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Neporo4naja [7]
3 years ago
5

Which of the following actions would require the company to obtain a permit for discharge from the EPA? a. Releasing water back

into a stream after it was used in production with the only change being its temperature. b. Releasing water back into a stream after it was used in production, but in a state cleaner than it entered the factory. c. Releasing water into a pond on company property. d. Releasing water from a power plant cooling tower into the ocean. e. All of the above would require an EPA permi
Physics
1 answer:
Fiesta28 [93]3 years ago
6 0

Answer:

e. All of the above would require an EPA permit

Explanation:

For a company to discharge their by-products or waste such as wastewater, the company will need to obtain a specific permit for the Environmental Protection Agency (EPA). The EPA was formed to ensure that the integrity of the environment is not compromised. Thus, all of the options in the given question requires an EPA permit.

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Sound waves travel in air at 343 m/s. The lowest frequency one can hear is 25.0 Hz; the highest frequency is 25.0 kHz. Find the
Slav-nsk [51]

Answer:

13.72 m and 0.01372 m respectively

Explanation:

Wavelength: This can be defined as the distance covered in one complete oscillation. The S.I unit of wavelength is meter (m).

The formula for the speed of a wave is given as

v = λf ............................. Equation 1

Where v = speed of the sound wave, λ = wavelength, f = frequency of the sound wave.

make λ the subject of the equation,

λ = v/f ......................... Equation 2

For the lowest frequency,

Given: f = 25 Hz, v = 343 m/s.

Substitute into equation 2

λ = 343/25

λ = 13.72 m.

For the highest frequency,

Given: f = 25 kHz = 25000 Hz, v = 343 m/s

Substitute into equation 2

λ = 343/25000

λ = 0.01372 m.

The wavelength of sound for 25 Hz and 25 kHz = 13.72 m and 0.01372 m respectively

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3 years ago
THIS IS MY NUMBER CALL me :::<br><br><br><br><br><br>PEHLI FURSAT MAIN NIKAL XD​
scZoUnD [109]

Answer:

nooo interested

8 0
3 years ago
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A slice of pizza has 500 kilocalories. If we could burn the pizza and use all the heat to warm a 50 liter container of cold wate
quester [9]

Answer:

10 degree C

Explanation:

Q = 500 kcal = 500 x 1000 x 4.186 J = 2.1 x 10^6 J

V = 50 liter

m = Volume x density = 50 x 10^3 x 1000 = 50 kg

Let ΔT be the rise in temperature.

Specific heat of water = 4186 J/kg C

Q = m x c x ΔT

2.1 x 10^6 = 50 x 4186 x ΔT

ΔT = 10 degree C

4 0
3 years ago
If you measured all the energy related to motion and all the stored energy in the particles of a substance, which would you be m
Nady [450]

If you measured all the energy related to motion and all the stored energy in the particles of a substance, you would be measuring the thermal energy of the particles. If there is movement of the particles, they are also releasing energy in the form of heat.

3 0
4 years ago
Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou
Gennadij [26K]

Answer:

a. The station is rotating at 1.496 \frac{rev}{min}

b. the rotation needed is 2.8502 \frac{rev}{min}

Explanation:

We know that the centripetal acceleration is

a_{c}= \omega ^2 r

where \omega is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed  is :

2.7 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.02454 \frac{rad^2}{s^2} }

\omega  = 0.1567 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 1.496 \frac{rev}{min}

<h3>b. </h3>

The rotational speed needed is :

9.8 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.08909 \frac{rad^2}{s^2} }

\omega  = 0.2985 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.2985 \frac{rev}{min}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 2.8502 \frac{rev}{min}

3 0
3 years ago
Read 2 more answers
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