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AlekseyPX
3 years ago
15

Chapter 3 skills and application

Engineering
1 answer:
OleMash [197]3 years ago
3 0

Answer:

what we have to answer please mention questions

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A treatment plant being designed for Cynusoidal City requires an equalization basin to even out flow and BOD variations. The ave
sineoko [7]

Answer:

The size of equalization basin is 6105.6 m³

Explanation:

The average flow is:

flow = ∑flow/n = 9.788/24 = 0.408 m³/s

Where n is the number or observations.

The inflow volume is:

V_{inflow} =Q*t

where t is the time interval

V_{inflow} =0.446(1*3600)=1605m^{3}

in the same way it is calculated the inflow volume for each observation

The outflow volume is:

V_{out} =Q_{out} *t=0.4*(1*3600)=1400m^{3}

The volume of flow is:

ds=V_{inflow} -V_{out} =1605-1400=205m^{3}

in the same way it is calculated the volume of flow for each observation. According to the file attach, the highest volume is 6105.6 m³

Download pdf
6 0
4 years ago
Which is the correct order for handwashing
cupoosta [38]

Answer:

Follow these five steps every time.

1.Wet your hands with clean, running water (warm or cold), turn off the tap, and apply soap.

2.Lather your hands by rubbing them together with the soap. Lather the backs of your hands, between your fingers, and under your nails.

3.Scrub your hands for at least 20 seconds. Need a timer? Hum the “Happy Birthday” song from beginning to end twice.

4.Rinse your hands well under clean, running water.

5.Dry your hands using a clean towel or air dry them.

8 0
3 years ago
Read 2 more answers
A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such
Harrizon [31]

Answer:

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

Rate of heat transfer = 6 W

k is the heat transfer coefficient = 152 W/m.K

A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152×4.9×10^-5×∆T/0.003

∆T = 6×0.003/152×4.9×10^-5 = 2.42 K

7 0
4 years ago
50. You are not permitted to work on any equipment or machinery at any time if the
dexar [7]
I assume this is a wood shop question:

You may not work on any equipment if the TEACHER is not present in the room.
7 0
4 years ago
An AM radio transmitter radiates 550 kW at a frequency of 740 kHz. How many photons per second does the emitter emit?
kifflom [539]

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = \frac{NE}{t}

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = \frac{NE}{t} = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}          [t = 1 s for per second]

or

550 × 10³ = \frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}

or

550 = N × 4903.24 × 10⁻³⁴

or

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

7 0
3 years ago
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