Category Archives: Physics

Problem 1-31: The lifetime of a human vs the mean life of an extremely unstable atomic nucleus

Advertisements
Advertisements

PROBLEM:

How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of  10-22 s .)

Advertisements
Advertisements

SOLUTION:

The lifetime of a human is 2×109 s, while the lifetime of an unstable atomic nucleus is 10-22 s.

Therefore, a human lifetime is much longer by

ThTn=2×109 sec10−22 sec=2×1031 times  (Answer) \frac{T_h}{T_n}=\frac{2\times 10^9\:\text{sec}}{10^{-22}\:\text{sec}}=2\times 10^{31}\:\text{times}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-30: The number of generations passed since the year 0 AD

Advertisements
Advertisements

PROBLEM:

A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?

Advertisements
Advertisements

SOLUTION:

We are looking for generations passed in history. The general assumptions are:

  • There are 1011 seconds in 1 history
  • In 1 generation, there is 1/3 of a lifetime.
  • In half a lifetime, there are 109 seconds

Therefore, the number of generations passed is

1 history=1 history×1011 sec1 history×1 generation13 lifetime×0.5 lifetime109 sec=150 generations  (Answer)\begin{align*} 1\ \text{history} & =1\:\text{history}\times \frac{10^{11}\:\text{sec}}{1\:\text{history}}\times \frac{1\:\text{generation}}{\frac{1}{3}\:\text{lifetime}}\times \frac{0.5\:\text{lifetime}}{10^9\:\text{sec}} \\ & =150\:\text{generations} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Advertisements
Advertisements

Problem 1-29: The number of heartbeats in a lifetime

Advertisements
Advertisements

PROBLEM:

How many heartbeats are there in a lifetime?

Advertisements
Advertisements

SOLUTION:

The general assumptions:

  • There is 1 heartbeat in 1 second
  • In half a lifetime, there are 109 seconds.

Therefore, in a lifetime, the number of heartbeats is

1 lifetime=(1 lifetime)(109 sec0.5 lifetime)(1 heartbeat1 sec)=2×109 heartbeats  (Answer)\begin{align*} 1\:\text{lifetime} & =\left(1\:\text{lifetime}\right)\left(\frac{10^9\:\text{sec}}{0.5\:\text{lifetime}}\right)\left(\frac{1\:\text{heartbeat}}{1\:\text{sec}}\right) \\ & =2\times 10^9\:\text{heartbeats} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Advertisements
Advertisements

Problem 1-28: Calculating the volume and its uncertainty of a car piston with dimensional uncertainties

Advertisements
Advertisements

PROBLEM:

A car engine moves a piston with a circular cross section of  7.500±0.002 cm diameter a distance of 3.250±0.001 cm  to compress the gas in the cylinder.

(a) By what amount is the gas decreased in volume in cubic centimeters?

(b) Find the uncertainty in this volume.

Advertisements
Advertisements

SOLUTION:

Part A

The average volume is

V=πr2h=π(7.5 cm2)2(3.25 cm)=143.5806 cm3  (Answer)\begin{align*} V & =\pi r^2h \\ & =\pi \left(\frac{7.5\:\text{cm}}{2}\right)^2\left(3.25\:\text{cm}\right) \\ & =143.5806\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part B

Solve for the percent uncertainties of each dimension

% uncr=0.002 cm7.500 cm×100%=0.027%% unch=0.001 cm3.25 cm×100%=0.031%\begin{align*} \%\:unc_r & =\frac{0.002\:\text{cm}}{7.500\:\text{cm}}\times 100\%=0.027\% \\ \%\:unc_h & =\frac{0.001\:\text{cm}}{3.25\:\text{cm}}\times 100\%=0.031\% \\ \end{align*}

The percent uncertainty in the volume is the combined effect of the uncertainties of the dimensions

% uncvol=0.027%+0.031%=0.058%\text{\%\:unc}_{vol}=0.027\%+0.031\%=0.058\%

The uncertainty in the volume is

δvol=0.058100×143.5806=0.083 cm3  (Answer) \delta _{vol}=\frac{0.058}{100}\times 143.5806=0.083\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-27: Calculating the area and its uncertainty of a room with a given dimensional uncertainties

Advertisements
Advertisements

PROBLEM:

The length and width of a rectangular room are measured to be 3.955 ±0.005 m and 3.050 ± 0.005 m . Calculate the area of the room and its uncertainty in square meters.

Advertisements
Advertisements

SOLUTION:

The average area of the room is

A=l×w=3.955 m×3.050 m=12.06 m2  (Answer)\begin{align*} A & =l\times w \\ & =3.955\:\text{m}\times 3.050\:\text{m} \\ & =12.06\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Compute for the percent uncertainties of each dimension.

% uncwidth=0.005 m3.050 m×100%=0.1639%% unclength=0.005 m3.955 m×100%=0.1264 %\begin{align*} \text{\%\:unc}_{width} & =\frac{0.005\:\text{m}}{3.050\:\text{m}}\times 100\%=0.1639\% \\ \text{\%\:unc}_{length} & =\frac{0.005\:\text{m}}{3.955\:\text{m}}\times 100\%=0.1264\:\% \end{align*}

The percent uncertainty in the area is the combined effect of the uncertainties of the length and width.

% uncarea=0.1639%+0.1264%=0.2903%\text{\%\:unc}_{area}=0.1639\%+0.1264\%=0.2903\%

The uncertainty in the area is

δarea=0.2903 %100 %×12.06 m2=0.035 m2  (Answer)\delta _{area}=\frac{0.2903\:\%}{100\:\%}\times 12.06\:\text{m}^2=0.035\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the area is

A=12.06±0.035 m2  (Answer)A=12.06\pm 0.035\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-26: Uncertainty and percent uncertainty of a pound-mass (lbm) unit

Advertisements
Advertisements

PROBLEM:

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1 lbm=0.4539 kg.

(a) If there is an uncertainty of 0.0001 kg in the pound-mass unit, what is its percent uncertainty?

(b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?

Advertisements
Advertisements

SOLUTION:

Part A

The percent uncertainty of the lbm is

% uncertaintylbm=0.0001 kg0.4539 kg×100%=0.022%  (Answer)\text{\%\:uncertainty}_{\text{lbm}}=\frac{0.0001\:\text{kg}}{0.4539\:\text{kg}}\times 100\%=0.022\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

For uncertainty of 1 kg, the corresponding lbm is

lbm=δlbm% uncertaintylbm×100%=1 kg0.02 %×1 lbm0.04539 kg×100%=11015.64 lbm  (Answer)\begin{align*} \text{lbm} & =\frac{\delta_{\text{lbm}}}{\text{\%\:uncertainty}_{\text{lbm}}}\times 100\% \\ \\ & =\frac{1\:\text{kg}}{0.02\:\%}\times \frac{1\:\text{lbm}}{0.04539\:\text{kg}}\times 100\% \\ \\ & =11015.64\:\text{lbm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Advertisements
Advertisements

Problem 1-25: Solving for the uncertainty of the volume of a box with dimensional uncertainties

Advertisements
Advertisements

PROBLEM:

The sides of a small rectangular box are measured to be 1.80±0.01 cm, 2.05±0.02 cm, and 3.1±0.1 cm long. Calculate its volume and uncertainty in cubic centimeters.

Advertisements
Advertisements

SOLUTION:

The average volume of the box is

Volume=l×w×h=(1.80 cm)(2.05 cm)(3.1 cm)=11.4 cm3  (Answer)\begin{align*} \text{Volume} & =l\times w\times h \\ & =\left(1.80\:\text{cm}\right)\left(2.05\:\text{cm}\right)\left(3.1\:\text{cm}\right) \\ &= 11.4\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

The percent uncertainty for each of the dimensions:

1.80±0.01 cm → 0.01 cm1.80 cm×100%=0.556%2.05±0.02 cm → 0.02 cm2.05 cm×100%=0.976%3.1±0.1 cm → 0.1 cm3.1 cm×100%=3.226%\begin{align*} 1.80\pm 0.01\:\text{cm}\:\rightarrow & \:\frac{0.01\:\text{cm}}{1.80\:\text{cm}}\times 100\%=0.556\% \\ \\ 2.05\pm 0.02\:\text{cm}\:\rightarrow & \:\frac{0.02\:\text{cm}}{2.05\:\text{cm}}\times 100\%=0.976\% \\ \\ 3.1\pm 0.1\:\text{cm}\:\rightarrow &\:\frac{0.1\:\text{cm}}{3.1\:\text{cm}}\times 100\%=3.226\% \\ \end{align*}

The percent uncertainty in the volume of the box is calculated by adding the percent uncertainties of the dimensions.

% uncertaintyvolume=0.556%+0.976%+3.226%=4.758%\begin{align*} \%\:\text{uncertainty}_{\text{volume}} & =0.556\%+0.976\%+3.226\% \\ & =4.758\% \end{align*}

The uncertainty of the volume is

δvolume=0.04758×11.4 cm3=0.54 cm3  (Answer)\begin{align*} \delta _{\text{volume}} & =0.04758\times 11.4\:\text{cm}^3 \\ & =0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the volume is

Volume=11.4±0.54 cm3  (Answer)\text{Volume}=11.4\pm 0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-24: Calculating uncertainties in distance, time and speed of a marathon runner

Advertisements
Advertisements

PROBLEM:

A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time.

(a) Calculate the percent uncertainty in the distance.

(b) Calculate the uncertainty in the elapsed time.

(c) What is the average speed in meters per second?

(d) What is the uncertainty in the average speed?

Advertisements
Advertisements

SOLUTION:

Part A

The percent uncertainty in the distance is

% uncertaintydistance=25 m42.188 km×1 km1000 m×100%=0.0593%  (Answer)\begin{align*} \text{\%\:uncertainty}_{\text{distance}} & =\frac{25\:\text{m}}{42.188\:\text{km}}\times \frac{1\:\text{km}}{1000\:\text{m}}\times 100\% \\ & =0.0593\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part B

The uncertainty in time is

% uncertaintytime=1 s9012 s×100%=0.0111%  (Answer)\begin{align*} \text{\%\:uncertainty}_{\text{time}} & =\frac{1\:\text{s}}{9012\:\text{s}}\times 100\% \\ & =0.0111\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part C

The average speed is

average speed=42.188 km9012 s×1000 m1 km=4.681 m/s  (Answer)\begin{align*} \text{average speed} & =\frac{42.188\:\text{km}}{9012\:\text{s}}\times \frac{1000\:\text{m}}{1\:\text{km}} \\ & = 4.681\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part D

The percent uncertainty in the speed is the combination of uncertainties of distance and time.

% uncertaintyspeed=% uncertaintydistance+% uncertaintytime=0.0593%+0.0111%=0.0704%\begin{align*} \text{\%\:uncertainty}_{\text{speed}} & =\text{\%\:uncertainty}_{\text{distance}}+\text{\%\:uncertainty}_{\text{time}} \\ & =0.0593\%+0.0111\% \\ & =0.0704\% \\ \end{align*}

Therefore, the uncertainty in the speed is

δspeed=0.0704%100%×4.681 m/s=0.003 m/s  (Answer)\begin{align*} \delta _{speed} & =\frac{0.0704\%}{100\%}\times 4.681\:\text{m/s} \\ & = 0.003\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
Advertisements
Advertisements

Problem 1-23: Calculating for the time it takes a marathon runner to finish 26.22 miles given a speed of 9.5 mi/h

Advertisements
Advertisements

PROBLEM:

If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?

Advertisements
Advertisements

SOLUTION:

The relationship of the velocity (or speed), time, and distance is

v=dtv=\frac{d}{t}

So, by rearranging the equation, we can equate time to

t=dvt=\frac{d}{v}

By direct substitution

t=26.22 mi9.5 mi/hr=2.76 hours  (Answer)t=\frac{26.22\:\text{mi}}{9.5\:\text{mi/hr}}=2.76\:\text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-22: Solving for the area of a circle with a given diameter

Advertisements
Advertisements

PROBLEM:

What is the area of a circle 3.102 cm in diameter?

Advertisements
Advertisements

SOLUTION:

The area of a circle can be computed using the formula below when the radius is given.

A=πr2A=\pi r^2

We also know that the radius is half the diameter, so the area can be calculated using the formula,

A=π(d2)2A=\pi \left(\frac{d}{2}\right)^2

So, by direct substitution

A=π(3.102 cm2)2=7.557 cm2  (Answer)A=\pi \left(\frac{3.102\:\text{cm}}{2}\right)^2=7.557\:\text{cm}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The area of the circle is 7.557 square centimeters.

Advertisements
Advertisements