Unit 3B

Unit 3B

Putting Numbers Into Perspective

Scale of the Universe 2

http://sciencenetlinks.com/tools/scale-universe-2/

Open Ended

29–32: Perspective Through Estimation. Use estimation to make the following comparisons. Discuss your conclusion.

32. Which is greater: the amount you spend in a year on transportation or the amount you spend in a year on food?

Fill in the Blanks

39. How many average candy bars would you have to eat to ­supply the energy needed for 8 hours of running?

•  $\frac{1\ candy\ bar}{1\ \times10^6\ joules}$  

•  $\frac{1\ hour\ of\ running}{4\ \times10^6\ joules}$  

•  $\frac{1\ candy\ bar}{1\ \times10^6\ joules}\cdot\frac{4\times10^6\ joules}{1\ hour\ run}=\frac{4\ candy\ bars}{\text{1 hour run}}$  

•  $\frac{4\ candy\ bar}{1\ hour\ run}\cdot\ 8\ hours\ run\ =\ 32\ candy\ bars$  

Fill in the Blanks

40. How many liters of oil are required to supply the electrical energy needs of an average home for a year?

•  $\frac{1\ liter\ of\ oil}{1.2\times10^7\ joules}\cdot\frac{5\times10^7joules}{1\ house\ \left(daily\right)}$  

•  $\frac{5\ liters\ of\ oil}{1.2\ houses\ \left(daily\right)}=4.16666666\ liters\ per\ house\ daily$  

• 4.16666 liters per day x 365 days = 1520.83 liters

Fill in the Blanks

43. If you could generate energy by fusing the hydrogen in water, how much water would you need to generate the electrical energy used daily by 10,000 typical homes? (Round to the nearest thousandths.)

•  $\frac{5\times10^7\ joules}{1\ house}\cdot10,000\ house=5\times10^{11}\ joules\ needed$  

•  $5\times10^{11}\ joules\ \cdot\frac{1\ liter\ of\ water}{6.9\times10^{13}\ joules}$  

•  $\frac{5}{6.9}\cdot10^{11-13}\ joules\ =0.724\times10^{-2}\ joules$  

• 0.007 joules

55. There were approximately 2.69 million deaths in the United States in 2015. Express this quantity in deaths per minute.

•  $\frac{2,690,000\ deaths}{1\ year}\cdot\frac{1\ years}{365\ days}\cdot\frac{1\ day}{24\ hours}\cdot\frac{1\ hour}{60\ \min}\approx5.11796\ deaths\ per\ \min$  

a. Assume that an average cell has a diameter of 6 micro­meters (6×10^−6  meter), which means it has a volume of 100 cubic micrometers. How many cells are there in a cubic centimeter?

•  $1cm\times1cm\times1cm\times\frac{1\ meter}{10^2\ cm}\times\frac{1\ meter}{10^2\ cm}\times\frac{1\ meter}{10^2\ cm}$  

•  $\frac{1\ meter\times meter\times meter}{10^6}\times\frac{1\ micrometer}{10^{-6}\ meter}\times\frac{1\ micrometer}{10^{-6}\ meter}\times\frac{1\ micrometer}{10^{-6}\ meter}$  

•  $\frac{1\ cubic\ micrometers}{10^{-12}}\times\frac{1\ cell}{100\ cubic\ micrometers}$  

•  $\frac{1}{10^{-10}}\ cells\ =\ 10^{10}\ cells$  

b. Estimate the number of cells in a liter, using the fact that a cubic centimeter equals a milliliter. (1 cubic cm is 10^10 cells)

 $1\ liter\ \times\frac{1,000\ milliliters}{1\ liter}\times\frac{1\ cubic\ centimeter}{1\ milliliter}\times\frac{10^{10}\ cells}{1\ cubic\ centimeter}$  

c. Estimate the number of cells in a 70-kilogram (154-pound) person, assuming that the human body is 100% water (actually it is about 60–70% water) and that 1 liter of water weighs 1 kilogram.

 $70\ kg\ \times\frac{1\ liter}{1\ kg}\times\frac{10^{13}\ cells}{1\ liter}$  

4.5 trillion gallons—could supply all U.S. households for five months.(Assume 100 million households.)

4.5 trillion gallons / 5 months = 0.9 trillion gallons per month

 $\frac{900,000,000,000\ gallons}{100,000,000\ houses}=9,000\ gallons\ per\ house$