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algol [13]
3 years ago
7

What are the advantages of(a) parallel combination of loads?​

Physics
1 answer:
nexus9112 [7]3 years ago
6 0

Answer:

i really hope this helps "One of the advantages of parallel circuits is that they ensure all components in the circuit have the same voltage as the source. For instance, all bulbs in a string of lights have the same brightness."

Explanation:

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The answer is 5 meters per second.

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From far away, Saturn looks like it has seven large rings. Each large ring is named for a letter of the alphabet. The rings were named in the order they were discovered.

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Scientist A believes she has discovered a new procedure that is better than a procedure currently in use by physicians. She test
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Read 2 more answers
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
Calculate the volume of 1280 kilograms of aluminium if the density is 2700kg/m3
hjlf

Answer:

0.47m3

Explanation:

Volume = Mass / Density

In this case:

Mass - 1280 kg

Density - 2700kg/m3

1280 / 2700 = 0.4740741m3

When this is rounded off ( 2 d.p ):

0.47 m3

HOPE THIS HELPED

6 0
3 years ago
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