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Softa [21]
3 years ago
8

Three resistors, of 100, 200, and 300 ohms are connected in parallel. What is their equivalent resistance?

Physics
2 answers:
satela [25.4K]3 years ago
5 0

Answer:

1/R = 1/100 + 1/200 + 1/300

1/R = (6 + 3 + 2)/600

Explanation:

garri49 [273]3 years ago
3 0

Answer:

Equivalent resistanfe in parallel (R):

1/R = 1/100 + 1/200 + 1/300

1/R = (6 + 3 + 2)/600

1/R = 11/600

R = 600/11 = 54.55 ohms

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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

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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

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6 0
3 years ago
A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r2 because A = πr2. Determine
Salsk061 [2.6K]

Answer:

a. T=r^{3/2}

b. K=\frac{1}{r}

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d. v=\sqrt{r}

Explanation:

To make analysis about the satellite circular earth the depends or r and A

T^2=\frac{4\pi}{GM}*r^3

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a.

T^2=r^3

T=r^{3/2}

b.

K=\frac{GM*m}{2}*\frac{1}{r}

K=\frac{1}{r}

c.

K=\frac{1}{2}*m*v^2

v^2=\frac{2*K}{m}=\frac{2*Gm*m}{2*m*r}

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d.

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The correct answer is elastic potential energy.

I hoped this helped!
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