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Novay_Z [31]
3 years ago
14

Describe Ohm's law and its limitations;

Physics
1 answer:
siniylev [52]3 years ago
5 0
<span>The Ohm's Law is a powerful and simple tool to analyze electric circuits. It shows the relationship between voltage (V) and current (I) in a conductor. But, in order to properly apply to circuits (real circuits), its limitations must be understood.   Most conductors, resistance is greatly unaffected by current or voltage (resistance is stable property).
Limitation: this law does not apply to rectifiers or semiconductors or to the conduction of electricity through gases.</span>
Formula: current (I) is equal to voltage (V) over resistance (R)


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The answer is 1/2 * base * height
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Please help! Will give brainly, 50 points!! I don't understand! I just need this to give me a boost on my other questions.
harina [27]

Answer:

The answer to your question is: c = 353 mi/h; Ф = 7.3°

Explanation:

You can see the picture below

To solve this problem, just imagine a right rectangle which the legs is 350 mi/h and 45 mi/h and we are going to find the hypotenuse.

                    c² = a² + b²

                    c² = 350² + 45²

                   c² = 122500 + 2025

                   c² = 124525

                   c = 352.9 ≈ 353 mi/h

tanФ = 45/ 350

tan Ф = 0.12

Ф = 7.3°

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Why do you think compound machines are sometimes called complex machines
Maksim231197 [3]

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4 years ago
1. A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1.00 x 103 J of energy from the highe
Alex73 [517]

Answer:

\Delta S_u=2.1429\ J.K^{-1}

W_c=416.67\ J

Explanation:

Given:

temperature of source reservoir, T_H=600\ K

temperature of sink reservoir, T_L=350\ K

energy absorbed from the source, Q_{in}=1000\ J

work done, W=250\ J

a.

<u>Now change in entropy of the surrounding:</u>

\Delta S_u=\frac{dQ_L}{T_L}

<em>Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.</em>

\Delta S_u=\frac{Q_H-W}{T_L}

\Delta S_u=\frac{1000-250}{350}

\Delta S_u=2.1429\ J.K^{-1}

b.

<u>We know Carnot efficiency is given as:</u>

\eta_c=1-\frac{T_L}{T_H}

\eta_c=1-\frac{350}{600}

\eta_c=0.4167

<u>Now the Carnot work done:</u>

W_c=Q_H\times \eta_c

W_c=1000\times 0.4167

W_c=416.67\ J .......................(1)

c.

From eq. (1) we have the Carnot work, so the difference:

\Delta W=W_c-W

\Delta W=416.67-250

\Delta W=166.67\ J

Now, we find:

T_L.\Delta S_u=350\times 2.1429

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