Unit conversions involving Pascals

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In summary: I will need a very basic lesson to understand this .No problem! There's plenty of resources online to get you started. Take a look at some of these links:https://math.stackexchange.com/questions/514070/what-does-the-exponentiation-of-negative-numbers-meanhttps://math.stackexchange.com/questions/214553/what-is-the-meaning-of-the-exponentiation-of-negative-numbers
  • #1
chriscarson
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Summary:: Pascal units digits

Do somebody have a chart that converts pascals , mega pascals etc to units to know how many digits or zeros there are after the point please ?

Thanks
 
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  • #2
Theres' nothing special about pascals. Kilo is 1000, mega is 1000000 etc.
 
  • #3
So for example when you have the result of 15 625 000 N m 2 how you put in pascals ? 15 625 kPa ?

Thanks
 
  • #4
chriscarson said:
So for example when you have the result of 15 625 000 N m 2 how you put in pascals ? 15 625 kPa ?

Thanks

Yes, ##1 \text{ Pa} = 1 \text{ N} \text{m}^{-2}## by definition. Like @Vanadium 50 alluded to, the SI prefixes are general.
 
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  • #7
etotheipi said:
Yes, ##1 Pa = 1 Nm^{-2}## by definition. Like @Vanadium 50 alluded to, the SI prefixes are general.

So 1 Pa = 0.01Nm with tha little -2 ?
 
  • #8
chriscarson said:
So 1 Pa = 0.01Nm with tha little -2 ?

No, ##\text{N} \text{m}^{-2}## is equivalent to ##\frac{\text{N}}{\text{m}^{2}}##! It has no relevance to the prefix whatsoever!
 
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  • #9
etotheipi said:
No, ##Nm^{-2}## is equivalent to ##N/m^{2}##! It has no relevance to the prefix whatsoever!
Ok
Thanks . Have to study more about these to understand.
 
  • #10
chriscarson said:
So for example when you have the result of 15 625 000 N m 2 how you put in pascals ? 15 625 kPa ?

Thanks

You seemed like you had it here! You can think of units sort of like algebraic quantities. To do the conversion, you could write down

##15625000 \text{ N}\text{m}^{-2} = 15625 \times 10^{3} \text{ N}\text{m}^{-2} = 15625 \text{ kN}\text{m}^{-2} = 15625 \text{ kPa}##

just like you obtained. Once you get the hang of it, you'll find that you won't really need to think at all/write all of that junk out!
 
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  • #11
etotheipi said:
You seemed like you had it here! You can think of units sort of like algebraic quantities. To do the conversion, you could write down

##15625000 Nm^{-2} = 15625 \times 10^{3} Nm^{-2} = 15625 kNm^{-2} = 15625 kPa##

just like you obtained. Once you get the hang of it, you'll find that you won't really need to think at all/write all of that junk out!

I notice you made always a -2 on the m .
 
  • #12
chriscarson said:
I notice you made always a -2 on the m .

##m^{-2}=\frac{1}{m^2}##
 
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  • #13
Mister T said:
##m^{-2}=\frac{1}{m^2}##

It s ok I give up . But thanks anyway for your help .
 
  • #14
etotheipi said:
You seemed like you had it here! You can think of units sort of like algebraic quantities. To do the conversion, you could write down

##15625000 Nm^{-2} = 15625 \times 10^{3} Nm^{-2} = 15625 kNm^{-2} = 15625 kPa##

just like you obtained. Once you get the hang of it, you'll find that you won't really need to think at all/write all of that junk out!
And it's very important to typeset units in roman (upright), it should read
$$1 \, \text{Pa}=1 \, \text{N} \, \text{m}^{-2}=1 \, \frac{\text{N}}{\text{m}^2}$$
etc.
 
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  • #15
vanhees71 said:
And it's very important to typeset units in roman (upright), it should read
$$1 \, \text{Pa}=1 \, \text{N} \, \text{m}^{-2}=1 \, \frac{\text{N}}{\text{m}^2}$$
etc.

Ah that's useful, never knew \text{} was a thing! My latex is dreadful...
 
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  • #16
Mister T said:
##m^{-2}=\frac{1}{m^2}##
chriscarson said:
It s ok I give up . But thanks anyway for your help .
Have you never seen negative exponents used to indicate reciprocals? $$10^{-2}=\frac 1 {10^2} = \frac 1 {100}$$ $$x^{-3} = \frac 1 {x^3}$$ etc.
 
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  • #17
jtbell said:
Have you never seen negative exponents used to indicate reciprocals? $$10^{-2}=\frac 1 {10^2} = \frac 1 {100}$$ $$x^{-3} = \frac 1 {x^3}$$ etc.
No . I finished school early now I m taking a course .
 
  • #18
chriscarson said:
No . I finished school early now I m taking a course .
The meaning for negative exponents follows naturally from the law of exponents:$$x^{a+b}=x^a \times x^b$$
If you have an exponent ##-a##, it then follows that:$$x^{-a} \times x^a = x^{-a+a} = x^0$$ By definition(*), ##x^0=1## so we can write: $$x^{-a} \times x^a = 1$$ If we divide through by ##x^a## that yields: $$x^{-a} = \frac{1}{x^a}$$

(*) One might quibble about the grounding definitions for exponentiation. But I like to start with the idea that an empty product yields the multiplicative identity (1) just like an empty sum yields the additive identity (0).
 
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  • #19
jbriggs444 said:
The meaning for negative exponents follows naturally from the law of exponents:$$x^{a+b}=x^a \times x^b$$
If you have an exponent ##-a##, it then follows that:$$x^{-a} \times x^a = x^{-a+a} = x^0$$ By definition(*), ##x^0=1## so we can write: $$x^{-a} \times x^a = 1$$ If we divide through by ##x^a## that yields: $$x^{-a} = \frac{1}{x^a}$$

(*) One might quibble about the grounding definitions for exponentiation. But I like to start with the idea that an empty product yields the multiplicative identity (1) just like an empty sum yields the additive identity (0).
I will need a very basic lesson to understand this . I started from the middle of the subject. but thanks
 
  • #20
chriscarson said:
I will need a very basic lesson to understand this . I started from the middle of the subject. but thanks
You could start with Wiki. Though a textbook might be better.
 
  • #21
jbriggs444 said:
You could start with Wiki. Though a textbook might be better.
I will but I m focusing on what the exams will be about and we stopped to work out stress , strain, and young modulus because it s an assistant technician course.
 
  • #22
chriscarson said:
So for example when you have the result of 15 625 000 N m 2 how you put in pascals ? 15 625 kPa ?

First of all it would be 15 625 000 N/m². That's 15 625 000 Newtons of force on each square meter of area. This would be, by definition, 15 625 000 Pa. And since there are 1000 pascals in a kilopascal, it would be equivalent to 15 625 kPa.
 
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  • #23
this one it s ok i fully understood it
 
  • #24
chriscarson said:
I will but I m focusing on what the exams will be about and we stopped to work out stress , strain, and young modulus because it s an assistant technician course.

Yes, but they will expect you to understand unit prefixes and exponents. What you are learning builds upon them. Knowledge is cumulative. If you have a gap, it will come up again and again until it's filled.
 
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  • #25
Vanadium 50 said:
Yes, but they will expect you to understand unit prefixes and exponents. What you are learning builds upon them. Knowledge is cumulative. If you have a gap, it will come up again and again until it's filled.

Yes it s true
 
  • #26
Some thing more I met and can t find the mistake is,when finding the area of a circle with 25 mm radius.
When calculating in mm the result is 1964 mm
When calculating in m the result is 0.00196375 m

When converting 0.00196375 m to mm it gives me 1.96375 m not as the first result of 1964 mm
 
  • #27
chriscarson said:
Some thing more I met and can t find the mistake is,when finding the area of a circle with 25 mm radius.
When calculating in mm the result is 1964 mm
When calculating in m the result is 0.00196375 m

When converting 0.00196375 m to mm it gives me 1.96375 m not as the first result of 1964 mm
An area should be expressed using a unit of area, such as square meters or square millimeters.

The conversion factor between square millimeters and square meters is 1,000,000.
 
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  • #28
jbriggs444 said:
An area should be expressed using a unit of area, such as square meters or square millimeters.

The conversion factor between square millimeters and square meters is 1,000,000.

jbriggs444 said:
An area should be expressed using a unit of area, such as square meters or square millimeters.

The conversion factor between square millimeters and square meters is 1,000,000.

So it s true to have different result ?
 
  • #29
chriscarson said:
So it s true to have different result ? And the 0.00196375 m squared is ok in exams ?
 
  • #30
chriscarson said:
Some thing more I met and can t find the mistake is,when finding the area of a circle with 25 mm radius.
When calculating in mm the result is 1964 mm
When calculating in m the result is 0.00196375 m

Try it this way: ##\pi r^2 = \pi (0.025 \ \text{m})^2##.
 
  • #31
Mister T said:
Try it this way: πr2=π(0.025 m)2πr2=π(0.025 m)2.

That s exactly what I did
 
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  • #32
chriscarson said:
That s exactly what I did
But you messed up the units. Note that @Mister T suggested you carry the units in the calculation.$$\pi \times 0.025 \text{m} \times 0.025 \text{m} = 0.0019635 \text{m}^2$$Which is different from$$0.0019635\text{m}$$
 
  • #33
jbriggs444 said:
But you messed up the units. Note that @Mister T suggested you carry the units in the calculation.$$\pi \times 0.025 \text{m} \times 0.025 \text{m} = 0.0019635 \text{m}^2$$Which is different from$$0.0019635\text{m}$$

yes apart of the units , the answer is 0.00196375 m 2 , is it acceptable in an exam ?
 
  • #34
chriscarson said:
yes apart of the units , the answer is 0.00196375 m 2 , is it acceptable in an exam ?
Yes, almost certainly. The units I would expect area to be reported in would be square meters.

However... The rules for significant figures suggest that less precision should be reported. And the calculated result you give is incorrect in the last two digits.
 
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  • #35
jbriggs444 said:
Yes, almost certainly. The units I would expect area to be reported in would be square meters.

Ok . thanks again
 
<h2>1. What is the standard unit for pressure in the metric system?</h2><p>The standard unit for pressure in the metric system is the Pascal (Pa). It is defined as one Newton per square meter (N/m²).</p><h2>2. How do I convert Pascals to other units of pressure?</h2><p>To convert Pascals to other units of pressure, you can use the conversion factor of 1 Pa = 0.00001 bar, 1 Pa = 0.000145 psi, or 1 Pa = 0.01 millibar. Simply multiply the number of Pascals by the appropriate conversion factor to get the equivalent value in the desired unit.</p><h2>3. How do I convert other units of pressure to Pascals?</h2><p>To convert other units of pressure to Pascals, you can use the conversion factor of 1 bar = 100,000 Pa, 1 psi = 6,894.76 Pa, or 1 millibar = 100 Pa. Simply multiply the number of the given unit by the appropriate conversion factor to get the equivalent value in Pascals.</p><h2>4. What are some common examples of pressure measurements in Pascals?</h2><p>Some common examples of pressure measurements in Pascals include atmospheric pressure (101,325 Pa), blood pressure (120,000 Pa), and tire pressure (200,000 Pa).</p><h2>5. Why is it important to use the correct units when measuring pressure in Pascals?</h2><p>It is important to use the correct units when measuring pressure in Pascals because it allows for accurate and consistent comparisons between different pressure measurements. Using the wrong units can lead to errors and misunderstandings in scientific data and calculations.</p>

1. What is the standard unit for pressure in the metric system?

The standard unit for pressure in the metric system is the Pascal (Pa). It is defined as one Newton per square meter (N/m²).

2. How do I convert Pascals to other units of pressure?

To convert Pascals to other units of pressure, you can use the conversion factor of 1 Pa = 0.00001 bar, 1 Pa = 0.000145 psi, or 1 Pa = 0.01 millibar. Simply multiply the number of Pascals by the appropriate conversion factor to get the equivalent value in the desired unit.

3. How do I convert other units of pressure to Pascals?

To convert other units of pressure to Pascals, you can use the conversion factor of 1 bar = 100,000 Pa, 1 psi = 6,894.76 Pa, or 1 millibar = 100 Pa. Simply multiply the number of the given unit by the appropriate conversion factor to get the equivalent value in Pascals.

4. What are some common examples of pressure measurements in Pascals?

Some common examples of pressure measurements in Pascals include atmospheric pressure (101,325 Pa), blood pressure (120,000 Pa), and tire pressure (200,000 Pa).

5. Why is it important to use the correct units when measuring pressure in Pascals?

It is important to use the correct units when measuring pressure in Pascals because it allows for accurate and consistent comparisons between different pressure measurements. Using the wrong units can lead to errors and misunderstandings in scientific data and calculations.

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