Respuesta :
Answer:
a) [tex](\frac{dP}{dt} )_{x=1}=20\times 1=20\ Pa.s^{-1}[/tex]
b) [tex]\frac{dP}{dt}>20x[/tex]
c) The answers to parts a) and b) are different because the rate of change of pressure is a function of position and in the two cases the position changes. In the later case the position is greater than 1 and the former case it is equal to 1.
Given:
expression of the speed of gas flow, [tex]v=5x\ m.s^{-1}[/tex]
expression for the pressure of gas, [tex]P=10x^2\ Pa[/tex]
a)
Time rate of change of pressure at x=1:
[tex]\frac{d}{dt} P=\frac{d}{dt} (10x^2)[/tex]
[tex]\frac{dP}{dt} =20x[/tex] ............................(1)
now at point x=1
[tex](\frac{dP}{dt} )_{x=1}=20\times 1=20\ Pa.s^{-1}[/tex]
b)
From equation (1):
[tex]\frac{dP}{dt} =20x[/tex]
When x>1 then the rate of change in pressure:
[tex]\frac{dP}{dt}>20x[/tex]
c)
The answers to parts a) and b) are different because the rate of change of pressure is a function of position and in the two cases the position changes. In the later case the position is greater than 1 and the former case it is equal to 1.
