So the correct SI units for the quantity A are meters (m).

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In summary, units in scientific measurement are standardized quantities used to express the magnitude of a physical quantity, allowing for consistent and universal communication and comparison. It is important to work out units in scientific calculations to ensure accuracy and avoid errors. Conversion factors or dimensional analysis can be used to convert units. Common units in scientific measurements include meters, seconds, kilograms, and liters, which are part of the SI and globally recognized. It is crucial to never mix units in scientific calculations, as this can lead to incorrect results and confusion. Consistent units should always be used and converted if necessary.
  • #1
tommyboo
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I am having a problem finding the correct SI unitsfor the quantity A?

In the equation

A=√(R/TY)

That is A equals the square root of R divided by TY

(not to good showing workings on the computer sorry)

, the SI units of the quantity R are kg m^3 s^–2, the SI units of the quantity T are kg and the SI units of the quantity Y are m s^–2. What are the correct SI units for the quantity A?
 
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  • #2
tommyboo said:
I am having a problem finding the correct SI unitsfor the quantity A?

In the equation

A=√R/TY

That is A equals the square root of R divided by TY
Both your notation and explanation are ambiguous.

Is the expression on the right side this?
[tex]\sqrt{\frac{R}{TY}}[/tex]
or this?
[tex]\frac{\sqrt{R}}{TY}[/tex]
tommyboo said:
(not to good showing workings on the computer sorry)

, the SI units of the quantity R are kg m^3 s^–2, the SI units of the quantity T are kg and the SI units of the quantity Y are m s^–2. What are the correct SI units for the quantity A?
 
  • #3
The first one R/ty all square root. Do apologise for the bad format
 
  • #4
Then, the units of A are meters.
 
  • #5
To clarify gsal's answer...

You have the expression
[tex]\sqrt{\frac{R}{TY}}[/tex]

Simply, insert the units for each variable (in place of the variables):
[tex]\sqrt{\frac{\frac{kg\cdot m^3}{s^2}}{(kg)(\frac{m}{s^2})}}[/tex]

and simplify...
[tex]\sqrt{\left(\frac{kg \cdot m^3}{s^2}\right) \left(\frac{s^2}{kg \cdot m}\right)}[/tex]

kg and s2 cancel out, leaving
[tex]\sqrt{\frac{m^3}{m}}[/tex]

which is
[tex]\sqrt{m^2}[/tex]

or, more simply m
 

1. What are units in scientific measurement?

Units in scientific measurement are standardized quantities used to express the magnitude of a physical quantity. These units allow scientists to communicate and compare measurements in a consistent and universal way.

2. Why is it important to work out the units in scientific calculations?

Working out the units in scientific calculations is crucial because it ensures that the final result is in the correct unit and is consistent with the given data. This helps to avoid errors and allows for accurate interpretation and communication of the results.

3. How do you convert units in scientific measurements?

To convert units in scientific measurements, you can use conversion factors or dimensional analysis. This involves multiplying the given quantity by a conversion factor that relates the two units you are converting between.

4. What are some common units used in scientific measurements?

Some common units used in scientific measurements include meters (m) for length, seconds (s) for time, kilograms (kg) for mass, and liters (L) for volume. These are part of the SI (International System of Units) and are globally recognized and used.

5. Can you mix units in scientific calculations?

No, you should never mix units in scientific calculations. This can lead to incorrect results and confusion. It is important to always use consistent units throughout a calculation and convert them if necessary.

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