Tag Archives: College Physics

Problem 1-15: Solving for the number of heartbeats of a person with the correct significant figure

Advertisements
Advertisements

PROBLEM:

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y?
(b) In 2.00 y?
(c) In 2.000 y?

Advertisements
Advertisements

SOLUTION:

Part a

This is a problem on rounding off. Take note that 72.0 has three significant figures, 2.0 has two significant figures, 2.00 has three significant figures, and 2.000 has four significant figures.

Convert 72 beats/min to beats/year.

\begin{align*}
72 \ \text{beats/min} & = \left(\frac{72.0\:\text{beats}}{1\:\bcancel{\text{min}}}\right)\left(\frac{60.0\:\bcancel{\text{min}}}{1.00\:\bcancel{\text{hr}}}\right)\left(\frac{24.0\:\bcancel{\text{hr}}}{1.00\:\bcancel{\text{day}}}\right)\left(\frac{365.25\:\bcancel{\text{days}}}{1\:\text{year}}\right)\\
& =3.7869\times 10^7\:\text{beats/year}
\end{align*}

In 2 years, the number of beats of an average person is

\begin{align*}
& = 3.7869 \times 10^{7} \ \text{beats/year} \times 2 \ \text{years} \\
& = 7.5738 \times 10^7 \ \text{beats} 
\end{align*}

For part a, the answer is limited to 2.0 which is 2 significant figures. The answer is 7.6 \times 10^{7} \ \text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part b

For item letter b), the answer is limited by 2.00, which is 3 significant figures. The answer is 7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Part c

For item letter c), the answer is limited by 72.0, which is 3 significant figures. The answer is  7.57\times 10^7\:\text{beats} \ \color{DarkOrange} \left( \text{Answer} \right).

Advertisements
Advertisements

Problem 1-14: Solving for the percent uncertainty of 130±5 beats/min

Advertisements
Advertisements

PROBLEM:

An infant’s pulse rate is measured to be 130±5 beats/min. What is the percent uncertainty in this measurement?

Advertisements
Advertisements

SOLUTION:

The percent uncertainty can be calculated as

\begin{align*}
\%\:\text{uncertainty}\:& =\frac{\delta _A}{A}\times 100\% \\
&  =\frac{5\:\text{beats/min}}{130\:\text{beats/min}}\times 100\% \\
& =3.85\:\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Therefore, the percent uncertainty is 3.85%.

Advertisements
Advertisements

Problem 1-13: Computing for the range of possible speeds given 90 km/h and 5% uncertainty

Advertisements
Advertisements

PROBLEM:

(a) A car speedometer has a  5.0% uncertainty. What is the range of possible speeds when it reads 90 km/h?

(b) Convert this range to miles per hour. (1 km=0.6214 mi)

Advertisements
Advertisements

SOLUTION:

Part A

The uncertainty in the velocity of the car is computed as

\begin{align*}
\delta _v & =\frac{5.0\:\%}{100\:\%}\times 90.0\:\text{km/hr} \\
& = 4.5\:\text{km/hr}
\end{align*}

Therefore, the range of the possible speeds is 

\begin{align*}
\text{Range} & =90.0\:\pm 4.5\:\text{km/hr} \\
\text{Range} & :85.8\:\text{km/hr}\:-\:94.5\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

So the range of the possible speeds is 85.5 km/hr to 94.5 km/hr.

Part B

Convert the range to mi/h

For 85.5 km/hr

\begin{align*}
85.5\:\text{km/hr}=\left(85\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=53.13\:\text{mi/hr}
\end{align*}

For 94.5 km/hr

94.5\:\text{km/hr}=\left(94.5\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=58.72\:\text{mi/hr}

Therefore, the range can be represented as 53.13 mi/hr to 58.72 mi/hr.

\text{Range} : 53.13 \ \text{mi/hr} - 58.72 \ \text{mi/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-12: Solving for the percent uncertainty of a 20-m tape that is off by 0.50 cm

Advertisements
Advertisements

PROBLEM:

A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?

Advertisements
Advertisements

SOLUTION:

The percent uncertainty can be calculated by the uncertainty divided by the total measure, but the units should be the same. That is,

\begin{align*}
\%\:\text{uncertainty} & =\frac{\delta _l}{l}\times 100\% \\
\\
& =\frac{0.5\:\text{cm}}{20\:\text{m}\left(\frac{100\:\text{cm}}{1\:\text{m}}\right)}\times 100\% \\
\\
& =\frac{0.5\:\text{cm}}{2000\:\text{cm}}\times 100\% \\
\\
&=0.025\:\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the percent uncertainty is 0.025%.

Advertisements
Advertisements

Problem 1-11: Solving for the uncertainty given mass and percent uncertainty

Advertisements
Advertisements

PROBLEM:

Suppose that your bathroom scale reads your mass as 65 kg with a 3%  uncertainty. What is the uncertainty in your mass (in kilograms)?

Advertisements
Advertisements

SOLUTION:

The uncertainty is computed as 

\delta _m=\frac{3\:\%}{100\:\%}\times 65\:\text{kg}=1.95\:\text{kg} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the uncertainty in the given mass is 1.95 kg.

Advertisements
Advertisements

Problem 1-10: The average speed of the earth in its orbit in kilometers per second and in meters per second

Advertisements
Advertisements

PROBLEM:

a) The average distance between the Earth and the Sun is  108 km. Calculate the average speed of the Earth in its orbit in kilometers per second. 

b) What is this is meters per second?

Advertisements
Advertisements

SOLUTION:

Part A

We assume that the earth revolves around the sun in a circular manner. Therefore, the distance between the earth and the sun will be the radius of its orbit.

The total distance traveled the earth in full revolution is

\begin{align*}
d & =2\pi r \\
& = 2\pi \left(10\right)^8\:\text{km} \\
\end{align*}

The total time of travel is

\begin{align*}
t & =365.25\:\text{days}\left(\frac{24\:\text{hr}}{1\:\text{day}}\right)\left(\frac{3600\:\text{s}}{1\:\text{hr}}\right) \\
& = 3.15576\times 10^7\:\text{sec} \\
\end{align*}

Therefore, the average speed is

\begin{align*}
\text{speed} & =\frac{\text{distance}}{\text{time}}=\frac{d}{t} \\
 & = \frac{2\pi \left(10\right)^8\:\text{km}}{3.15576\times 10^7\:\text{s}} \\
& = 19.91\:\text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the average speed of the earth is 19.91 km/s.

Part B

Convert 19.91 km/s to m/s.

\begin{align*}
s & =\left(19.91\:\frac{\text{km}}{\text{s}}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right) \\
& =19\,910\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Therefore, the velocity is 19 910 m/s. 

Advertisements
Advertisements

Problem 1-9: The distance moved in 1 sec and the speed in kilometers per million years of tectonic plates

Advertisements
Advertisements

PROBLEM:

Tectonic plates are large segments of the Earth’s crust that move slowly. Suppose that one such plate has an average speed of 4.0 cm/year.
(a) What distance does it move in 1.0 s at this speed?
(b) What is its speed in kilometers per million years?

Advertisements
Advertisements

SOLUTION:

Part A

We know that the formula for the distance traveled, d, if we are given rate, r, and time, t is given by d=r\times t.

The rate in cm/s is 

\begin{align*}
r & =\left(4\:\frac{\text{cm}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{year}}}{365\:\frac{1}{4}\:\bcancel{\text{days}}}\times \frac{1\:\bcancel{\text{day}}}{24\:\bcancel{\text{hours}}}\times \frac{1\:\bcancel{\text{hour}}}{60\:\bcancel{\text{mins}}}\times \frac{1\:\bcancel{\text{min}}}{60\:\text{s}}\right) \\
& = 1.2675 \times 10^{-7} \  \text{cm/s}
\end{align*}

The distance traveled is

\begin{align*}
d & = r \times t \\
d & = \left( 1.2675\times 10^{-7} \ \text{cm/s} \right)\left( 1.0  \ \text{s} \right) \\
d & = 1.3\times 10^{-7} \ \text{cm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We need to convert 4.0 cm/year to km/My.

\begin{align*}
4.0 \ \frac{\text{cm}}{\text{year}} & = 4.0\:\frac{\bcancel{\text{cm}}}{\bcancel{\text{year}}}\times \frac{1\:\bcancel{\text{m}}}{100\:\bcancel{\text{cm}}}\times \frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\times \frac{1\:000\:000\:\bcancel{\text{years}}}{1\:\text{My}} \\ \\
& =40\:\text{km/My} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the speed of the tectonic plates is 40 km/My. 

Advertisements
Advertisements

Problem 1-8: The speed of sound in km/h

Advertisements
Advertisements

PROBLEM:

The speed of sound is measured to be  342 m/s on a certain day. What is this in km/h?

Advertisements
Advertisements

SOLUTION:

We know that 1 km=1000 m and 1 hr=3600 sec.

Convert 342 m/s.

\begin{align*}
342\:\text{m/s} & =\left(342\:\frac{\bcancel{\text{m}}}{\bcancel{\text{s}}}\right)\left(\frac{1\:\text{km}}{1000\:\bcancel{\text{m}}}\right)\left(\frac{3600\:\bcancel{\text{s}}}{1\:\text{hr}}\right) \\
& =1231.2\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the speed of sound in km/hr is 1231.2 km/hr.

Advertisements
Advertisements

Problem 1-7: The height of Mount Everest in kilometers

Advertisements
Advertisements

PROBLEM:

Mount Everest, at 29,028 feet, is the tallest mountain on the Earth. What is its height in kilometers? (Assume that 1 kilometer equals 3,281 feet.)

Advertisements
Advertisements

SOLUTION:

Convert 29028 ft to km

29,028\:\text{ft}=\left(29\:028\:\text{ft}\right)\left(\frac{1\:\text{km}}{3281\:\text{ft}}\right)=8.847\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the height of Mount Everest is 8.847 kilometers.

Advertisements
Advertisements

Problem 1-6: The height of a 6 ft 1 in tall person in meters

Advertisements
Advertisements

PROBLEM:

What is the height in meters of a person who is 6 ft 1.0 in. tall? (Assume that 1 meter equals 39.37 in.)

Advertisements
Advertisements

SOLUTION:

First, convert 6 ft to inches

6\:\text{ft}=\left(6\:\text{ft}\right)\left(\frac{12\:\text{in}}{1\:\text{ft}}\right)=72\:\text{in}

Add the 1.0 inch

72\:\text{in}\:+1\:\text{in}=73\:\text{inches}

So, the total height of the person is 73 inches. We convert this to meters to come up with the desired unit.

73\:\text{in}=\left(73\:\text{in}\right)\left(\frac{1\:\text{m}}{39.37\:\text{in}}\right)=1.85\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

So, the height of the person is 1.85 meters.

Advertisements
Advertisements