Formulas/Models

Section 1.3 Formulas/Models

Many principles of sciences and other fields can be expressed using mathematical notation. These are often called formulas or mathematical models. In this section we will practice reading formulas and calculating values from them.

When we have a formula and we know enough of the values we can calculate others. Note in the following examples how the provided values are inserted into the formula before performing the arithmetic.

Example 1.3.1.

Ohm’s Law relates three properties of electricity: voltage, current, and resistance. Voltage, measured in volts (V), is analagous to the amount of pressure to move the electrons. Current, measured in amperes (amps), is how much electricity is moving. Resistance measured in Ohms (\(\Omega\)), is, as it sounds, the resitance of a material to letting electricity flow.

The relationship is

\begin{equation*} V = IR \end{equation*}

where \(V\) is voltage, \(I\) is current, and \(R\) is the resistance.

If we know that the current is 3 amps and the resistance is 8 ohms then we can calculate

\begin{equation*} V = 3 \cdot 8 = 24 \end{equation*}

Thus in this system there must be 24 volts.

Similarly if we know that the current is 1.7 amps and the resistance is 6 ohms, then we can calculate

\begin{equation*} V = 1.7 \cdot 6 = 10.2 \end{equation*}

Thus in this system there must be 10.2 volts.

Example 1.3.2.

The lift equation relates lift (\(L\)), air density (\(\rho\)), surface area of wing (\(s\)), something called the coefficient of lift (\(C_L\)), and velocity (\(v\)). Lift is the force that keeps aircraft in the air. Air density is the amount of air per volume; you may see this as highs and lows on a weather map and is related to pressurizing high flying aircraft. The lift equation is

\begin{equation*} L = \frac{1}{2}\rho s C_L v^2. \end{equation*}

If we know that air density is 0.002378 slugs per cubic feet, surface area is 125 ft², \(C_L=1.5617\text{,}\) and velocity is 84.4 ^ft⁄_s, we can calculate the lift.

\begin{equation*} L=\frac{1}{2} \cdot 0.002378 \cdot 125 \cdot 1.5617 \cdot 84.4^2 \approx 1653. \end{equation*}

Example 1.3.3.

The ideal gas law is relationship between volume, pressure, and temperature.

\begin{equation*} \frac{P_1 V_1}{T_1+273} = \frac{P_2 V_2}{T_2+273}. \end{equation*}

\(P_1\text{,}\) \(V_1\text{,}\) and \(T_1\) are the initial pressure, volume, and temperature. \(P_2\text{,}\) \(V_2\text{,}\) and \(T_2\) are pressure, volume, and temperature at another time. Temperature is in degrees Celsius. Pressure and volume can be in any SI units but must be the same on both sides of the equation.

Suppose the initial conditions are \(P_1=101.3 \text{ Pa}\text{,}\) \(V_1=0.125 \text{ m}^3\text{,}\) and \(T_1=10.2^\circ \text{ C}\text{.}\) Also \(V_2=0.125 \text{ m}^3\) and \(T_2=50.7^\circ \text{ C}\text{.}\) We can calculate the new pressure.

\begin{align*} \frac{101.3 \cdot 0.125}{10.2+273} \amp = \frac{P_2 \cdot 0.125}{50.7+273}\\ 0.0447 \amp \approx 0.000386 P_2\\ \frac{0.0447}{0.000386} \amp \approx \frac{0.000386 P_2}{0.000386}\\ 115.8 \amp \approx P_2 \end{align*}

Checkpoint 1.3.4.

Calculate the voltage if the current is \(I=3 \text{ amps}\text{,}\) \(R=4 \Omega\text{?}\)

Answer.

\(12\)

Solution.

Using \(V=IR\text{:}\) \(V = 3 \cdot 4 = 12\) volts.

Checkpoint 1.3.5.

Using the ideal gas law to calculate the new pressure (\(P_2\)) if \(P_1=101.3\text{,}\) \(V_1=0.250\text{,}\) \(T_1=11.3\text{,}\) and \(V_2=0.250\text{,}\) \(T_2=14.5\text{?}\)

Answer.

\(102.4\)

Solution.

\begin{equation*} \begin{aligned} \frac{101.3 \cdot 0.250}{11.3+273} \amp = \frac{P_2 \cdot 0.250}{14.5+273}\\ 0.0891 \amp \approx 0.000870 P_2\\ \frac{0.0891}{0.000870} \amp; \approx \frac{0.000870 P_2}{0.000870}\\ 102.4 \amp \approx P_2 \end{aligned} \end{equation*}

Checkpoint 1.3.6.

Using the ideal gas law calculate the new volume (\(V_2\)) if \(P_1=101.3\text{,}\) \(V_1=0.250\text{,}\) \(T_1=11.3\text{,}\) and \(P_2=98.7\text{,}\) \(T_2=14.5\text{?}\)

Answer.

\(0.0259\)

Solution.

\begin{equation*} \begin{aligned} \frac{101.3 \cdot 0.250}{11.3+273} \amp = \frac{98.7 \cdot V_2}{14.5+273}\\ 0.0891 \amp \approx 0.343 P_2\\ \frac{0.0891}{0.343} \amp \approx \frac{0.343 P_2}{0.343}\\ 0.0259 \amp \approx P_2 \end{aligned} \end{equation*}

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