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Ivan
3 years ago
5

Engineers design for everyone and consider all design challenges opportunities to problem-solve. The roller coaster in this phot

o is the largest wooden coaster in East China. It stands 32 meters tall, 1,060 meters long, and is capable of speeds of up to 91 kilometers per hour. Is the need to build a roller coaster should be considered an engineering design problem?
Engineering
1 answer:
AlexFokin [52]3 years ago
5 0

Answer:

No

Explanation:

The "need" to build a roller coaster would not be considered an engineering design problem. This would be more of a management/accounting problem because they are the ones that analyze numbers and decide what the amusement park would need in order to maintain/increase profitability by attracting more customers. Therefore, if they "need" a new roller coaster to do so then it becomes their problem. For it to be an engineering design problem the statement should be "the need to design a roller coaster with certain specifics" or something along those lines.

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Limited time only for christmas give yourself free 100 points YES YES muhahahahhaha
Setler79 [48]
Answer:

Thank you so much and may god bless you.
6 0
2 years ago
Read 2 more answers
A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.
Lena [83]

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, \mu = 2.501 V

standard deviation, \sigma = 0.113 V

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, P(X\leq 2.70):

z = \frac{x - \mu}{\sigma} = \frac{2.70 - 2.501}{0.113} = 1.761

Now, from the Probability table of standard normal distribution:

P(z\leq 1.761) = 0.9608

Now,

P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%

Now, for the expected no. of readings greater than 2.70 V:

P(X\geq 2.70) = \frac{No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V}{Total\ no.\ of\ readings}

No. of readings expected to be greater than 2.70 V = P(X\geq 2.70)\times Total\ no.\ of\ readings

No. of readings expected to be greater than 2.70 V = 0.0392\times 60 = 2.352 ≈ 2

7 0
3 years ago
Tech A says that bleeding an electronic brake control system is just like bleeding a non-electronic brake control system. Tech B
PtichkaEL [24]

Answer:

Tech A is correct.

Explanation:

An electric brake controller is a device that sends a signal to the trailer via vehicle's brakes. This reduces the wear and tear on the vehicle brakes. As a result, the vehicle stops.

Tech A says that bleeding an electronic brake control system is just like bleeding a non-electronic brake control system.

So,

Tech A is correct.

3 0
3 years ago
When something is plumb, it is
GrogVix [38]

Answer:

  C. exactly vertical

Explanation:

"Plumb" means "exactly vertical."

5 0
3 years ago
Read 2 more answers
The connection is made using a bolt and nut and two washers. If the allowable bearing stress of the washers on the boards is (sb
Yuki888 [10]

Answer:

P = 0.490 kip

Explanation:

given data

allowable bearing stress = 2 ksi

allowable tensile stress = 18 ksi

diameter = 0.31 in

outer diameter = 0.75 in

inner diameter (hole) = 0.50 in

solution

we find here cross section area of shank that is express as

Area = \frac{\pi }{4} \times d^2      ..................1

area = \frac{\pi }{4} \times 0.31^2

area  = 0.0754 in²  

and

now we get here allowable load in bolt will be

\sigma = \frac{P}{A}     ...................2

P = \sigma \times A  

P = 18 × 10³ × 0.0754

P = 1357.2 = 1.357 kip

and

now find here area of washer is

Area = \frac{\pi }{4} \times (d^2-d1^2)     .......................3

put here value

Area = \frac{\pi }{4} \times (0.75^2-0.5^2)  

area = 0.2454 in²

so now we get here allowable load of washer will be

\sigma = \frac{P}{Area}      .....................4

P = 2 × 10³ × 0.245

P = 490 = 0.490 kip

6 0
3 years ago
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