Unit analysis - using measurements as variables - correct (?)

In summary, the homework states that if you want to find the side length of a square garden, you should square the area and then divide by the square root of the area.
  • #1
Wolfowitz
12
0

Homework Statement


Currently, I'm expected to find the side length of a square garden whose area is 25cm2. Of course, we're talking about a square here, and since the area of a rectangle is l * w, and, when talking about a square, l = w, the area of a square is S2. Of course, I know each side length is going to be 5cm - that's rather obvious. Instead, I'm asking as to whether I should treat the actual measurement, "cm", as a variable.

(e.g.

25cm^2 = squareroot[25cm^2] * squareroot[25cm^2]
25cm^2 = squareroot[25] * squareroot[25] * squareroot[cm^2] * squareroot[cm^2]
25cm^2 = 5 * 5 * cm * cm
25cm^2 = 25*cm^2
25cm^2 = 25cm^2.)

Please pay attention to how I'm treating cm as a variable - is it mathematically correct to do this? That's my question.

If not, my question would be this:

How do you algebraically find the final unit of measurement in which your final answer will be presented, in this question, of course.
 
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  • #2
No. "cm" is a unit of measurement. Length, which you called, l, is a variable. Width, which you called, w, is a varaible. Both the numbers, l and w, count centimeters.
 
  • #3
Yes, in essence, a physical quantity consists of a number and a unit. The unit can be treated as something that is multiplied by the number (since you have 5 units of 1 cm). As a result, when you square the quantity, you square both the number and the unit. (The square of a product of two numbers is equal to the product of the square of those two numbers).

(5 cm)2 = 52 cm2 = 25 cm2

In reverse, it is as follows:

(25 cm2)1/2 = (25)1/2 (cm2)1/2

= 5 cm

I guess that's basically what you wrote.
 
  • #4
cepheid said:
Yes, in essence, a physical quantity consists of a number and a unit. The unit can be treated as something that is multiplied by the number (since you have 5 units of 1 cm). As a result, when you square the quantity, you square both the number and the unit. (The square of a product of two numbers is equal to the product of the square of those two numbers).

(5 cm)2 = 52 cm2 = 25 cm2

In reverse, it is as follows:

(25 cm2)1/2 = (25)1/2 (cm2)1/2

= 5 cm

I guess that's basically what you wrote.

Thanks a bunch!
 

Related to Unit analysis - using measurements as variables - correct (?)

What is unit analysis?

Unit analysis is a process used in science to check the correctness and consistency of measurements and equations. It involves examining the units of measurement used in an equation or calculation to ensure they are appropriate and consistent with the desired outcome.

Why is unit analysis important?

Unit analysis is important because it helps to identify and prevent errors in calculations. It ensures that the units of measurement used in an equation or calculation are consistent, which is crucial for obtaining accurate and reliable results in scientific experiments.

How do you perform unit analysis?

To perform unit analysis, you must first identify the units of measurement used in the equation or calculation. Then, check if the units are consistent throughout the equation. If not, use conversion factors to convert the units to the desired form. Finally, perform the calculation and make sure the units in the final answer are correct and make sense in the context of the problem.

What are some common mistakes in unit analysis?

Some common mistakes in unit analysis include using the wrong units, not converting units correctly, and forgetting to include units in the final answer. It is also important to be aware of the difference between units of measurement (e.g. meters, seconds) and quantities (e.g. distance, time) to avoid confusing them in calculations.

Can unit analysis be used in all types of measurements and calculations?

Yes, unit analysis can be used in all types of measurements and calculations, including those involving length, mass, time, volume, and temperature. It is a universal tool for ensuring the accuracy and consistency of scientific measurements and calculations.

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