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2 years ago

15

Which is the correct standard notation for 9.3x10^5

1 answer:

horsena [70]2 years ago

7 0

Explanation:

9.3x105

Simplifies to:

9.3x100000

=9.3x100000

Hope that helps

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Where is the most of the liquid freshwater on earth located?

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Which sequence occurs from fastest to slowest on the earth's surface?

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Discuss Joule-Thompson effect with relevant examples and formulae.

Delicious77 [7]

Answer:

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

Explanation:

Joule -Thompson effect

Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process

Let

$P_1,T_1$ Pressure and temperature at inlet and

$P_2,T_2$ Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

$\mu _j=\left(\frac{\partial T}{\partial p}\right)_h$

Now from T-ds equation

dh=Tds=vdp

So

$Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp$

⇒ $dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

So Joule -Thompson coefficient

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

$\dfrac{\partial v}{\partial T}=\dfrac{v}{T}$

So by putting the values in

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

$\mu _j=0$ For ideal gas.

6 0

3 years ago

Please help with any of them!!!!!!

Law Incorporation [45]

Its to blury for me to see it

5 0

4 years ago

Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the

Mila [183]

Answer: $d=\frac{v^2}{8a}$

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

$v=at$

$t=\frac{v}{a}$

Now loosing car is accelerating with $\frac{a}{4}$

Distance traveled by loosing car in time t

$s_1=ut+\frac{at^2}{2\cdot 4}$

$s_1=0+\frac{a}{8}\times (\frac{v}{a})^2$

$s_1=\frac{v^2}{8a}$

Thus distance d traveled by loosing car is given by $d=\frac{v^2}{8a}$

5 0

4 years ago