All categories

3 years ago

______________ is one of four characteristics of hazardous waste.

Engineering

1 answer:

kherson [118]3 years ago

6 0

Answer:

My guess is toxicty from gas fumes. Hope this helps

Explanation:

You might be interested in

A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted

alexira [117]

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - $D^{2}$ /Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf = free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

= 240v/h

q = Dj ( U - $U^{2}$ /Uf)

2100 = 240 ( U - $U^{2}$ /55)

Solve for U

U = 44m/h

5 0

3 years ago

Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

Katyanochek1 [597]

Answer:

$K_{I}=5.21 MPa\sqrt{m}$

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

Y = $\frac{0.8792}{0.89}$

Y = 0.9878

so put here value in equation 1

$K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}$

$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

$K_{I}$ = 5210.45 × 10³ $Pa\sqrt{m}$

$K_{I}$ = 5.21 MPa $\sqrt{m}$

5 0

3 years ago

The insulator is the connection between the grounded circuit conductor and the equipment grounding conductor at the service.

Arada [10]

The answer to your question is correct/ true

4 0

3 years ago

When Andrew began doing more residential home inspections than commercial building inspections, he found that his knowledge tran

inessss [21]

Answer:

what are the answers

Explanation:

7 0

4 years ago

A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round

Harlamova29_29 [7]

Answer:

$\frac{Q_{square}}{Q_{circle}} = 0.785$

Explanation:

given data

types of drinking straws

square cross-sectional shape
round shape

solution

we know that both perimeter of the cross section are equal

so we can say that

perimeter of square = perimeter of circle

4 × S = π × D

here S is length and D is diameter

S = $\frac{\pi D}{4}$ ....................1

and

ratio of flow rate through the square and circle is here

$\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}$

$\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}$

$\frac{Q_{square}}{Q_{circle}} = 0.785$

4 0

4 years ago