College Physics Problem 1.28


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.


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College Physics Problem 1.27


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.


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College Physics Problem 1.26


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?


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College Physics Problem 1.25


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.


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College Physics Problem 1.24


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?


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College Physics Problem 1.22


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


Solution:

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

\displaystyle A=\pi r^2

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

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

So, by direct substitution

\displaystyle A=\pi \left(\frac{3.102\:cm}{2}\right)^2=7.557\:cm^2          ☚

The area of the circle is 7.557 square centimetre.


College Physics Problem 1.21


A person measures his or her heart rate by counting the number of beats in 30 s. If 40±1  beats are counted in 30±0.5 s, what is the heart rate and its uncertainty in beats per minute?


Solution:

In order to compute for the heart rate in beats per minute, we need to solve for the base. The base is

\displaystyle \frac{40\:beats}{30\:sec\:}\times \frac{60\:sec}{1\:min}=80\:beats/min

Then we compute for the percent uncertainty by combining the uncertainties of the number of beats and time. That is

\displaystyle \%\:uncertainty=\frac{1\:beat}{40\:beats}\times 100\%+\frac{0.5\:s}{30.0\:s}\times 100\%

\displaystyle \%\:uncertainty=2.5\%+1.7\%

\displaystyle \%\:uncertainty=4.2\%

Based on this percent uncertainty, we compute for the tolerance

\displaystyle \delta _A=\frac{\%\:uncertainty}{100\:\%}\times A

\displaystyle \delta _A=3.3\:beats/min

Therefore, the heart rate is

\displaystyle 80\pm 3\:beats/min          ☚


College Physics Problem 1.20


(a) A person’s blood pressure is measured to be 120±2 mmHg. What is its percent uncertainty?

(b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mmHg?


Solution:

Part a

The percent uncertainty is computed as

\displaystyle \%\:uncertainty=\frac{2\:mmHg}{120\:mmHg}\times 100\%=1.7\%

Part b

The uncertainty in the blood pressure is

\displaystyle \delta _{bp}=\frac{1.7\:\%}{100\:\%}\times 80\:mmHg=1.3\:mmHg


College Physics Problem 1.19


(a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h, what is the percent uncertainty?

(b) If it has the same percent uncertainty when it reads 60 km/ h, what is the range of speeds you could be going?


Solution:

Part a

The percent uncertainty is computed as

\displaystyle \%\:uncertainty=\frac{2.0\:km/hr}{90\:km/hr}\times 100\%=2.2\%

Part b

The tolerance of the velocity is

\displaystyle \delta _v=\frac{2.2\:\%}{100\:\%}\times 60\:km/hr=1\:km/hr

Therefore, the range of the velocity is 60±1km/h, or that is 59 to 61 km/h.