$y(z) = x(z) - x(z) {z}^{ - 2} - \frac{1}{4} y(z) {z}^{ - 2} \\ y(z) + \frac{1}{4} y(z) {z}^{ - 2} = x(z) - x(z) {z}^{ - 2} \\ y(z) (1 + \frac{1}{4}{z}^{ - 2}) = x(z)(1 - {z}^{ - 2}) \\ h(z) = \frac{y(z)}{x(z)} = \frac{(1 + \frac{1}{4}{z}^{ - 2})}{(1 - {z}^{ - 2})}$

The rest is straightforward...

6 0

2 years ago

Bài 3: Cho cơ cấu culít (hình 3.5) với các kích thước động lAB = 0,5lAC = 0,1m. Khâu 3 chịu tác dụng của mô men M3 = 500 N. Cơ c

lesya [120]

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3 0

2 years ago

What is the first thing a person must do before anything can be done today corroded area

melamori03 [73]

Answer:

Explanation:

We have learned that three things are required for the anodic and cathodic steps of corrosion to occur: an electrolyte, an exposed metal surface, and an electron acceptor. It follows, then, that we can prevent corrosion by removing one of these essential conditions.

8 0

2 years ago

Using the AASHTO procedure, determine the thickness required for a base and a surface layer over existing subgrade. The structur

garri49 [273]

Answer:

<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>

Explanation:

structural number = 4.5

stone base course material coefficient = 0.13

hma material layer coefficient = 0.40

drainage coefficient = 0.90

we will use layered analysis procedure to get thickness

D1 >= sN1/a1

when we cross multiply,

sN1 = a1D1 >=sN1

D2 >= -sN2-sN1/a2m2

sN2* + sN1* >= sN2

D3 >= sN3-(sN1*+sN2*)/a2m2

where a1,a2,a3 = layer coefficient

d1 d2 d3 = actual thickness

m2,m3 = coefficient of base

a1 = 0.4

a1 = 0.13

sN = 4.5

m2 = 0.9

D1 >= sN1/a1 = 4.5/0.4

= 11.25

thickness of surface = 12 inches

a1D1 = 0.4x12 = 4.8

we have value of sN2 = 5.5

(5.5 -4.8)/(0.13*0.9)

= 0.7/0.117

= 5.9829 inches

approximately 6 inches

so the pavement will have 12inch HMA surface and 6 inches granular base layer.

7 0

3 years ago