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pashok25 [27]
2 years ago
10

Explain how the objects in a battery work with its components

Engineering
1 answer:
kotykmax [81]2 years ago
3 0

Answer: Batteries have three parts

Explanation:

Batteries have three parts, an anode , a cathode , and the electrolyte. The cathode and anode the positive and negative sides at either end of a traditional battery are hooked up to an electrical circuit. The chemical reactions in the battery causes a build up of electrons at the anode.

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The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated
madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss  :Due to friction of pipe surface

2.Minor loss  :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and  it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

Losses=K\dfrac{V^2}{2g}

8 0
3 years ago
What are the functions of each computer program
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3 years ago
How do I get my son to do his work?
valentinak56 [21]

Answer:

Explanation:

Reward him if he does his homework/work

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3 years ago
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In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe
KIM [24]

Answer:

\eta=0.60

Explanation:

Given :Take \gamma=1.4 for air

      P_1=100 KPa  ,T_1=300K

  \frac{V_1}{V_2}=r ⇒ r=16

As we know that  

   T_2=T_1(r^{\gamma-1})

So T_2=300\times 16^{\gamma-1}

  T_2=909.42K

Now find the cut off ration \rho

      \rho=\frac{V_3}{V_2}  

         \frac{V_3}{V_2}=\frac{T_3}{T_2}

\rho=\frac{2031}{909.42}

 \rho=2.23

So efficiency of diesel engine

\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}

Now by putting the all values

\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}

So \eta=0.60

So the efficiency of diesel engine=0.60

     

7 0
3 years ago
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