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brilliants [131]
4 years ago
13

Power boosters can leak internally or externally.

Engineering
2 answers:
cupoosta [38]4 years ago
7 0

Answer: True.

Explanation: Damage can be inflicted upon power boosters, which would most likely lead to the leak (internally, externally).

DedPeter [7]4 years ago
4 0

Answer:

True, power boosters can leak internally or externally.

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What do you enjoy most and least about engineering?
melisa1 [442]

Answer:

“I really love the design work in engineering, the face-to-face interaction with clients, and the opportunity to see projects come to life. But if I had to pick one thing that I don’t enjoy as much, I would have to say it’s contract preparation.”

5 0
3 years ago
Read 2 more answers
Two 2.30 cm × 2.30 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC . Part A What is the electric field
Alchen [17]

Answer:

A. E = 1.512\times 10^{5}\ V

B. 151.2 V

C. E = 1.512\times 10^{5}\ V

D. V = 302.4 V

Solution:

As per the question:

Area of the plates of the parallel plate capacitors, A = 2.30\times 2.30 = 5.29\ cm^{2} = 5.29\times 10^{- 4}\ m^{2}

Charge on the plates of the capacitor, Q_{c} = \pm 0.708\ nC = \pm 0.708\times 10^{- 9} \C

Now,

(A) To calculate the electric field strength, E when the separation distance, d = 1.00 mm = 10^{- 3}\ m:

E = \frac{Q}{\epsilon_{o}A}

E = \frac{0.708\times 10^{- 9}}{8.85\times 10^{- 12}\times 5.29\times 10^{- 4}} = 1.512\times 10^{5}\ N/C

(B) To calculate potential difference between the plates:

V = Ed = 1.512\times 10^{5}\times 10^{- 3} = 151.2V

(C) Electric field strength when spacing is 2 mm, i.e., 2\times 10^{- 3}\ m:

E = \frac{Q}{\epsilon_{o}A}

Since, the above expression of the electric field shows that it does not depend on the separation distance between the plates thus it will remain same, i.e., 1.512\times 10^{5}\ V

(D) Potential difference across the capacitor when d = 2\times 10^{- 3}\ m:

V = Ed = 1.512\times 10^{5}\times 2\times 10^{- 3} =302.4\ V

7 0
4 years ago
It is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20 °C?
vovangra [49]

Answer:

Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.

5 0
4 years ago
As a general rule of thumb, the ratio of the rate of etch-product formation to the flow rate of etch gas should be greater than
g100num [7]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

8 0
3 years ago
A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t
Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa

The design factor for metals is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}

The design factor for plastic and ceramics is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5

This material will break.

3 0
3 years ago
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