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Westkost [7]
3 years ago
12

Why Elastic Modulus is important and on what it depends?

Engineering
1 answer:
Bad White [126]3 years ago
5 0

Answer:

The Young's Modulus of a material is a fundamental property of every material that cannot be changed. It is dependent upon temperature and pressure however. The Young's Modulus (or Elastic Modulus) is in essence the stiffness of a material. In other words, it is how easily it is bended or stretched.

Explanation:

Have a great day

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A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above
creativ13 [48]

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation N_{o} = N_{i} * \frac{E_{f}-E_{i}  }{KT}

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, N_{d}  = 1.65*10^{16}  - 10^{16} \\           = 6.5 * 10^{15} per cm cube

5 0
3 years ago
Read 2 more answers
An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0
3 years ago
How do you put air knight in slingshock mode.
Nesterboy [21]

Answer:

I am not sure I am understanding plz more context

Explanation:

3 0
3 years ago
In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap
Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0
3 years ago
The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find
Nataly_w [17]

The change in annual cost when Q is increased from 340 to 341 is -1.23 and the instantaneous rate of change when Q = 340 is -1.25

<h3>How to find the Instantaneous rate of change?</h3>

The annual inventory cost C for a manufacturer is given as;

C = (1012000/Q) + 7.5Q

where Q is the order size when the inventory is replenished.

Now, the change in C can be calculated by evaluating the cost function at Q = 340 and Q = 341

Change in C = [1,012,000/341 + 7.5*341] - [1,012,000/340 + 7.5*340] ≈ -1.23

Instantaneous rate of change in C is first order derivative C':

C'(Q) = -1,012,000/(Q²) + 7.5

C'(340) = -1,012,000/(340²) + 7.5 ≈ -1.25

Read more about Instantaneous rate of change at; brainly.com/question/14666106

#SPJ1

8 0
1 year ago
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