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Lelechka [254]
3 years ago
7

The difference between ideal voltage source and the ideal current source​

Engineering
2 answers:
DENIUS [597]3 years ago
5 0
An ideal voltage source provides no energy when it is loaded by an open circuit (i.e. an infinite impedance), but approaches infinite energy and current when the load resistance approaches zero (a short circuit). ... An ideal current source has an infinite output impedance in parallel with the source.
Margarita [4]3 years ago
4 0
The ideal voltage source maintains a fixed voltage, and zero internal resistance. A current source is supposed to support any amount of potential difference across its terminals, and supply the same amount of current into the nodes it is connected to. No physical current source is ideal, just as ideal voltage sources.
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A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at
BartSMP [9]

Answer:

a) C= 1/120

b) P(X>=5) = 0.333

Explanation:

The attached file contains the explanation for the answers

7 0
3 years ago
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In what type of automobile is a transaxle most commonly found?
user100 [1]

Answer: vehicles with a front engine and FWD or a rear engine and RWD.

Explanation But the transaxle can also be integrated into the rear axle on cars with a front engine and rear-wheel drive. The transaxle is in the rear where the differential would be rather than beside the engine.

7 0
3 years ago
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
White raven [17]

Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
3 years ago
What are the disadvantages of having a liquid cooled engine?
Feliz [49]
One notable disadvantage of liquid cooling over air cooling is that it is considerably costly to set up. Cooling fans are prevalent in the market, and this overabundance of supply means they are cheap. The components of a liquid cooling system can be expensive.
5 0
2 years ago
True or false tensile forces are smaller in arch bridges
Ivan

Answer:

True

Explanation:

The tensile forces are small in most arches and usually negligible.

4 0
2 years ago
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