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Sedaia [141]
3 years ago
10

In the 1960s through 1980s, a medical filter manufacturer in Ann Arbor discharged 1,4-dioxane (an industrial solvent) directly i

nto the environment, where it ultimately leached into underlying groundwater aquifers. A plume of 1,4-dioxane has been spreading under Ann Arbor and could pose water quality risks if it reaches Barton Pond, which supplies 85% of the city's drinking water.
a) Barton Pond contains 1.5 billion gallons (5.68 million m3) of water and initially contains no 1,4-dioxane. Assume the pond is perfectly mixed with one stream flowing in and out at a rate of 1 m3/s. Suppose the plume enters the pond at a rate of 0.1 m3/s and contains 50 mg/L of 1,4-dioxane, which has a first-order decay constant of 6.4*105 s1. What is the steady-state concentration of 1,4-dioxane in the pond?

b) Local water authorities find a way to divert the plume such that 1,4-dioxane is no longer entering the pond. How long will it take for the 14-dioxane concentration in the pond to reach the Michigan Department of Environmental Quality's new allowable drinking water level of 7.2 ug/L?

c) The EPA classifies 1,4-dioxane as a probable carcinogen with a potency factor of 0.011 (mg/kg/d)1. If the water contains the EPA's allowable limit of 7.2 μg/L of 1,4-dioxane, calculate the carcinogenic risk based on the default EPA exposure factors over a lifetime for ingestion of potable water in an average adult. Is the water safe to drink based on the "rule of thumb" discussed in class?
Engineering
1 answer:
stepladder [879]3 years ago
5 0

Answer:

<u>Option-</u> C

  • In the 1960s through 1980s, a medical filter manufacturer in Ann Arbor discharged 1,4-dioxane (an industrial solvent) directly into the environment, where it ultimately leached into underlying groundwater aquifers. A plume of 1,4-dioxane has been spreading under Ann Arbor and could pose water quality risks if it reaches Barton Pond, which supplies 85% of the city's drinking water. So, we can evaluate the given information by concluding that the  EPA classifies 1,4-dioxane as a probable carcinogen with a potency factor of 0.011 (mg/kg/d)1. If the water contains the EPA's allowable limit of 7.2 μg/L of 1,4-dioxane, calculate the carcinogenic risk based on the default EPA exposure factors over a lifetime for ingestion of potable water in an average adult. Is the water safe to drink based on the "rule of thumb" discussed in class?                                      

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A negative pressure respirator brings fresh air to you through a hose<br>A) True<br>B)False​
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Answer:FALSE

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Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.

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P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si
klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V  0.184 W

2.499 mV  125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\*5

    = 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:  

V_o = A_voc*V_i*(R_o/R_l+R_o)

      = 4.545*(100/100+50)

      = 3.03 V  

Now we can determine delivered power:  

P_L = V_o^2/R_L

      = 0.184 W

Apply voltage-divider principle to the circuit:  

V_o = (R_o/R_o+R_s)*V_s

       = 50/50+100*10^3*5

       = 2.499 mV

Now we can determine delivered power:  

P_l = V_o^2/R_l

     = 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.  

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What is an OTG USB? how is it useful ​
antoniya [11.8K]

Answer:

An OTG or On The Go adapter (sometimes called an OTG cable, or OTG connector) allows you to connect a full sized USB flash drive or USB A cable to your phone or tablet through the Micro USB or USB-C charging port

Explanation:

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