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3 years ago

Select the correct answer.

Engineering

2 answers:

Law Incorporation [45]3 years ago

7 0

Answer:

C. Contour farming

Explanation:

Because contour farming is the practice of tilling lines on sloped lands to improve water distribution and decrease land loss.

boyakko [2]3 years ago

6 0

I think the answer is b

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You are an engineer at company XYZ, and you are dealing with the need to determine the maximum load you can apply to a set of bo

qwelly [4]

Answer:

$F_i = 28.8 kN$

Alternating and mean forces on the bolt are 0.85 kN and 29.65 kN respectively.

Mean and alternating stresses in the bulk of the bolt are 378 MPa and 10 MPa.

Safety factor = 5.5

Explanation:

Basic Dimension of Isometric Screw thread" is considered for analysis.

At Nominal diameter, d = 12 mm , $A_t = 84.3mm^{2}$ , $F_i = 0.9A_tS_p$ , $S_p = 380 MPa$

Rolled threads; $K_f = 2.2$

8 0

2 years ago

Does anyone know how to fix this? It's a chromebook and project where I have to try to fix it

Zielflug [23.3K]

Answer:

Hey I found this on You Tube it may help u a lot

Explanation:

https://youtu.be/ab9BdT6fYJ8

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2 years ago

simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The

Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

$q(x)=\frac{x}{L} q_{0}$

σ = 120 MPa = 120*10⁶ Pa

$L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\$

We can see the pic shown in order to understand the question.

We apply

∑MB = 0 (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6 (↑)

Then, we apply

$v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x$

Then, we can get the maximum bending moment as follows

$M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m$

then we get

$M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}$