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2 years ago

A book of the Law was found in the Temple, which was being repaired, during the___

Engineering

2 answers:

Sergio [31]2 years ago

7 0

Answer:

Its 1 eighteenth

Explanation:

According to an account in 2 Kings and 2 Chronicles, Hilkiah was a kohen gadol (High Priest) of the Temple of Jerusalem during the reign of King Josiah of Judah (639-609 BC) and the discoverer of "The Book of the Law" in the Temple, in the 18th year of Josiah's Reign (622 BC).

grigory [225]2 years ago

5 0

Answer:

Explanation:

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1. Describe simply what will happen to an airplane in flight in the following conditions:

notka56 [123]

Answer:

For the two you haven't answered: (Drag greater than thrust, lift greater than weight) It will accelerate backwards (decelerate) and upwards

(Lift greater than weight, thrust greater than drag) accelerate upwards and forwards.

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3 years ago

What's the difference between accuracy and percision in measuring and gaging?

lidiya [134]

Answer:

The term Accuracy means that how close our result to the original result.

Suppose we do any experiment in laboratory and we calculate mass = 7 kg but answer is mass = 15 kg then our answer is not accurate.

And the term Precision means how likely we get result like this.

Suppose we do any experiment in laboratory and we calculate mass five times and each time we get mass = 7 kg then our answer is precised but not accurate.

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3 years ago

A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal

Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X $_{L}$ , which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX $_{L}$

X $_{L}$ = $\frac{V}{I}$ ................ 2

Given that; V = 240V, f = 50Hz, $\pi$ = $\frac{22}{7}$ , I = 15A, so that;

From equation 2,

X $_{L}$ = $\frac{240}{15}$

= 16Ω

To determine the inductance of the inductor,

X $_{L}$ = 2 $\pi$ fL

L = $\frac{X_{L} }{2 \pi f}$

= $\frac{16}{2*\frac{22}{7}*50 }$

= 0.05091

The inductance of the inductor is 0.051H.

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3 years ago

What is the core domain for Accenture’s Multi-party Systems practice?

12345 [234]

Answer:

Explanation:

digital identity is the answer

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3 years ago

Air in a large tank at 300C and 400kPa, flows through a converging diverging nozzle with throat diameter 2cm. It exits smoothly

-Dominant- [34]

Answer:

The answer is " $3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}$ "

Explanation:

Given data:

Initial temperature of tank $T_1 = 300^{\circ}\ C= 573 K$

Initial pressure of tank $P_1= 400 \ kPa$

Diameter of throat $d* = 2 \ cm$

Mach number at exit $M = 2.8$

In point a:

calculating the throat area:

$A*=\frac{\pi}{4} \times d^2$

$=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2$

Since, the Mach number at throat is approximately half the Mach number at exit.

Calculate the Mach number at throat.

$M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4$

Calculate the exit area using isentropic flow equation.

$\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}$

Here: $\gamma$ is the specific heat ratio. Substitute the values in above equation.

$\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm$

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

$\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K$

Calculate the velocity at exit.

$V*=M*\sqrt{ \gamma R T*}$

Here: R is the gas constant.

$V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}$

Calculate the density of air at inlet

$\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}$

Calculate the density of air at throat using isentropic flow equation.

$\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}$

Calculate the mass flow rate.

$m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}$

5 0

2 years ago