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Ira Lisetskai [31]

3 years ago

13

Describe with an example how corroded structures can lead to environment pollution?

1 answer:

Reil [10]3 years ago

8 0

An example to describe how it can lead to environment pollution is littering into the oceans , that’s one example how it can help lead to environment pollution , Hope this helps !

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If a law enforcement officer stops a vehicle for a traffic violation and finds a front-seat passenger 18 years of age or older n

Serggg [28]

Safety belt violation would be charged to the front seat passenger.

Answer: Option A.

<u>Explanation:</u>

Essential safety belt laws permit law implementation officials to ticket a driver or traveler for not wearing a safety belt, with no other traffic offense occurring. 11 states incorporate back seats as auxiliary implementation. Five of these are states with essential safety belt laws for front seat tenants.

In 15 of the 50 expresses, the safety belt law is viewed as an optional offense, which implies that a cop can't stop and ticket a driver for the sole offense of not wearing a safety belt. (One special case to this is Colorado, where kids not appropriately limited is an essential offense and brings an a lot bigger fine.)

6 0

3 years ago

Cimmaan08 for you! Thank you again :)

fomenos

Answer:

Your welcome!!! I hope that helped!!!

Explanation:

5 0

2 years ago

Read 2 more answers

A rigid bar ABCD is pinned at A and supported by two steel rods connected at B and C, as shown. There is no strain in the vertic

mylen [45]

Answer:

See attached picture.

Explanation:

4 0

3 years ago

A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4

Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

A.

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

B.

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

C.

gd = Y/(1 + w)

Or

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0

3 years ago

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

<h3>Flow rate in pipe B is = 0.3094 $\frac{m^{3} }{s}$ </h3>

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

$Q_{B} = \frac{\pi }{4} (0.3)^{2} \times 4.38$

$Q_{B} = 0.3094$ $\frac{m^{3} }{s}$

4 0

3 years ago