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3 years ago

Give three examples of how engineering has made human life better in your opinion.

Engineering

1 answer:

omeli [17]3 years ago

3 0

Answer:

This is in your opinion. If you truly wish, you may comment on this and tell me that you need something.

Explanation:

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How can you store energy with a rubber band

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Most obviously and effectively by stretching it. Heating it up, lifting it up or throwing it across the room will all store a small amount of energy in it for variously short spans of time

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3 years ago

The net work done by a net force acting on an object is equal to the change in the

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3 years ago

Find support reactions at A and B and then calculate the axial force N, shear force V, and bending moment M at mid-span of AB. L

storchak [24]

Answer:

$A=198N$ , $B=-38.33N$ , $N=0$ , $V=118N$ , $M=-37.33\ Nm$

Explanation:

It is given that $L=4m, \ q_{0}=160N/m,\ P=200N,\ M_{0}=380 N-m$ .

Thus taking equilibrium in x-direction (horizontal)

∑ $F_{x}=0$ ⇒ $B_{x}=-\frac{3}{5}* P=-\frac{3}{5} *200 N=-120N$

Taking Equilibrium moments for point A.

Giving ∑ $M_{A} =0$

Thus

$B_{y}*L+\frac{4}{5}*P*(L+\frac{L}{2})=M_{0}+\frac{1}{2}*q_{0}*L*\frac{L}{3}$

So it can be written as

$B_{y}=\frac{1}{L}*(M_{0}+\frac{1}{2}*q_{0}*L*\frac{L}{3} -\frac{4}{5} *P*(L +\frac{L}{2}))$

$B_{y}=\frac{1}{4}*(380+\frac{1}{2}*160*4*\frac{4}{3} -\frac{4}{5} *200*(4 +\frac{4}{2}))$

$B_{y}=-38.33N$

Now taking Equilibrium in y-direction (Vertical).

∑ $F_{y} =0$

thus it becomes as

$A_{y}+B_{y}=-\frac{4}{5}*P+\frac{1}{2} *q_{0}*L$

it can be written as

$A_{y}=-\frac{4}{5}*P+\frac{1}{2} *q_{0}*L-B_{y}$

$A_{y}=-\frac{4}{5}*200+\frac{1}{2} *160*4-(-38.33)$

$A_{y}=198N$

Now, As for equilibrium in x=direction

∑ $F_{x}=0$ ⇒ $N=0$

And for equilibrium in y- direction

∑ $F_{y} =0$

$A_{y}=V+\frac{1}{2}*\frac{q_{0} }{2} *\frac{L}{2}$

it can be written as

$V=A_{y}-\frac{1}{2}*\frac{q_{0} }{2} *\frac{L}{2}$

$V=198-\frac{1}{2}*\frac{160}{2} *\frac{4}{2}$

$V=118N$

Now taking equilibrium of moments about A,

∑ $M_{A} =0$

$M_{0}+M=\frac{1}{2}*\frac{q_{0}*L }{4} *\frac{2*L}{3*2}+V*\frac{L}{2}$

it can be written as

$M=\frac{1}{2}*\frac{q_{0}*L }{4} *\frac{2*L}{3*2}+V*\frac{L}{2}-M_{0}$

$M=\frac{1}{2}*\frac{160*4}{4} *\frac{2*4}{3*2}+118*\frac{4}{2}-380$

$M=-37.33\ Nm$

4 0

3 years ago

Define ways in which you would go about networking to explore opportunities in your career field and obtain more information for

kkurt [141]

Answer:

mmmmmm

Explanation:

ffmfmfmmfmfmmfmfmfmfmfmffmfmfmfmfmfmfmfmfmfmfmfmfffmmfmffmmmfmfmfmfmfmfmfmfffmfmfmfmfmfmmfmfmmfmfmfmfmfmfffmfmfffmfmfmfmffmmfmfmffmfmfmfmfffmfmfmfmfmmmmfmfmfmfmfmfmfmmmfmfmfmfmfmfmfmfmmfmfmfmfmfmfmfmfmfmfmffmfmfmfmfmfmmfmfmmmmmfmfmfmffmfmfmfmffmffmfm

6 0

3 years ago

A 0.82-in-diameter aluminum rod is 5.5 ft long and carries a load of 3000 lbf. Find the tensile stress, the total deformation, t

Lelu [443]

Answer:

Tensile stress = 0.1855Kpsi

Total deformation = 0.0012243 in

Unit strain = 1.855 *10^-5 or 18.55μ

Change in the rod diameter = 5.02 * 10^ -6 in

Explanation:

Data given: D= 0.82 in

L = 5.5 ft * 12 = 66 in

load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm

Area = (π/4) D² = (π/4) 0.82² = 0.502655 in²

∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²

Rt = 0.1855 Kpsi

∴ Total deformation = PL / AE = Rt * L/ Eal

= 0.1855 * 10³ * 66 / 10000 * 10³

= 0.0012243 in

∴the unit strains = total deformation / L = 0.0012243/ 66

=0.00001855 = 1.855 *10^-5

= 18.55μ

∴ Change in rod Δd/ d = μ ΔL/L

= (0.33) 1.855 *10^-5 * 0.82

= 5.02 * 10^ -6 in

4 0

3 years ago