What do you understand by the term phase angle?<br>

. A two-dimensional fluid motion is represented by a point vortex of strength Γ set at distance c from an infinite straight soli

Lostsunrise [7]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

8 0

3 years ago

What is the magnitude of the speed of a

Mars2501 [29]

The equatorial diameter of the earth is 12,756 km.

Therefore the equatorial radius is

r = 12756/2 = 6378 km.

The angular velocity of the earth is

ω = (2π radians)/(24 hours) = 0.2618 rad/h

The tangential velocity of a person at the equator is

v = rω

= (6378 km)*( 0.2618 rad/h)

= 1670 km/h

Answer: 1670 km/h

I hope this helps you.

Brainly, thanks you, rating are all appreciated greatly!

Smile and have an amazing day ; )

6 0

2 years ago

A semiconductor diode has the following parameters: Is = 4.0×10-13 Amps, n = 1.35, and is operated at a temperature of T = 270K.

andrew-mc [135]

Answer:

$V_T=0.02328V\\I=7.77\times 10^{-5} A \\V_D=0.77V$

Explanation:

Let's use Shockley ideal diode equation which relates the current intensity and the potential difference:

$I=I_S (e^{\frac{V_D}{nV_T} } -1)$

Where:

$I=Diode\hspace{3}current\\I_S=Reverse\hspace{3}bias\hspace{3}saturation\hspace{3}current\\V_D=Voltage\hspace{3}across\hspace{3}the\hspace{3}diode\\V_T=Thermal\hspace{3}voltage\\n=Ideality\hspace{3}factor$

Thermal voltage at any temperature it is a known constant defined by:

$V_T=\frac{kT}{q}$

Where:

$k= Boltzmann\hspace{3}constant \approx1.38\times 10^{-23} \\T=Absolute\hspace{3}temperature\\q=Charge\hspace{3}of\hspace{3}an\hspace{3}electron \approx 1.6\times 10^{-19} C$

(a)

Using the data provided:

$V_T=\frac{(1.38\times 10^{-23})*(270) }{1.6\times 10^{-19} }= 0.0232875V$

(b)

Using the data provided and Shockley ideal diode equation

$I=(4\times 10^{-13} )*(e^{\frac{0.6}{1.35*0.0232875} }-1)=7.773505834\times10^{-5}\approx 7.77 \times10^{-5}A\\I\approx0.0777mA$

(c) Let's isolate $V_D$ from Shockley ideal diode equation:

$I=I_S (e^{\frac{V_D}{nV_T} } -1)\\\\Multiply\hspace{3}both\hspace{3}sides\hspace{3}by\hspace{3}I_S\\\\\ \frac{I}{I_S} = e^{\frac{V_D}{nV_T} } -1\\\\Add\hspace{3}1\hspace{3}both\hspace{3}sides\\\\$

$\frac{I}{I_S} +1 =e^{\frac{V_D}{nV_T} } \\\\Take\hspace{3}the\hspace{3}natural\hspace{3}logarithm\hspace{3}of\hspace{3}both\hspace{3}sides\\\\\frac{V_D}{nV_T} =log(\frac{I}{I_S} +1) \\\\Multiply\hspace{3}both\hspace{3}sides\hspace{3}by\hspace{3}nV_T\\\\V_D=nV_T*log(\frac{I}{I_S} +1)$

Finally, using the data provided:

$V_D=(1.35)(0.0232875)*log(\frac{20\times 10^{-3} }{4\times 10^{-13} }+1)=0.77448729\approx 0.77V$

7 0

3 years ago

Which of the following is not a method of heat transfer? A. Conduction B. Convection C. Injection D. Radiation is desirable.

Alex Ar [27]

Answer:

C) Injection

Explanation:

Injection is a molding process, not a heat transfer mechanism.

8 0

3 years ago

The radiation meter is showing radiation 2x as much as background. Is this a hot zone? If so why or why not?

kakasveta [241]

Answer:

This is not a clear indication of the hot zone as the information of the radioactivity of the background is not provided clearly.

Explanation:

According to IAEA as well as NRCP, the hot area is defined on the basis of the radioactivity reading it shows instead of contrast or comparative reading from the background. The value of radiation activity which will be required to declare an area as hot zone is if it is greater than 0.1 mSv/h or $1.5091\times 10^{29} kg^{-1} s^{-1}$ .

5 0

2 years ago

What do you understand by the term phase angle?​

What do you understand by the term phase angle?