Yellow wood glue will typically also work with metal, glass, and adhesives

Assume a factory releases a continuous flow of wastewater into a local stream, resulting in an in-stream carcinogen concentratio

AlladinOne [14]

Answer:

Calculate the individual residential lifetime cancer risk.

Risk = Potential factor x CDI …… (1)

Calculate the value of C

C = $C_{o} * e^{-kt}$

t = d / v

t = 150 miles / 2 mile per hrs.

t = 75 hrs

t = 75/24

t = 3.13 days

Substitute the obtained value in (2).

C = $C_{o} * e^{-kt}$

C = 0.72 x e^(-0.1*3.13)

C = 0.72 x 0.7313

C = 0.526 mg/L

Substitute the obtained value in (1).

Risk = Potential factor x CDI

Risk = 0.30kg.d/mg x 0.526mg/L x 2L/d x 350day/365days

Risk = 0.3026

8 0

3 years ago

A 0.3 m3 rigid tank initially contains refrigerant-134a at 14°C. At this state, 55% of the mass is in the vapor phase, and the r

Oksanka [162]

Answer:

I am a girl want private photos

Explanation:

8 0

3 years ago

University administrators have developed a Markov model to simulate graduation rates at their school. Students might drop out, r

abruzzese [7]

Solution :

The percentage of the students who have a chance of repeating their current year = 3%

The drop out students for the first year and the sophomores = 6%

Drop out rate of first year and the seniors = 4%

Now for the state space :

S = { first year(1), sophomores(2), juniors(3), seniors(4), graduates(G), Dropouts(D) }

Therefore

the first year students are indicated as '1'

Sophomores are indicated as '2'

Juniors are indicated as '3'

Seniors are indicated as '4

Graduates are indicated as 'G'

Dropouts are indicated as 'D'

The transition diagram is attached below.

The probability of the students who have the chance of repeating their current year = 3/100 = 0.03

Probability of first year dropouts and sophomores = 6/100 = 0.06

Probability of dropout rate of juniors and seniors = 4/100 = 0.04

Therefore, the probability matrix can be made as :

1 2 3 4 G D

$\begin{matrix}1\\ 2\\ 3\\ 4\\ G\\ D\end{matrix}$ $\begin{bmatrix}0.03 & 0.91 & 0 & 0 & 0 & 0.06\\ 0& 0.03 & 0.91 & 0 & 0 & 0.06\\ 0& 0 & 0.03 & 0.93 & 0 & 0.04\\ 0& 0 & 0 & 0.03 & 0.93 & 0.04\\ 0& 0 & 0 & 0 & 1 & 0\\ 0& 0 & 0 & 0 & 0 & 1\end{bmatrix}$

Here, G represents 'graduates' and D represents 'Dropouts.'

5 0

3 years ago

This test should be performed on all cord sets, receptacles that aren't part of a building or structure's permanent wiring, and

vova2212 [387]

Answer:

A continuity test

Explanation:

A continuity test is used to verified that current will flow in an electrical circuit, it performed by placing a small voltage across the chosen path. continuity test ensure that the equipment grounding conductor is electrically continuous and this test is perform on all the cord sets, receptacles that aren't part of a building or structure's permanent wiring, and cord-and-plug connected equipment required to be grounded. example of equipment used in testing current flow in continuity test are Analog multi-meter, voltage/continuity tester etc.

Continuity test and terminal connection test are the two test required by OSHA on all electrical equipment

8 0

3 years ago

Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 85 vol% carbide

velikii [3]

Answer:

Explanation:

k_max = 26.9 w/mk

k_min = 22.33 w/mk

Explanation:

a) the maximum thermal conductivity is given as

K_MAX = k_m v_m + k_p v_p

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.15

v_p = proportion of carbide in the cement = 0.85

$K_MAX = k_m v_m + k_p v_p$

= 66*0.15 + 20*0.85

k_max = 26.9 w/mk

b) the minimum thermal conductivity is given as

$k_min = \frac{ k_{carbide} *k_{metal}}{k_{metal} v_{carbide} +k_{carbide} v_{metal}}$

= \frac{20*66}{20*0.15 +66*0.85}

k_min = 22.33 w/mk

5 0

3 years ago