Problem 97 - Elementary Algebra

Question

\(P=\frac{3.5 \times 10^{2} \text { joule }}{2.5 \times 10^{-1} \mathrm{~s}}\)

Step-by-Step Solution

Answer

Power (P) = 1.4 x 10^3 watts.

Step 1 - Identify the Variables

First, identify the given values in the expression. Here, we have energy (or work) denoted as 3.5 x 10^2 joules and time denoted as 2.5 x 10^-1 seconds.

Step 2 - Write the Power Formula

Recall that power (P) is defined as the energy (work) done per unit time. The formula for power is: \[ P = \frac{Energy (J)}{Time (s)} \]

Step 3 - Substitute the Values

Substitute the given values into the power formula: \[ P = \frac{3.5 \times 10^{2} \text{ joule}}{2.5 \times 10^{-1} \text{ s}} \]

Step 4 - Simplify the Expression

To simplify this expression, divide 3.5 by 2.5 and subtract the exponent in the denominator from the exponent in the numerator: \[ P = \frac{3.5}{2.5} \times 10^{2 - (-1)} \] This simplifies to: \[ P = 1.4 \times 10^{3} \]

Step 5 - Final Answer

The power is calculated as: \[ P = 1.4 \times 10^{3} \text{ watts} \]

Key Concepts

scientific notation

Scientific notation is a way to express very large or very small numbers more conveniently. It makes calculations easier and helps keep track of significant figures. Scientific notation uses the form:

• a × 10ⁿ, where a is a number between 1 and 10.
• n is an integer, representing the power of 10.

In the given exercise, we have 3.5 x 10² joules and 2.5 x 10^-1 seconds. By using scientific notation, complex values are simplified, allowing for straightforward multiplication and division.

power formula

The power formula is essential in physics and engineering. It is defined as: \[ P = \frac{Energy (J)}{Time (s)} \] where:

• P: Power, usually measured in watts (W).
• Energy: Work done or energy transferred, measured in joules (J).
• Time: Time taken to do the work, measured in seconds (s).

In our exercise, power is calculated by dividing the energy (3.5 x 10² J) by the time (2.5 x 10^-1 s). This yields a straightforward understanding of the rate at which work is done or energy is transferred.

division of exponents

When dealing with scientific notation, you often need to divide numbers with exponents. The rule for dividing numbers with the same base is to subtract the exponent of the denominator from the exponent of the numerator. This can be expressed as: \[ \frac{a \times 10^{m}}{b \times 10^{n}} = \frac{a}{b} \times 10^{m-n} \]In the given problem:\[ P = \frac{3.5 \times 10^{2}}{2.5 \times 10^{-1}} \] you divide 3.5 by 2.5, which gives you 1.4. Then, subtract the exponents: 2 - (-1) = 2 + 1 = 3. Thus, the result is: \[P = 1.4 \times 10^{3} \] This technique simplifies complex calculations, making it easier to handle larger numbers.

unit conversion

Unit conversion is critical in scientific calculations to ensure that the units of measurement are consistent. In the given problem, the units used are joules (J) for energy and seconds (s) for time. Power is then measured in watts (W), where: 1 watt = 1 joule/second (1 W = 1 J/s). This conversion helps to understand and compare different energies and times more easily. To summarize, the exercise involves converting energy and time into power measured in watts. By following these steps and understanding the fundamental concepts, you can tackle similar problems with ease and confidence.