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2 years ago

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:

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Can someone please answer this, it will be very appreciative

Lisa [10]

Answer:

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Explanation:

7 0

2 years ago

An incoming signal is at a frequency of 500kHz. This signal needs to be acquired and all other signals attenuated. Design a pass

Finger [1]

Answer:

C_h = 0.166 nF

C_L = 0.153 nF

Explanation:

Given:

- Ideal frequency f_o = 500 KHz

- Bandwidth of frequency BW = 40 KHz

- The resistance identical to both low and high pass filter = 2 Kohms

Find:

Design a passive band-pass filter to do this by cascading a low and high pass filter.

Solution:

- First determine the cut-off frequencies f_c for each filter:

f_c,L for High pass filter:

f_c,L = f_o - BW/2 = 500 - 40/2

f_c,L = 480 KHz

f_c,h for Low pass filter:

f_c,h = f_o + BW/2 = 500 + 40/2

f_c,h = 520 KHz

- Now use the design formula for R-C circuit for each filter:

General design formula:

f_c = 1 /2*pi*R*C_i

C,h for High pass filter:

C_h = 1 /2*pi*R*f_c,L

C_h = 1 /2*pi*2000*480,000

C_h = 0.166 nF

C,L for Low pass filter:

C_L = 1 /2*pi*R*f_c,h

C_L = 1 /2*pi*2000*520,000

C_L = 0.153 nF

7 0

3 years ago

You are traveling upstream on a river at dusk. You see a buoy with the number 5 and a flashing green light . What should you do?

nalin [4]

Answer:

please give brainliest my brother just got the corona virus

Explanation:

this is my brothers account he wants to get 5 brainliest

8 0

3 years ago

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a)

Sonbull [250]

Answer:

Explanation:

(a)

$\frac{354 mg \, 45 km}{0.0356 kN} = 354 mg \times \frac{1 kg}{10^6 mg} \times 45 km \times \frac{10^3m}{1 km} \times \frac{1}{0.0356 kN} \times \frac{1 kN}{10^3 N} = 0.447 \frac{kg \, m}{N}$

(b)

$0.00453 Mg \times 201 ms = 0.00453 Mg \times \frac{10^3 kg}{1 Mg} \times 201 ms \times \frac{1 s}{10^3 ms} = 0.911 kg \, s$

(c)

$\frac{435 MN}{23.2 mm} = 435 MN \times \frac{10^6 N}{1 MN} \times \frac{1}{23.2 mm} \times \frac{10^3 mm}{1 m} = 18.75 \times 10^9 \frac{N}{m} = 18.75 \frac{GN}{m}$

7 0

4 years ago

Disadvantages of bezier curve and b-spline

Gnesinka [82]

Answer:

The main disadvantage of the Bezier-curves is the global influence of the control points on the whole curve.

This has two major drawbacks: (1) every change of a control point (insert, move, delete) changes the look of the curve in all points of the curve, and (2) the computation time needed for a big set of control points is comparatively high. The reason is the weighting-function form.

hope it helps,pls mark me as brainliest

5 0

3 years ago