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san4es73 [151]
3 years ago
15

Answer the following questions about your own experience in the labor force.

Engineering
1 answer:
liubo4ka [24]3 years ago
4 0

Answer:

Following are the solution to this question:

Explanation:

In point a:

This takes me six weeks for both the took ideas that I was searching for but it continued for 3 years (12 weeks) as it's an intern.  

In point b:

Finding job:

\to f =  \frac{1}{6} = 0.166 jobs weekly  

Separation of jobs:

\to \frac{ 1}{12}=0.083 employment per week.  

In point c:

Its natural rate of unemployment is:  \frac{U}{L} = s+(s \times f).  

The normal level of employment for that community I represent, once we add up from that preceding section, is as follows:

\to \frac{U}{L} = 0.083+ (0.083\times 0.166) = 0.096

If on average, it requires six weeks to find another job or the work lasted 12 weeks, the group's unemployment level is 0.096 \ \%.

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Water flows through two smooth pipes with the same diameter and length as shown below. ipe is twice that through the first-pipe.
Alisiya [41]

Answer:

The right answer is (d)

Explanation:

The pressure drop in smooth pipes with laminar flow is determined by the resistance of the fluid to flow, which is controlled by Reynolds's number. At higher Re, higher drop. But in this case, we have no information about the speed of the fluid in each pipe, and the ipe is not related to that speed. therefore we don´t have the information to say that any of the other options are the right ones

3 0
3 years ago
If a car travels 8 miles in 15 minutes, what is the speed of the car in miles per hour
mestny [16]

Answer:

32 miles per hour

Explanation:

if 8 miles is in 15 minutes then multiply 8 by 4 to get miles per hour.

8 0
3 years ago
What are the units or dimensions of the shear rate dv/dy (English units)? Then, what are the dimensions of the shear stress τ= μ
swat32

Answer:

1) Dimensions of shear rate is [T^{-1}] .

2)Dimensions of shear stress are [ML^{-1}T^{-2}]

Explanation:

Since the dimensions of velocity 'v' are [LT^{-1}] and the dimensions of distance 'y'  are [L] , thus the dimensions of \frac{dv}{dy} become

\frac{[LT^{-1}]}{[L]}=[T^{-1}] and hence the units become s^{-1}.

Now we know that the dimensions of coefficient of dynamic viscosity \mu are [ML^{-1}T^{-1}] thus the dimensions of shear stress can be obtained from the given formula as

[\tau ]=[ML^{-1}T^{-1}]\times [T^{-1}]\\\\[\tau ]=[ML^{-1}T^{-2}]

Now we know that dimensions of momentum are [MLT^{-1}]

The dimensions of Area\times time are [L^{2}T]

Thus the dimensions of \frac{Moumentum}{Area\times time}=\frac{MLT^{-1}}{L^{2}T}=[MLT^{-2}]

Which is same as that of shear stress. Hence proved.

4 0
3 years ago
Type the correct answer in the box. Spell all words correctly.
sesenic [268]

Answer:

Chemical Engineers use chemistry, math and physics to design and use to make chemical products. The fibers in clothing are designed by chemical engineers.

6 0
3 years ago
Read 2 more answers
Guys, can you rate this toolbox, please. It is my DT project, made for car trips, what do you think of the idea/design/usability
Maurinko [17]

Answer:

Honestly overall i think it looks fantastic

Explanation:

It looks like some really nice clean craftsmanship and i love the use of some different colors for some drawers to make it pop. the only con that i can possibly think of is that with it being wood and you moving it from place to place, some rubber feet or something that would prevent it from scratching/damaging anything else if it doesn't already (cant really see under it). other then that one thing i think it looks really good. well done.

3 0
3 years ago
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