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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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A_____ transducer is a device that can convert an electronic controller output signal into a standard pneumatic output. A. pneum
makkiz [27]

Answer:

The correct answer is

option C. current to pneumatic (V/P)

Explanation:

A current to pneumatic controller is  basically used to receive an electronic signal from a controller and converts it further into a standard pneumatic output signal which is further used to operate a positioner or control valve. These devices are reliable, robust and accurate.

Though Voltage and current to pressure transducers are collectively called as electro pneumatic tranducers and the only electronic feature to control output pressure in them is the coil.

6 0
3 years ago
1. A thin plate of a ceramic material with E = 225 GPa is loaded in tension, developing a stress of 450 MPa. Is the specimen lik
mina [271]

Answer:

fracture will occur as the value is less than E/10 (= 22.5)

Explanation:

If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}

=  2\times 750 (\frac{\frac{0.2mm}{2}}{0.001 mm}})^{1/2}

                 = 15 GPa

fracture will occur as the value is less than E/10 = 22.5

7 0
3 years ago
While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),
MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2} (1)

Where:

y_{o}, y - Initial and final vertical position, measured in meters.

v_{o} - Initial speed, measured in meters per second.

\theta - Launch angle, measured in sexagesimal degrees.

g - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that y_{o} = 2\,m, y = 0\,m, v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and g = -9.807\,\frac{m}{s^{2}}, then the time taken by the ball is:

-4.904\cdot t^{2}+13.482\cdot t +2 = 0 (2)

This second order polynomial can be solved by Quadratic Formula:

t_{1} \approx 2.890\,s and t_{2} \approx -0.141\,s

Only the first root offers a solution that is physically reasonable. That is, t \approx 2.890\,s.

The vertical velocity of the ball is calculated by this expression:

v_{y} = v_{o}\cdot \sin \theta +g\cdot t (3)

Where:

v_{o,y}, v_{y} - Initial and final vertical velocity, measured in meters per second.

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ}, g = -9.807\,\frac{m}{s^{2}} and t \approx 2.890\,s, then the final vertical velocity is:

v_{y} = -14.860\,\frac{m}{s}

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball (x) is determined by the following expression:

x = (v_{o}\cdot \cos \theta)\cdot t (4)

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and t \approx 2.890\,s, then the distance covered by the ball is:

x = 32.695\,m

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground (v), measured in meters per second, is determined by the following Pythagorean identity:

v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}} (5)

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and v_{y} = -14.860\,\frac{m}{s}, then the magnitude of the velocity of the ball is:

v \approx 18.676\,\frac{m}{s}.

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta_{o} = 50^{\circ} and v_{y} = -14.860\,\frac{m}{s}, the angle of the total velocity of the ball just before hitting the ground is:

\theta \approx -52.717^{\circ}

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

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