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prohojiy [21]
3 years ago
10

An individual eats fish from a river contaminated with benzene. What concentrations of benzene in water (mg/L) would produce a l

ifetime cancer risk of 10-6 to an individual who eats 2 meals of fish per week for 30 years if the BCF for benzene is 10+3?
Engineering
1 answer:
avanturin [10]3 years ago
5 0

This question is incomplete, the complete question is;

An individual eats fish from a river contaminated with benzene. What concentrations of benzene in water (mg/L) would produce a lifetime cancer risk of 10⁻⁶ to an individual who eats 2 meals of fish per week for 30 years if the BCF for benzene is 10⁺³?

Answer:

0.021 mg/L concentration of benzene in water  would produce a lifetime cancer risk 10⁻⁶ to an individual who eats 2 meals of fish per week for 30 years if the BCF for benzene is 10⁺³

Explanation:

Chronic daily intake (CDI) =(C/W) (Intake rate/ lifetime) ( Exposure)

values from EPA Exposure Factors Table)

CDI = [ (5.2 L/d × C mg/L  0.054 kg/d) / 70 kg] [30/70] [350/365]

CDI = 1.65 × 10⁻³C

Now concentration of benzene;  ( RISK) = CDI × pF

= 10⁻⁶ = [2.9× 10⁻² / (MG/kg-d)] ( 1.65 × 10⁻³C)

C = 0.021 mg/L

Therefore 0.021 mg/L concentration of benzene in water  would produce a lifetime cancer risk 10⁻⁶

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Answer: A capacitor connected across the output allows the AC signal to pass through it and blocks the DC signal, thus acting as a high pass filter. The output across the capacitor is thus an unregulated filtered DC signal. This output can be used to drive electrical components like relays, motors, etc.

Explanation:

4 0
3 years ago
Technician A says that a fully charged battery is less likely to freeze than a discharged battery. Technician B says that the st
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Answer : Technician B is correct
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g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor
zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

= (5.65*10^{12} )^{0.5} = 2377.35 km

4 0
3 years ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
3 years ago
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