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2 years ago

Excepted packaging refers to fiberboard boxes, sturdy wooden boxes, or steel crates that are designed to transport:

1 answer:

5 0

Excepted packaging refers to fiberboard boxes, sturdy wooden boxes, or steel crates that are designed to transport materials with extremely low levels of radioactivity.

<h3>What is excepted packaging?</h3>

Excepted packaging can be defined as a type of box that is designed and developed for the transportation of materials with extremely low levels of radioactivity.

This ultimately implies that, a type of box which is strictly used for the transportation of materials with extremely low levels of radioactivity in order to prevent hazard.

Read more on packaging here: brainly.com/question/14985175

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For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration

Jobisdone [24]

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

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3 years ago

We can process oil into a lot of useful fuels to run our cars, trucks, and even airplanes. Oil is used for making lots of other

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Answer:

Explanation:

Products of oil in our everyday life:

(1) Petro-Chemical Feedstock: These are by product of Refining of Oil which it is used extensively to make PET bottles, Paints, Polyester Shirts, Pocket combs e.t.c

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(4) Lubricating Oil/Grease : This is another product from crude oil Fractional Distillation.

(5) Propane/ Cooking Gas: This is also a product from oil which is used in our everyday life for cooking, grilling etc.

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3 years ago

Is EPA is the organization that was formed to ensure safe and health working conditions for workers by setting and enforcing sta

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2 years ago

Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$