Orders of Magnitude, Applications Section 3.5c

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Orders of Magnitude, Applications Section 3.5c. Orders of Magnitude. Order of Magnitude – the common logarithm of a positive quantity. Examples. 10. Mercury is about 5.79 x 10 meters from the Sun. 12. Pluto is about 5.9 x 10 meters from the Sun. - PowerPoint PPT Presentation

Transcript of Orders of Magnitude, Applications Section 3.5c

Page 1: Orders of Magnitude, Applications Section 3.5c
Page 2: Orders of Magnitude, Applications Section 3.5c

Orders of MagnitudeOrder of Magnitude – the common logarithm of a positivequantity.

Examples

Mercury is about 5.79 x 10 meters from the Sun10

Pluto is about 5.9 x 10 meters from the Sun12

Find the common logarithm of both numbers, and thencalculate the difference between these values…

Pluto’s distance from the Sun is 2 orders ofPluto’s distance from the Sun is 2 orders of magnitude greater than Mercury’smagnitude greater than Mercury’s

Page 3: Orders of Magnitude, Applications Section 3.5c

Orders of MagnitudeOrder of Magnitude – the common logarithm of a positivequantity.

Determine how many orders of magnitude the quantities differ.

1. A kilometer and a meter 332. A $1 dollar bill and a penny 223. A horse weighing 400 kg and a mouse weighing 40 g 44

4. NYC with 7 million people and Earmuff Junction with a population of 7 66

Examples

Page 4: Orders of Magnitude, Applications Section 3.5c

The level of sound intensity in decibels (dB) is

Application: DecibelsApplication: Decibels

0

10log II

where (beta) is the number of decibels, I is the soundintensity in W/m , and I = 10 W/m is the threshold ofhuman hearing (the quietest audible sound intensity).

2

0–12 2

Page 5: Orders of Magnitude, Applications Section 3.5c

The level of sound intensity in decibels (dB) is

Application: DecibelsApplication: Decibels

0

10log II

The sound of a subway train is 100 dB, and the sound

of a soft whisper is 10 dB. By how many orders ofmagnitude do these quantities differ?

11

0

10log 100II

22

0

10log 10II

We seek the logarithm of the ratio 1 2I I

Page 6: Orders of Magnitude, Applications Section 3.5c

Application: DecibelsApplication: Decibels1

10

10log 100II

22

0

10log 10II

1 21 2

0 0

10log 10log 100 10I II I

1

2

10log 90II

1

2

log 9II

The two sound intensitiesThe two sound intensities

differ by 9 orders of magnitude!!!differ by 9 orders of magnitude!!!

Page 7: Orders of Magnitude, Applications Section 3.5c

Application: Richter ScaleApplication: Richter ScaleThe Richter Scale magnitude R of an earthquake is

log aR BT

where a is the amplitude in micrometers of the vertical groundmotion at the receiving station, T is the period of the associatedseismic wave in seconds, and B accounts for the weakening ofthe seismic wave with increasing distance from the epicenterof the earthquake.

Page 8: Orders of Magnitude, Applications Section 3.5c

Application:Application:Richter ScaleRichter Scale

How many more times severe was the 2001 earthquake itGujarat, India (R = 7.9) than the 1999 earthquake in Athens,Greece (R = 5.9)?

log aR BT

12

11 log 7.9aR B

T 2

2 log 5.9aR BT

We seek the ratio of severities 1 2a a1 2

1 2 log loga aR R B BT T

Page 9: Orders of Magnitude, Applications Section 3.5c

Application:Application:Richter ScaleRichter Scale

log aR BT

1 21 2 log loga aR R B B

T T

1 2log log 7.9 5.9a aT T

1

2

log 2aa

21 2 10 100a a

A Richter scale difference of 2A Richter scale difference of 2corresponds to an amplitude ratiocorresponds to an amplitude ratio

of 100 of 100 The Gujarat quake was 100 The Gujarat quake was 100times as severe as the Athens quake!!!times as severe as the Athens quake!!!

Page 10: Orders of Magnitude, Applications Section 3.5c

Application:Application:Chemical AcidityChemical Acidity

The measure of acidity is pH, the opposite of the common logof the hydrogen-ion concentration of a solution:

pH log H Note: More acidic solutions have higher hydrogen-ionconcentrations, and therefore have lower pH values…

Page 11: Orders of Magnitude, Applications Section 3.5c

Application:Application:Chemical AcidityChemical Acidity

Some especially sour vinegar has pH of 2.4, and a box ofbaking soda has a pH of 8.4.

pH log H

(a) What are their hydrogen-ion concentrations?

Vinegar: log H 2.4 log H 2.4 2.4H 10

33.981 10 molesper liter

Baking Soda: log H 8.4 log H 8.4 8.4H 10

93.981 10 molesper liter

Page 12: Orders of Magnitude, Applications Section 3.5c

Application:Application:Chemical AcidityChemical Acidity

Some especially sour vinegar has pH of 2.4, and a box ofbaking soda has a pH of 8.4.

pH log H

(b) How many times greater is the hydrogen-ion concentration of the vinegar than that of the baking soda?

H of vinegar

H of baking soda

2.4

8.4

1010

610 timesgreater

Page 13: Orders of Magnitude, Applications Section 3.5c

Application:Application:Chemical AcidityChemical Acidity

Some especially sour vinegar has pH of 2.4, and a box ofbaking soda has a pH of 8.4.

pH log H

(c) By how many orders of magnitude do the concentrations differ?

The hydrogen-ion concentration of the vinegar is 6 ordersThe hydrogen-ion concentration of the vinegar is 6 ordersof magnitude greater than that of the baking soda…of magnitude greater than that of the baking soda…

This is exactly the difference in their pH values!!!This is exactly the difference in their pH values!!!

Page 14: Orders of Magnitude, Applications Section 3.5c

Application:Application:Newton’s Law of CoolingNewton’s Law of Cooling

An object that has been heated will cool to the temperature ofthe medium in which it is placed (such as the surrounding airor water). The temperature T of the object at time t can bemodeled by Newton’s Law of Cooling:

0kt

m mT t T T T e Temp. of thesurrounding

mediumInitial Temp.of the object

A constant

Page 15: Orders of Magnitude, Applications Section 3.5c

Application:Application:Newton’s Law of CoolingNewton’s Law of Cooling

A hard-boiled egg at temperature 96 C is placed in 16 C waterto cool. Four minutes later the temperature of the egg is 45 C.Use Newton’s Law of Cooling to determine when the egg willbe 20 C.

0kt

m mT t T T T e

Identify terms: 0 96T 16mT Plug into equation: 16 80 ktT t e Use the point (4, 45) to solve for k:

44 16 80 45kT e

Page 16: Orders of Magnitude, Applications Section 3.5c

Application:Application:Newton’s Law of CoolingNewton’s Law of Cooling

A hard-boiled egg at temperature 96 C is placed in 16 C waterto cool. Four minutes later the temperature of the egg is 45 C.Use Newton’s Law of Cooling to determine when the egg willbe 20 C.

0kt

m mT t T T T e

44 16 80 45kT e 4 29 80ke

4 ln 29 80k

ln 29 804

k

0.253 Save this value inyour calculator!!!

Page 17: Orders of Magnitude, Applications Section 3.5c

Application:Application:Newton’s Law of CoolingNewton’s Law of Cooling

A hard-boiled egg at temperature 96 C is placed in 16 C waterto cool. Four minutes later the temperature of the egg is 45 C.Use Newton’s Law of Cooling to determine when the egg willbe 20 C.

0kt

m mT t T T T e

20 16 80 kte

1 20kte

ln 1 20kt

Finally, solve this equation: ln 1 20

tk

11.809

The temp. of the egg will beThe temp. of the egg will be20 C after about 11.8 minutes20 C after about 11.8 minutes