Problem 49 - Elementary Algebra

Question

Use numerical evaluation on the equations. \(E=m c^{2} . \quad\) Find \(E\) if \(m=2\) and \(c=186,000\)

Step-by-Step Solution

Answer

Answer: 69,192,000,000 Joules

Step 1: Write down the given equation and variables

The equation given is: \(E=mc^2\). We are given the values of mass, \(m=2\) and the speed of light, \(c=186,000\) miles per second. We need to find the energy, \(E\).

Step 2: Plug the given values into the equation

Replace the variables \(m\) and \(c\) in the equation with their given values: \(E=2 \cdot (186,000)^2\)

Step 3: Evaluate the expression

Calculate the square of \(c\) and then multiply it by \(m\) to find the value of \(E\): \(186,000^2 = 34,596,000,000\) Now multiply this by \(m=2\): \(E=2 \cdot 34,596,000,000\)

Step 4: Calculate E

Finish calculating the value of \(E\): \(E=69,192,000,000\) So, the energy (E) is \(69,192,000,000\) Joules.

Key Concepts

E=mc^2 Equation

The equation \( E=mc^2 \) is one of the most famous formulas in physics, developed by Albert Einstein as part of his theory of special relativity. This equation is a statement about the equivalence of energy (\( E \)) and mass (\( m \)), with the speed of light (\( c \)) squared acting as the conversion factor.

In simple terms, it tells us that a small amount of mass can be converted into a very large amount of energy, reflecting the fundamental principle that mass and energy are interchangeable. The speed of light, (\( c \)), is a constant in vacuum with a value of approximately 299,792,458 meters per second. In exercises and practical applications, it's common to use different units, like miles per second, as seen in the given exercise.

How to apply the equation in a numerical evaluation?

• Identify the given values for mass (\( m \)) and the speed of light (\( c \)).
• Replace \( m \) and \( c \) in the equation with their numerical values.
• If necessary, carry out any conversions between units to ensure consistency.
• Perform the calculations to solve for energy (\( E \)).

In our problem, you would multiply the given mass by the square of the speed of light to convert the mass into energy.

Substitution Method in Algebra

The substitution method is a fundamental technique in algebra used to solve systems of equations. This method involves replacing one variable with another equivalent expression to simplify the problem.

The general steps to perform substitution are:

1. Begin with two equations that are set equal to one another.
2. Solve one of the equations for one of its variables.
3. Take the expression found for the variable and substitute it into the other equation.
4. Simplify and solve the resulting equation for the remaining variable.
5. Use the value of the found variable to solve for the other variable in one of the original equations.

In the context of numerical evaluations, you might not be solving for variables but simply replacing variables with the numerical values given. This effectively 'substitutes' the abstract algebraic expression with concrete numbers, making it possible to calculate a definitive answer.

Solving Equations

The process of solving equations is at the heart of algebra. Whether dealing with simple linear equations or more complex forms, the goal is to isolate the variable on one side of the equation. When solving an equation, we often perform a series of operations to both sides of the equation to maintain balance, ultimately finding the value of the unknown.

Here are some general strategies to solve equations:

• Perform inverse operations to move terms from one side of the equation to the other.
• Combine like terms.
• Factor when possible to simplify expressions.
• Check to ensure the solutions are valid by plugging them back into the original equation.

In the numerical evaluation of the \( E=mc^2 \) scenario, the process includes substituting the known values, squaring the speed of light, and then multiplying it by the mass to find the energy. This is a more straightforward example of solving an equation where there's only one variable to solve for and no need for complex algebraic manipulations.