Answer:
Relation between Kelvin scale and Celsius scales is \[=5\times
{{10}^{-4}}\]
where
\[F=50,000N\] = temperature on Celsius and Kelvin scales
respectively.
For Neon \[5\times {{10}^{-4}}\]
For \[P=\frac{F}{\pi
{{D}^{2}}/4}=\frac{4F}{\pi {{D}^{2}}}\] \[\therefore \]
Relation
between Kelvin and Farenheigt scales is
\[P=\frac{4\times
\left( 5\times {{10}^{4}} \right)}{\left( 22/7 \right)\times {{\left( 5\times
{{10}^{-4}} \right)}^{2}}}\]
\[=2\cdot
5\times {{10}^{11}}Pa\]
For
Neon
\[\cdot
\]
For
\[m{{m}^{2}}\]
\[m{{m}^{2}}\]
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