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When comparing solids to fluids, the following is true: for elastic solids, the stress must be normal. For Newtonian fluids, the

Nat2105 [25]

Answer: D

Find the answer in the explanation

Explanation:

Elastic solid will obey Hooke's law which state that the force applied is proportional to the extension provided the elastic limit is not exceeded.

Examples of Newtonian fluid are water, glycerol, honey and all organic solvents

When comparing solids to fluids, the below statement is true

We can therefore conclude that

For elastic solids, stress is linearly related to strain, and for Newtonian fluids, stress is linearly related to strain rate

4 0

2 years ago

Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

vichka [17]

Answer:

Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

updated information on model changes that do not affect the technician. Who is correct? the answer is c

4 0

2 years ago

Problem 5) Water is pumped through a 60 m long, 0.3 m diameter pipe from a lower reservoir to a higher reservoir whose surface i

kap26 [50]

Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2

where P_1 = P_2 = 0

V1 AND V2 =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

4 0

2 years ago

The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting

Rama09 [41]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

AC = $\frac{5L}{3}T, CD = \frac{4L}{3} C$

B) AB = AD = BC = BD = 0

AC = $\frac{5L}{3} T, CD = \frac{4L}{3} C$

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = $\frac{5L}{3} T, CD = \frac{4L}{3}C$

7 0

2 years ago

Outline an algorithm in **pseudo code** for checking whether an array H[1..n] is a heap and determine its time efficiency.

svlad2 [7]

Answer:

Condition to break: $H[j] \geq max {H[2j] , H[2j+1]}$

Efficiency: O(n).

Explanation:

Previous concepts

Heap algorithm is used to create all the possible permutations with K possible objects. Was created by B. R Heap in 1963.

Parental dominance condition represent a condition that is satisfied when the parent element is greater than his children.

Solution to the problem

We assume that we have an array H of size n for the algorithm.

It's important on this case analyze the parental dominance condition in order to the algorithm can work and construc a heap.

For this case we can set a counter j =1,2,... [n/2] (We just check until n/2 since in order to create a heap we need to satisfy minimum n/2 possible comparisions $and we need to check this:Break condition: [tex]H[j] \geq max {H[2j] , H[2j+1]}$

And we just need to check on the array the last condition and if is not satisfied for any value of the counter j we need to stop the algorithm and the array would not a heap. Otherwise if we satisfy the condition for each $j =1,2,.....,[n/2]p$ then we will have a heap.

On this case this algorithm needs to compare 2*(n/2) times the values and the efficiency is given by O(n).

3 0

2 years ago