km}{1000 \space m}\right)$

$\boxed{C = 4.0 \times 10^4 \space km}$

Surface Area of the Earth:

$A = 4 \times \pi R^2$

$A = 4 \times 3.14 \times (6.37 \times 10^6 \space m)^2$

$A = 12.56 \times 40.5769 \times 10^8 \space m^2$

$A = 50964586400 \space m^2$

$A = 50964586400 \space m^2 \times \frac{1 \space km}{1000 \space m}$

$A = 50964586,40 \space m \times 1 \space km$

$A = 50964586,40 \space m \times \space km$

$A = 50964586,40 \times km \times \frac{1 \space km}{1000 \space m}$

$A = 50964586,40 \times km \times km$

$\boxed{A = 5,09 \times 10^7 \space km^2}$ (?)

$ V = \frac{4}{3} \times \pi \times R^3 $

$ V = \frac{12.56}{3} \times (6.37 \times 10^6m)^3 $

$ V = 418.66 \times (6.37 \times 10^6m)^3 $

$ V = 418.66 \times 2584748,53 \times 10^14m^3 $

$ V = 108213081956,98 \times 10^9m^3 $

$ V = 108213081956,98 \times 10^9m^3 \times \frac{1 km}{1000 m} $

$ V = 108213081956,98 \times 10^6m^2 \times km $

$ V = 108213081956,98 \times 10^6m^2 \times km \times \frac{1 km}{1000 m} \times \frac{1 km}{1000m} $

$ V = 108213081956,98 \times 10^(6-6) \times km \times km \times km $

$ V = 108213081956,98 km^3 $

$ V = 1,08 \times 10^11 km^3 $ (?)

Review by Gemini AI

I showed my attempt abovementioned to Google’s Gemini AI and it corrected me on the unit conversion. In my attempt, I used meters in the calculation and employed conversion at the end of the calculation. This is considered inefficient by Gemini. It then suggested me to use unit conversion at the beginning of the calculation process instead of at the end. Also, since the questions are asking for kilometers, Gemini further emphasized that I should convert from meter to kilometer.

This is a new insight for me and I think it is logical and cool. I found it quite hard to calculate properly when I convert from meter to kilometer at the end of the calculation. I made repeated mistake here and there. So I tried to do what Gemini suggested and it worked like a charm. Below is my second attempt into exercising Measurement (unit conversion).

Second Attempt

$ R_{earth} = 6.37 \times 10^6 \space m $

$ R_{earth} = 6.37 \times 10^{(6-3)} \space km $

$ R_{earth} = 6.37 \times 10^3 \space km $

Circumference

$ C_{earth} = 2 \pi R $

$ C_{earth} = (2) (3.14) (6.37 \times 10^3 \space km) $

$ C_{earth} = (6.28) (6.37 \times 10^3 \space km) $

$ C_{earth} = 40.0036 \times 10^3 \space km $

$ C_{earth} = 40.00 \times 10^3 \space km $ (round the decimal into two places)

$ C_{earth} = 4.00 \times 10^4 \space km $ (move decimal place by one to the left and increase the magnitude by 1)

\[\boxed{C_{earth} = 4.00 \times 10^4 \space km}\]

Surface Area

$ A_{earth} = 4 \pi R^2 $

$ A_{earth} = (4) (3.14) (6.37 \times 10^3 \space km)^2 $

$ A_{earth} = (12.56) (6.37)^2 \times 10^{(3+2)} \space km^2 $

$ A_{earth} = (12.56) (40.57) \times 10^5 \space km^2 $

$ A_{earth} = 509.55 \times 10^5 \space km^2 $

$ A_{earth} = 5.09 \times 10^{(5+2)} \space km^2 $ (unifying with format in the question. One significant and two decimal places)

\[\boxed{A_{earth} = 5.09 \times 10^7 \space km^2}\]

Volume

$ V_{earth} = \frac{4}{3} \pi R^3 $

$ V_{earth} = (\frac{4}{3}) (3.14) (6.37 \times 10^3 \space km)^3 $

$ V_{earth} = (4.18) (6.37)^3 \times 10^{(3+3)} \space km^3 $

$ V_{earth} = (4.18) (257.23) \times 10^6 \space km^3 $

$ V_{earth} = 1075.2214 \times 10^6 \space km^3 $

$ V_{earth} = 1.0752214 \times 10^{(6+3)} \space km^3 $ (unifying the format to that of the question)

$ V_{earth} = 1.07 \times 10^9 \space km^3 $ (then round the decimal to two places)

\[\boxed{V_{earth} = 1.07 \times 10^9 \space km^3}\]

Problem 2

$ 1 \space gry = \frac{1}{10} \space line $

$ 1 \space line = \frac{1}{12} \space inch $

$ 1 \space point = \frac{1}{72} \space inch $

How much is $0.5 \space gry^2$ in $points^2$?

First Attempt

In essence, we need to convert all of the ancient units into one uniform unit. Since the question asked in points, I think it would be good to convert everything into points unit.

$ 1 \space gry = \frac{10 \space gry}{1 \space line} $

$ \frac{1 \space \cancel{gry}}{\cancel{gry}} = (\frac{10 \space \bcancel{gry}}{1 \space line}) \div \bcancel{gry} $

$ \frac{1 \space gry}{gry} = \frac{10 \space gry}{1 \space line} $