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3 years ago

If the Zener diode is connectedwrong polarity the voltageacross the load is?

Engineering

1 answer:

o-na [289]3 years ago

6 0

Answer:

i think so

Explanation:

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An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th

sp2606 [1]

Answer:

- 14.943 W/m^2K ( negative sign indicates cooling )

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

Calculate the overall heat loss coefficient

Note : heat lost by water = heat loss through convection

m*Cp*dT = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp ( integrate the relation )

In ( $\frac{49-8}{60-8}$ ) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

= - 14.943 W/m^2K ( heat loss coefficient )

7 0

3 years ago

Provide an argument justifying the following claim: The average (as defined here) of two Java ints i and j is representable as a

ahrayia [7]

Answer:

public static int average(int j, int k) {

return (int)(( (long)(i) + (long)(j) ) /2 );

}

Explanation:

The above code returns the average of two integer variables

Line 1 of the code declares a method along with 2 variables

Method declared: average of integer data type

Variables: j and k of type integer, respectively

Line 2 calculates the average of the two variables and returns the value of the average.

The first of two integers to average is j

The second of two integers to average is k

The last parameter ensures average using (j+k)/2

3 0

3 years ago

A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The

AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = $\frac{60+180}{2}$ = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0

3 years ago

Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)

IceJOKER [234]

Answer:

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa) Number- Average Molecular Weight (g/mol)

82 12,700

156 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

SOLUTION:

We know that :

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

where;

$T_S$ = Tensile Strength

$T_{S \infty}$ = Tensile Strength (Infinity)

$M_n$ = Number- Average Molecular Weight (g/mol)

SO;

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

From equation (1) ; collecting the like terms; we have :

$T_{S \infty} =82+ \dfrac{A}{12700}$

From equation (2) ; we have:

$T_{S \infty} =156+ \dfrac{A}{28500}$

So; $T_{S \infty} = T_{S \infty}$

Then;

$T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}$

Solving by L.C.M

$\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}$

$\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}$

By cross multiplying ; we have:

$({4446000 + A})* {12700} ={28500} *({1041400 + A})$

$(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)$

Collecting like terms ; we have

$(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)$

$2.67843*10^{10} = 15800 \ A$

Dividing both sides by 15800:

$\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}$

A = 1695208.861

From equation (1);

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

Replacing A = 1695208.861 in the above equation; we have:

$82= T_{S \infty} - \dfrac{1695208.861}{12700}$

$T_{S \infty}= 82 + \dfrac{1695208.861}{12700}$

$T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{2736608.861 }{12700}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

From equation(2);

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

Replacing A = 1695208.861 in the above equation; we have:

$156= T_{S \infty} - \dfrac{1695208.861}{28500}$

$T_{S \infty}= 156 + \dfrac{1695208.861}{28500}$

$T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{6141208.861}{28500}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

Then:

$181 = 215.481- \dfrac{1695208.861 }{M_n}$

Collecting like terms ; we have:

$\dfrac{1695208.861 }{M_n} = 215.481- 181$

$\dfrac{1695208.861 }{M_n} =34.481$

1695208.861= 34.481 $M_n$

Dividing both sides by 34.481; we have:

$M_n$ = $\dfrac{1695208.861}{34.481}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

5 0

3 years ago

To read signs you need good focal vision

kow [346]

Answer:eyesight

Explanation:

7 0

3 years ago

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If the Zener diode is connectedwrong polarity the voltageacross the load is?​

If the Zener diode is connectedwrong polarity the voltageacross the load is?