Unit conversions involving Pascals

[chriscarson]

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

 

[Vanadium 50]

Theres' nothing special about pascals. Kilo is 1000, mega is 1000000 etc.

 

[chriscarson]

So for example when you have the result of 15 625 000 N m 2 how you put in pascals ? 15 625 kPa ?

Thanks

 

[etotheipi]

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.

 

[A.T.]

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 ?

https://en.wikipedia.org/wiki/Metric_prefix#List_of_SI_prefixes

 

[chriscarson]

https://en.wikipedia.org/wiki/Metric_prefix#List_of_SI_prefixes

Thank you

 

[chriscarson]

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 ?

 

[etotheipi]

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!

 

[chriscarson]

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.

 

[etotheipi]

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!

 

[chriscarson]

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 .

 

[Mister T]

I notice you made always a -2 on the m .

$m^{-2}=\frac{1}{m^2}$

 

[chriscarson]

$m^{-2}=\frac{1}{m^2}$

It s ok I give up . But thanks anyway for your help .

 

[vanhees71]

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.

 

[etotheipi]

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...

 

[jtbell]

$m^{-2}=\frac{1}{m^2}$

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.

 

[chriscarson]

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 .

 

[jbriggs444]

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).

 

[chriscarson]

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

 

[jbriggs444]

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.

 

[chriscarson]

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.

 

[Mister T]

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.

 

[chriscarson]

this one it s ok i fully understood it

 

[Vanadium 50]

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.

 

[chriscarson]

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

 

[chriscarson]

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

 

[jbriggs444]

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.

 

[chriscarson]

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.

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 ?

 

[chriscarson]

So it s true to have different result ? And the 0.00196375 m squared is ok in exams ?

 

[Mister T]

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: $\pi r^2 = \pi (0.025 \ \text{m})^2$.

 

[chriscarson]

Try it this way: πr2=π(0.025 m)2πr2=π(0.025 m)2.

That s exactly what I did

 

[jbriggs444]

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$$0.0019635\text{m}$$

 

[chriscarson]

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 ?

 

[jbriggs444]

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.

 

[chriscarson]

Yes, almost certainly. The units I would expect area to be reported in would be square meters.

Ok . thanks again