Author Archives: Engineering Math

Chemistry in Context| Essential Ideas| Chemistry| Openstax| Problem 6

According to one theory, the pressure of gas increases as its volume decreases, because the molecules in the gas have to move a shorter distance to hit the walls of the container. Does this theory follow a macroscopic or microscopic description of chemical behavior? Explain your answer.

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Chemistry in Context| Essential Ideas| Chemistry| Openstax| Problem 5

Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For any in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature.
(a) A certain molecule contains one H atom and one Cl atom.
(b) Copper wire has a density of about 8 g/cm3.
(c) The bottle contains 15 grams of Ni powder.
(d) A sulfur molecule is composed of eight sulfur atoms.

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Chemistry in Context| Essential Ideas| Chemistry| Openstax| Problem 4

Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For any in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature.
(a) The mass of a lead pipe is 14 lb.
(b) The mass of a certain chlorine atom is 35 amu.
(c) A bottle with a label that reads Al contains aluminum metal.
(d) Al is the symbol for an aluminum atom.

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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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College Physics by Openstax Chapter 2 Problem 48

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

Solution:

The known values are: y_0=2.20\:\text{m}; y=1.80\:\text{m}; v_0=11.0\:\text{m/s}; and a=-9.80\:\text{m/s}^2

We are going to use the formula

 \Delta y=v_0t+\frac{1}{2}at^2

Substituting the given values:

\begin{align*}
 \Delta y & =v_0t+\frac{1}{2}at^2 \\
1.80\:\text{m}-2.20\:\text{m} & =\left(11.0\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\
-0.40\:\text{m} & =\left(11.0\:\text{m/s}\right)t-\left(4.90\:\text{m/s}^2\right)t^2 \\
4.90t^2-11t-0.40 & =0
\end{align*}

Using the quadratic formula solve for t, we have

\begin{align*}
t & =\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)} \\
\end{align*}
 t=2.28\:\text{sec}\:\:\:\:\:\text{or}\:\:\:\:\:\:t=-0.04 \ \text{sec}

We can discard the negative time, so

t=2.28\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 2 Problem 47

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s.

(b) How long would it take to reach the ground if it is thrown straight down with the same speed?

Solution:

Part A

Refer to the figure below.

The known values are: t=2.35\:\text{s}; y=0\:\text{m}; v_0=+8.00\:\text{m/s}; and a=-9.8\:\text{m/s}^2

Based on the given values, the formula that we shall use is

y=y_0+v_0t+\frac{1}{2}at^2

Substituting the values, we have

\begin{align*}
y & =y_0+v_0t+\frac{1}{2}at^2 \\
0\: & =y_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2 \\
y_0 & =8.26\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Therefore, the cliff is 8.26 meters high.

Part B

Refer to the figure below

The knowns now are: y=0\:\text{m}; y_0=8.26\:\text{m}; v_0=-8.00\:\text{m/s}; and a=-9.80\:\text{m/s}^2

Based on the given values, we can use the formula 

y=y_0+v_0t+\frac{1}{2}at^2

Substituting the values, we have

\begin{align*}
y & =y_0+v_0t+\frac{1}{2}at^2 \\
0\:\text{m} & =8.26\:\text{m}+\left(-8.00\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\
4.9 t^2+8t-8.26 & =0 \\
\end{align*}

Using the quadratic formula to solve for the value of t, we have

\begin{align*}
t &=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)} \\
t &=0.717\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 46

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.

(a) How long are her feet in the air?

(b) What is her highest point above the board?

(c) What is her velocity when her feet hit the water?

Solution:

The known values are: y_0=1.80\:\text{m}, y=0\:\text{m}, a=-9.80\:\text{m/s}^2, v_0=4.00\:\text{m/s}.

Part A

Based from the knowns, the formula most applicable to solve for the time is \Delta y=v_0t+\frac{1}{2}at^2. If we rearrange the formula by solving for t, and substitute the given values, we have

\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a} \\
t & =\frac{-4.00\:\text{m/s}\pm \sqrt{\left(4.00\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.80\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t & =1.14\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We have the formula

\begin{align*}
\Delta y & =\frac{v^2-v_0^2}{2a} \\
\Delta y & =\frac{\left(0\:\text{m/s}\right)^2-\left(4.00\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)} \\
\Delta y & =0.816\: \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

The formula to be used is

v^2=v_0^2+2a\Delta y

Substituting the given values

\begin{align*}
v^2 & =v_0^2+2a\Delta y \\
v & =\pm \sqrt{v_0^2+2a\Delta y} \\
v & =\pm \sqrt{\left(4.00\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-1.80\:\text{m}\right)} \\
v & =\pm \sqrt{51.28\:\text{m}^2/\text{s}^2} \\
v &=\pm 7.16\:\text{m/s}
\end{align*}

Since the diver must be moving in the negative direction,

v=-7.16\:\text{m/s}  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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