Tag Archives: Solution Manual

Chemistry in Context| Essential Ideas| Chemistry| Openstax| Problem 3

Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.
(a) The pressure of a sample of gas is directly proportional to the temperature of the gas.
(b) Matter consists of tiny particles that can combine in specific ratios to form substances with specific properties.
(c) At a higher temperature, solids (such as salt or sugar) will dissolve better in water.

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Chemistry in Context| Essential Ideas| Chemistry| Openstax| Problem 2

Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.
(a) Falling barometric pressure precedes the onset of bad weather.
(b) All life on earth has evolved from a common, primitive organism through the process of natural selection.
(c) My truck’s gas mileage has dropped significantly, probably because it’s due for a tune-up.

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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 1

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PROBLEM:

Evaluate lim⁡x→2(x2−4x+3)\displaystyle \lim _{x\to 2}\left(x^2-4x+3\right).

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SOLUTION:

lim⁡x→2(x2−4x+3)=lim⁡x→2(x2)−lim⁡x→2(4x)+lim⁡x→2(3)=[lim⁡x→2(x)]2−4lim⁡x→2(x)+3=(2)2−4(2)+3=−1  (Answer)\begin{align*} \lim_{x\to 2}\left(x^2-4x+3\right)& = \lim_{x\to 2}\left(x^2\right)-\lim_{x\to 2}\left(4x\right)+\lim_{x\to 2}\left(3\right)\\ & =\left[\lim_{x\to 2}\left(x\right)\right]^2-4\lim_{x\to 2}\left(x\right)+3\\ & =\left(2\right)^2-4\left(2\right)+3\\ & =-1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

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PROBLEM:

If  f(x)=4x+3\displaystyle f\left(x\right)=\frac{4}{x+3} and  g(x)=x2−3\displaystyle \:g\left(x\right)=x^2-3 , find f[g(x)]\displaystyle f\left[g\left(x\right)\right] and g[f(x)]\displaystyle g\left[f\left(x\right)\right].

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SOLUTION:

Part A

f[g(x)]=4(x2−3)+3=4x2  (Answer)\begin{align*} f\left[g\left(x\right)\right] & =\frac{4}{\left(x^2-3\right)+3}\\ & =\frac{4}{x^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part B

g[f(x)]=(4x+3)2−3=16(x+3)2−3=16−3(x+3)2(x+3)2=16−3(x2+6x+9)(x+3)2=16−3x2−18x−27(x+3)2=−3x2−18x−11(x+3)2  (Answer)\begin{align*} g\left[f\left(x\right)\right] & =\left(\frac{4}{x+3}\right)^2-3\\ & =\frac{16}{\left(x+3\right)^2}-3\\ & =\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}\\ & =\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}\\ & =\frac{16-3x^2-18x-27}{\left(x+3\right)^2}\\ & =\frac{-3x^2-18x-11}{\left(x+3\right)^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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College Physics by Openstax Chapter 2 Problem 23

a) A light-rail commuter train accelerates at a rate of 1.35 m/s2. How long does it take to reach its top speed of 80.0 km/h, starting from rest?

b) The same train ordinarily decelerates at a rate of 1.65 m/s2. How long does it take to come to a stop from its top speed?

c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s2?

Solution:

Part A

We are given the following: a=1.35 m/s2a=1.35 \ \text{m/s}^2; vf=80.0 km/hv_f=80.0 \ \text{km/h}; and v0=0 m/sv_0=0 \ \text{m/s}.

From the formula vf=v0+atv_f=v_0+at, we can solve for tt as

t=vf−v0at=\frac{v_f-v_0}{a}

We need to convert 80.0 km/h to m/s first so that we have a unit uniformity for all the given values.

80 km/hr=(80 km/hr)(1000 m1 km)(1 hr3600 s)=22.2222 m/s\begin{align*} 80\:\text{km/hr} & =\left(80\:\text{km/hr}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right) \\ & =22.2222\:\text{m/s} \end{align*}

Substituting the given values into the formula, we have

t=vf−v0at=22.2222 m/s−0 m/s1.35 m/s2t=16.5 s  (Answer)\begin{align*} t & =\frac{v_f-v_0}{a} \\ t & =\frac{22.2222\:\text{m/s}-0\:\text{m/s}}{1.35\:\text{m/s}^2} \\ t & =16.5\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

For this problem, we still use the formula used in Part (a). This time, the values of the final and initial velocities interchange and the value of the given deceleration is negative of acceleration. The given values are a=−1.65 m/s2a=-1.65 \ \text{m/s}^2; v_0=22.2222 \ \text{m/s}[/katex]; and vf=0 m/sv_f=0 \ \text{m/s}

t=vf−v0at=0 m/s−22.2222 m/s−1.65 m/s2t=13.5 s  (Answer)\begin{align*} t & =\frac{v_f-v_0}{a} \\ t & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{-1.65\:\text{m/s}^2} \\ t & =13.5\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For this problem, we will use the formula

a=vf−v0Δt\begin{align*} a & =\frac{v_f-v_0}{\Delta t} \end{align*}

Substituting all the given values into the formula, we have

a=0 m/s−22.2222 m/s8.30 sa=2.68 m/s2  (Answer)\begin{align*} a & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{8.30\:\text{s}} \\ a & =2.68\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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