Physical Quantities and Units
Accuracy, Precision, and Significant Figures
Approximation
You can browse on the itemized questions with solutions of the College Physics by Openstax below. Also, you can buy the whole Complete Solution Manual here.
Chapter 14: Heat and Heat Transfer Methods
Chapter 15: Thermodynamics
Chapter 16: Oscillatory Motion and Waves
Chapter 17: Physics of Hearing
Chapter 18:
Electric Charge and Electric Field
Chapter 19:
Electric Potential and Electric Field
Chapter 20:
Electric Current, Resistance, and Ohm’s Law
Chapter 21: Circuits and DC Instruments
Chapter 22:
Magnetism
Chapter 23:
Electromagnetic Induction, AC Circuits, and Electrical Technologies
Chapter 24:
Electromagnetic Waves
Chapter 25: Geometric Optics
Chapter 26: Vision and Optical Instrument
Chapter 27: Wave Optics
Chapter 28: Special Relativity
Chapter 29: Introduction to Quantum Physics
Chapter 30: Atomic Physics
Chapter 31: Radioactivity and Nuclear Physics
Chapter 32: Medical Applications of Nuclear Physics
Chapter 33:
Particle Physics
Chapter 34: Frontiers of Physics
A straight substitution of x=0 leads to the indeterminate form 0\cdot 0 which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows
\begin{align*}
\lim\limits_{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)\right) & =\lim\limits_{x\to 0}\left(\frac{1\cdot \:\left(\frac{1}{3}-\frac{1}{\sqrt{x+9}}\right)}{x}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\frac{\sqrt{x+9}-3}{3\sqrt{x+9}}}{x}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{\sqrt{x+9}-3}{3x\sqrt{x+9}}\right)\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} \\ \\
& =\lim\limits_{x\to 0}\left(\frac{x}{3x\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\
& =\lim\limits_{x\to 0}\left(\frac{1}{3\left(\sqrt{x+9}+3\right)\sqrt{x+9}}\right) \\ \\
& =\frac{1}{3\left(\sqrt{0+9}+3\right)\sqrt{0+9}} \\ \\
& =\frac{1}{54} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Solution:
A straight substitution of x=3 leads to the indeterminate form \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows.
\begin{align*}
\begin{align*}
\lim\limits_{x\to \:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right) & =\lim\limits_{x\to \:\:3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)\cdot \frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}} \\
& =\lim\limits_{x\to 3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(\sqrt{x-2}-\sqrt{4-x}\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}\right] \\
& =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2x-6}\right]\\
& =\lim\limits_{x\to3}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{-x+4}\right)}{2\left(x-3\right)}\right] \\
& =\lim\limits_{x\to 3}\left[\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\right]\\
& =\frac{\sqrt{3-2}+\sqrt{4-3}}{2}\\
& =1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
\end{align*}SOLUTION:
A straight substitution of x=2 leads to the indeterminate form \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows
\begin{align*}
\lim\limits_{x\to 2}\left(\frac{x^3-8}{x^2-4\:}\right) & =\lim\limits_{x\to 2}\left[\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x-2\right)}\right] \\
&=\lim\limits_{x\to 2}\left[\frac{\left(x^2+2x+4\right)}{\left(x+2\right)}\right] \\
& =\frac{2^2+2\cdot 2+4}{2+2} \\
& =3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}A straight substitution of x=4 leads to the indeterminate form \frac{0}{0} which is meaningless.
Therefore, to evaluate the limit of the given function, we proceed as follows:
\begin{align*}
\\
\lim\limits_{x\to 4}\left(\frac{\frac{1}{x}-\frac{1}{4}}{x-4}\right)& =\lim\limits_{x\to 4}\left(\frac{\frac{4-x}{4x}}{x-4}\right)\\
\\
& =\lim\limits_{x\to 4}\frac{4-x}{4x\left(x-4\right)}\\
\\
&=\lim\limits_{x\to 4}\left(\frac{4-x}{-4x\left(4-x\right)}\right)\\
\\
& =\lim\limits_{x\to 4}-\frac{1}{4x}\\
\\
& =-\frac{1}{4\cdot 4}\\
\\
& =-\frac{1}{16} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\\
\end{align*}According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 20 pieces of cotton fiber is taken and the absorbency on each piece was measured. The following are the absorbency values:
| 18.71 | 21.41 | 20.72 | 21.81 | 19.29 | 22.43 | 20.17 |
| 23.71 | 19.44 | 20.50 | 18.92 | 20.33 | 23.00 | 22.85 |
| 19.25 | 21.77 | 22.11 | 19.77 | 18.04 | 21.12 |
a. Calculate the sample mean and median for the above sample values.
b. Compute the 10% trimmed mean.
c. Do a dot plot of the absorbency data.
The formula for the total distance traveled is
\Delta s=\Delta \theta \times r
Therefore, the total distance traveled is
\begin{align*}
\Delta s & =\left(200000\:\text{rotations}\:\times \frac{2\pi \:\text{radian}}{1\:\text{rotation}}\right)\left(\frac{1.15\:\text{m}}{2}\right) \\
\Delta s & =722566.3103\:\text{m} \\
\Delta s & =722.6\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Solution:
The formula for friction is
f=\mu _{k\:}NWhen we solve for the normal force, N, in terms of the other variables, we have
N=\frac{f}{\mu _k}The coefficient of kinetic friction is 0.04. Therefore, the normal force is
\begin{align*}
N & =\frac{f}{\mu _k} \\
N & =\frac{0.200\:\text{newton}}{0.04} \\
N & =5.00\:\text{newton} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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