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Sergeeva-Olga [200]
2 years ago
5

Describe the difference between technology based effluent standards and water quality based effluent standards under the Clean W

ater Act. Also, indicate which of these two different standards is likely to be controlling on a small stream designated as a cold water fishery and why.
Physics
1 answer:
harkovskaia [24]2 years ago
8 0

Explanation:

Technology-based:

1.As the name implies technology, no technology will be clarified about it. It only depends on the variables, which describes them.

2. That is based on a single facility's findings.

3. It takes into account the contaminants type and volume, and their equations to monitor them.

4. This is reserved for city or urban wastewater treatment plants only.

5. It takes into account the pH, need for oxygen and the suspended solids.

Water quality based on:

1.This is enforced if there is a need to apply stricter limits to pollutants that are not pleased with the limits of technology.

2. All basing on water quality was risk-based.

3. They placed some less importance on the technologies which is used in the technology based limit.

Water quality dependent restrictions are used as cool water fishery for streams.

The act on clean water will also include bodies of water belonging to wildlife, agriculture and others. The law also included that the physical chemical and biological variables of all the state water bodies must be controlled by these water quality based limits.

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When the temperature of the air is 25 degress C, the velocity of a sound wave traveling through the air is approximately
dusya [7]
Assuming an ideal gas, the speed of sound depends on temperature 
only.   Air is almost an ideal gas. 

Assuming the temperature of 25°C in a "standard atmosphere", the 
density of air is 1.1644 kg/m3, and the speed of sound is 346.13 m/s. 

The velocity can't be specified, since the question gives no information 
regarding the direction of the sound.
6 0
3 years ago
What is the net force on a truck if the force of friction is 31 N and the force of the engine is 79 N?
Evgen [1.6K]

Answer:

Fr = 48 [N] forward.

Explanation:

Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.

∑F = Fr

F_{engine}-F_{brake} =F_{r}\\F_{r}=79-31\\F_{r}=48[N]

The movement remains forward, since the force produced by the movement is greater than the braking force.

3 0
2 years ago
Snail 3, trying to keep up with Snail 2, managed to get to
Kipish [7]

Answer:

Acceleration is the change in velocity over the change in time = Δv/Δt.  To do these problems, you need to find out how much the speed changed and over what period of time it changed.

Snail 1 changes from 4 cm/min to 7 cm/min in 3 minutes.  Subtract the starting velocity (4 cm/min) from the ending velocity (7 cm/min) then divide by the time (3 min):

Snail 1 = (7 cm/min. - 4 cm/min)/(3 minutes) = ?    (remember to put down the units)

Snail 2 changed from 7 cm/min. down to 1 cm/min. in 3 minutes

Snail 2 = (1 cm/min. - 7 cm/min.)/(3 min.) = ?        (note that the acceleration is negative when you slow down)

I hope this helps you

6 0
3 years ago
A 0.0010-kg pellet is fired at a speed of 50.0m/s at a motionless 0.35-kg piece of balsa wood. When the 
aleksandr82 [10.1K]
P = m*v

conservation of momentum suggests

initial momentum equals final momentum

mv-initial = mv-final

(0.0010 kg)(50 m/s) = (0.0010 kg + 0.35 kg)v

thus:

v = (0.0010)(50)/(0.351) = 0.142 m/s
8 0
3 years ago
Read 2 more answers
The Earth is 1.5 × 1011 m from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It
katrin2010 [14]

Answer:

Part a)

v_{cm} = 2.98 \times 10^4 m/s

Part b)

K_{trans} = 2.68 \times 10^{33} J

Part c)

\omega = 7.27 \times 10^{-5} rad/s

Part d)

KE_{rot} = 2.6 \times 10^{29} J

Part e)

KE_{tot} = 2.68 \times 10^{33} J

Explanation:

Time period of Earth about Sun is 1 Year

so it is

T = 1 year = 3.15 \times 10^7 s

now we know that angular speed of the Earth about Sun is given as

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{3.15 \times 10^7}

now speed of center of Earth is given as

v_{cm} = r\omega

r = 1.5 \times 10^{11} m

v_{cm} = (1.5 \times 10^{11})(\frac{2\pi}{3.15 \times 10^7})

v_{cm} = 2.98 \times 10^4 m/s

Part b)

now transnational kinetic energy of center of Earth is given as

K_{trans} = \frac{1}{2}mv^2

K_{trans} = \frac{1}{2}(6 \times 10^{24})(2.98 \times 10^4)^2

K_{trans} = 2.68 \times 10^{33} J

Part c)

Angular speed of Earth about its own axis is given as

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{24 \times 3600}

\omega = 7.27 \times 10^{-5} rad/s

Part d)

Now moment of inertia of Earth about its own axis

I = \frac{2}{5}mR^2

I = \frac{2}{5}(6 \times 10^{24})(6.4 \times 10^6)^2

I = 9.83 \times 10^{37} kg m^2

now rotational energy is given as

KE_{rot} = \frac{1}{2}I\omega^2

KE_{rot} = \frac{1}{2}(9.83 \times 10^{37})(7.27 \times 10^{-5})^2

KE_{rot} = 2.6 \times 10^{29} J

Part e)

Now total kinetic energy is given as

KE_{tot} = KE_{trans} + KE_{rot}

KE_{tot} = 2.68 \times 10^{33} + 2.6 \times 10^{29}

KE_{tot} = 2.68 \times 10^{33} J

6 0
3 years ago
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