# Chapter 1

## Homework Problems

7,26,45; 11,28,46; 54
## Solutions

**[7]**

We assume an uncertainty of 0.2 s.

(a) % uncertainty = [(0.2 s)/(5 s)] 100 = 4%.

Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.

(b) % uncertainty = [(0.2 s)/(50 s)] 100 = 0.4%.

Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.

(c) % uncertainty = [(0.2 s)/(5 min)(60 s/min)] 100 = 0.07%.

Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure.

**[26]**

We assume that a good runner can run 6 mi/h (equivalent to a 10-min mile) for 5 h/day. Using 3000 mi for the distance across the United States, we have

time = (3000 mi)/(6 mi/h)(5 h/day) ~ 100 days.

**[45]**

If we assume there is 1 automobile for 2 persons and a U. S. population of 250 million, we have 125 million automobiles. We estimate that each automobile travels 15,000 miles in a year and averages 20 mi/gal. Thus we have

N = (250x10^6 people/US)(1 auto/2 people)(15,000 mi/yr/auto)(1 gal/20 mi) ~ 1x10^11 gal/yr/US.

**[11]**

(Method 1)

We compare the volume with the specified radius to the volume for the extreme radius.

vol = 4/3 pi R^3 = 1.33 pi(2.86 m)^3 = 98.0 m^3;

Vol = 4/3 pi (R+r)^3 = 1.33 pi (2.86 m + 0.08 m)^3 = 106.45 m^3,

so the uncertainty in the volume is dV =Vol - vol = 8.5 m^3; and the % uncertainty is

% uncertainty = [(8.5 m^3)/(98.0 m^3)] 100 = 9%.

(Method 2)

The formula for the volume is Vol =4/3 pi R^3 so calculus gives us the derivative
as dVol = 4 pi R^2 dR. Then the relative error is dVol/Vol = 3 dR/R = 3(0.08m)/2.86m= 8%

**[28]**

My resting heart rate is 80 beats/min. (I know, that's terrible, but if I exercise it goes up.)
The US average lifespan is 72 years, but for me, it will probably be 65. Using these numbers,

N = (80 beats/min)(65 yr)(365 day/yr)(24 h/day)(60 min/h) ~3e9

**[46]**

We let D represent the diameter of a gumball. Because there are air gaps around the gumballs, we estimate the volume occupied by a gumball as a cube with volume D3 . The machine has a square cross-section with sides equivalent to 10 gumballs and is about 14 gumballs high, so we have

N = volume of machine/volume of gumball = (14D)(10D)(10D)/D^3 ~1.4e3 gumballs.

**[54]**

We assume the oil slick is circular with diameter D and thickness d (like a very flat can, a right
circular cylinder). For the volume we have

V = d(pi r^2) = d pi/4 D^2;

Solving for the diameter D, we have

D = (4 V /d pi)^1/2 = [4 (1000 cm^3) / 2e-10m(1e2cm/m) pi]^1/2 ~ 3e5 cm. (or 3e3m, 3 km)

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