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mrs_skeptik [129]

2 years ago

4

A specimen of a ceramic material having an elastic modulus of 250 GPa is pulled in tension with a stress of 750 MPa. Will the sp

ecimen fail if its ""most severe flaw"" is an internal crack that has a length of 0.20 mm and a tip radius of curvature of 0.001 mm? Why or why not?

1 answer:

cestrela7 [59]2 years ago

7 0

Answer:Check the attached

Explanation:

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The following passage contains a fragment. Select the correct revision. Presley took the exuberance of gospel and added the freq

Semmy [17]

Answer:

A

Explanation:

Option A was the only option that did not have a sentence fragment. All the other choices were fragments because they interrupted the flow of the sentence. A sentence fragment means that if you started reading where the period is, you would understand everything about the sentence and it would be clear. You cannot have an incomplete sentence that does not have background information. An example would be: And that is how you install it. This leaves information unanswered, such as what the subject of the sentence is, or in other words, what you were installing. Another indicator that this was a sentence fragment was that the sentence began with the word "And." Sentence fragments very commonly start with the word and, because that word shows that it is a continuation of a topic.

7 0

2 years ago

) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p

andrey2020 [161]

Answer:

number of pulses produced = 162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = $\frac{linear\ displacement}{circumference}$ × No of pulses per revolution .................2

number of pulses produced = $\frac{200}{314.16}$ × 256

number of pulses produced = 162 pulses

5 0

2 years ago

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routin

juin [17]

Answer:

Answer for the question:

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routine that we covered in class on the quicksort slides. Consider another almost-best case for quicksort, in which the pivot always splits the arrays 1/3: 2/3, i.e., one third is on the left, and two thirds are on the right, for all recursive calls of Deterministic Quicksort. (a) Give the runtime recurrence for this almost-best case. (b) Use the recursion tree to argue why the runtime recurrence solves to Theta (n log n). You do not need to do big-Oh induction. (c) Give a sequence of 4 distinct numbers and a sequence of 13 distinct numbers that cause this almost-best case behavior. (Assume that for 4 numbers the array is split into 1 element on the left side, the pivot, and two elements on the right side. Similarly, for 13 numbers it is split with 4 elements on the left, the pivot, and 8 elements on the right side.)

is given in the attachment.

Explanation:

3 0

2 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.

<u>Explanation</u>:

<u>Given</u>:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

<u>To calculate</u>:

The magnitude of applied stress in the direction of [101] and [011].

<u>Formula</u>:

zcr=σ cosФ cosλ

<u>Solution</u>:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.

5 0

2 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

2 years ago