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3 years ago

6

True or false a critique of hazwoper incidents that have occurred in the past year should not be included in hazwoper 8 hour ref

resher training

1 answer:

Jobisdone [24]3 years ago

7 0

Answer:

False

Explanation:

No matter if something happened in the past year or so, it still should be included for safety reasons so it wont happen again

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(40%) A bank wants to store the account number of its customers (an 8-digit number) in encrypted form on magnetic stripe ATM car

9966 [12]

Answer:

Explanation:

Attached is the solution to the question

3 0

3 years ago

Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced

bekas [8.4K]

Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

<u>a) calculate the conversion exiting the first reactor </u>

CAo = 16.1 / 2 mol/dm^3

Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst

Vo = 7.24 dm^3/s

Vm = 800 gal = 3028 dm^3

hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins

next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = $\frac{Xai}{1-Xai}$ ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm --- ( 2 )

<em>where : K = 0.311 , Im = 6.97 ( input values into equation 2 )</em>

Xai = 0.684

<u>B) calculate the conversion exiting the second reactor</u>

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

Vo = 7.24 dm^3/s

To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below

XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )

<em> where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )</em>

XA2 = 0.90

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4 0

2 years ago

. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si

Over [174]

Answer:

y ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

$v =1/n R^{2/3} s^{1/2}$

where R is hydraulic radius

S = bed slope

$Q = Av =A 1/n R^{2/3} s^{1/2}$

$A = 1/2 \times (B+B+4y) \times y =(B+2y) y$

$R =\frac{A}{P}$

P is perimeter $= (B+2\sqrt{5} y)$

$R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}$

$Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}$

solving for y $100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}$

solving for y value by using iteration method ,we get

y ≈ 2.5

5 0

3 years ago

Problem 1: A catchment has the following Horton’s infiltration parameters: f0=280 mm/hr, fc=25 mm/hr and k = 2.5 hr-1 . For the

harkovskaia [24]

Answer:

Ponding will occur in 40mins

Explanation:

We say that the infiltration rate is the velocity or speed at which water enters into the soil. This often times is measured by the depth (in mm) of the water layer that can enter the soil in one hour. An infiltration rate of 15 mm/hour means that a water layer of 15 mm on the soil surface, will take one hour to infiltrate.

Consider checking attachment for the step by step solution.

6 0

3 years ago

Yuri [45]

Answer:

1

Explanation:

because every time you dived a number by its own number it is 1

6 0

2 years ago

Read 2 more answers