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In-s [12.5K]
3 years ago
9

What is stress corrosion cracking?

Engineering
1 answer:
aksik [14]3 years ago
6 0

Answer:

The growth of crack formation in a corrosive environment.

Explanation:

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youn need to use your hands

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Component(s) that only allow(s) electrons to flow in one direction. Mark all that apply
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A,D, and E

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If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed of
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Explanation:if you stretch the hose more tightly the speed of the pulse will reduce..

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3 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
3 years ago
What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?
nasty-shy [4]

Answer:

0.14\ lb/in^{3}

Explanation:

Density is defined as mass ler unit volume, expressed as

\rho=\frac {m}{v}

Where m is mass, \rho is density and v is the volume. For a sphere, volume is given as

v=\frac {4\pi r^{3}}{3}

Replacing this into the formula of density then

\rho=\frac {m}{\frac {4\pi r^{3}}{3}}

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}

4 0
3 years ago
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