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nadezda [96]
2 years ago
12

Phosphorus and nitrogen are included in which category of water pollutants?

Engineering
1 answer:
Evgesh-ka [11]2 years ago
3 0

Answer: Hello :)

They are in the <u>nutrient pollution</u> category.

Explanation:

You might be interested in
Find the phasor form of:
mixer [17]

Answer:

I=24.598\angle 50.377

Explanation:

A tension or current expressed in cosine form and with a positive sign can be converted directly into a phasor. This is done by indicating the tension and the offset angle:

Acos(10\omega t +\phi)=A\angle \phi

So:

i(t)=10cos(10t+63)+15cos(10t-42)=10\angle 63 + 15\angle42=I

You can sum the phasors simply using a calculator, however, let's do it manually:

Let's find the rectangular form of each phasor using the next formulas:

A=\sqrt{a^2+b^2} \\\phi=arctan(\frac{b}{a})

For 10\angle 63

63=arctan(\frac{b}{a} )\\\\tan(63)=\frac{b}{a} \\\\b=a*tan(63)

10=\sqrt{a^2+(a*tan(63))^2} \\\\10^2=a^2+a^2*(tan(63))^2\\\\Solving\hspace{3}for\hspace{3}a\\\\a=\sqrt{\frac{100}{1+tan(63)^2} } =4.539904997\\\\and\hspace{3}b\\b=\sqrt{100-a^2} =8.910065242

So:

Z_1=a+bj=4.539904997+8.910065242j

For 15\angle 42

42=arctan(\frac{b_2}{a_2} )\\\\tan(42)=\frac{b_2}{a_2} \\\\b_2=a_2*tan(42)

15=\sqrt{a_2^2+(a_2*tan(42))^2} \\\\15^2=a_2^2+a_2^2*(tan(42))^2\\\\Solving\hspace{3}for\hspace{3}a_2\\\\a_2=\sqrt{\frac{225}{1+tan(42)^2} } =11.14717238\\\\and\hspace{3}b_2\\\\b_2=\sqrt{225-a^2} =10.0369591

So:

Z_2=a_2+b_2j=11.14717238+10.0369591j

Hence:

Z_T=Z_1+Z_2=(4.539904997+11.14717238)+(8.910065242+10.0369591)j\\Z_T=15.68707738+18.94702434j

Finally:

I=\sqrt{15.68707738^2+18.94702434^2} \angle arctan (\frac{18.94702434}{15.68707738} )=24.598\angle 50.377

7 0
3 years ago
6. Construct a neural network that computes the XOR function of two inputs. Make sure to specify what sort of units you are usin
IgorC [24]

Answer:

Figure 1 shows the neural network that computes the XOR of two inputs.

Two Input unit: X₁ and X₂

One hidden layer with two hidden units: h₁ and h₂

Two Weights for hidden unit 1 = 20, 20

Bias for hidden unit 1 = -10

Two Weights for hidden unit 2 = -20, -20

Bias for hidden unit 2 = 30

Weights for output layer = 20, 20

Bias for output layer = -30

Activation function = sigmoid (σ)

Output at hidden unit and at the output unit Y is shown in Table attached in Fig 2

3 0
3 years ago
Which size of impurity atom, smaller impurity atom or larger impurity atom, when located near a dislocation, will nullify some o
svp [43]

Answer:

Smaller impurity atom will nullify some of the compressive strain of a dislocation in a crystal. Because, smaller impurity atoms located near a dislocation creates tensile strain on atoms around it thereby partially nullifying compressive strain at the dislocation.

4 0
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\pifL

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

  = \frac{16}{2*\frac{22}{7}*50 }

 = 0.05091

The inductance of the inductor is 0.051H.

4 0
4 years ago
A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p
Stella [2.4K]

Answer:

Cut-off ratio\dfrac{V_3}{V_2}=6.15

Cxpansion ratio\dfrac{V_4}{V_3}=3.25

The exhaust temperatureT_4=1997.5R

Explanation:

Compression ratio CR(r)=20

\dfrac{V_1}{V_2}=20

P_2=P_3=920 psia

T_1=520 R ,T_{max}=T_3,T_3=3200 R

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{520}=20^{1.4 -1}

T_2=1723.28R

\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}

\dfrac{V_3}{V_2}=\dfrac{3200}{520}

So cut-off ratio\dfrac{V_3}{V_2}=6.15

\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}

Now putting the values in above equation

\dfrac20=\dfrac{V_4}{V_3}\times 6.15

\dfrac{V_4}{V_3}=3.25

So expansion ratio\dfrac{V_4}{V_3}=3.25.

\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}

\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}

T_4=1997.5R

So the exhaust temperatureT_4=1997.5R

3 0
4 years ago
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