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Kitty [74]
3 years ago
14

What’s is the proper fastener to use to join two wires.

Engineering
1 answer:
xxMikexx [17]3 years ago
7 0
The answer would be C !
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Time, distance, damage
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Khái niệm về môi trường nhiệt nóng và môi trường nhiện lạnh ?
I am Lyosha [343]

Explanation:

उह्ह्नमजज्ल्ह्ह्बनुतनकुहक्जो

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3 years ago
Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of th
sleet_krkn [62]

Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, \sigma is given by

\sigma=\frac {F}{A} where F is force and A is area and area is given by \frac {\pi d^{2}}{4} hence

\sigma=\frac {4F}{\pi d^{2}} where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

\sigma=\frac {4*9970 N}{\pi 0.01m^{2}}=126941982.6 N/m^{2}

\sigma \approx 127 Mpa

From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

\epsilon=\frac {\triangle l}{l} where\epsilon is the strain, \triangle l is elongation and l is original length and making elongation the subject

\triangle l= \epsilon \times l and substituting strain with 0.0015 and length l with 100.6 mm then

\triangle l=0.0015\times 100.6=0.1509 mm

(b)

Lateral strain is given by

\epsilon_{lat}=\frac {\triangle d}{d} and substituting -v\epsilon for \epsilon_{lat} where v is poisson ratio then

-v\epsilon=\frac {\triangle d}{d} and making \triangle d the subject then

\triangle d=-vd\epsilon and substituting 0.35 for v, 0.0015 for strain and 10 mm for d

\triangle d=-(0.35)*10*0.0015=-0.00525 mm and the negative sign indicates decrease in diameter

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3 years ago
The convection heat transfer coefficient for a clothed person standing in moving air is expressed as h 5 14.8V0.69 for 0.15 , V
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Read 2 more answers
A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i
kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

                            = 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

   = 25050 ft

using super elevation formula

e + f = \dfrac{v^2}{rg}

0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}

r = \dfrac{95.33^2}{32.2\times 0.18}

  r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

L = \dfrac{\piR\Delta}{180}

L = \dfrac{\pi\times 1574\times 38}{180}

L = 1044 ft

Tangent of the curve calculation

  T = R tan(\dfrac{\Delta}{2})

  T = 1574 tan(\dfrac{38}{2})

      T = 542 ft

The station PC and PT are

 PC = PI - T

 PC = 25050 - 542

       = 24508 ft

       = 245 + 8 ft

PT = PC + L

     = 24508 + 1044

     =25552

     = 255 + 52 ft

the middle ordinate calculation

MO = R(1-cos\dfrac{\Delta}{2})

MO = 1574\times (1-cos\dfrac{38}{2})

     MO = 85.75 ft

degree of the curvature

D = \dfrac{5729.578}{R}

D = \dfrac{5729.578}{1574}

D = 3.64°

8 0
4 years ago
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