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Citrus2011 [14]
3 years ago
5

Limestone (calcium carbonate) particles

Engineering
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

606506 026216n ytj dtyjjtj t

Explanation:

tyjtyjtyjtyj ty

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A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici
Pavlova-9 [17]

Answer:

8.24μm

Explanation:

The theory of brittle fracture was used to solve this problem.

And if you follow through with the attachment made a the subject of the formula

Such that,

a = 2x(69x10⁹)x0.3/pi(40x10⁶)²

= 4.14x10¹⁰/5.024x10¹⁵

= 8.24x10^-06

= 8.24μm

This is the the maximum length of the surface flaw

4 0
3 years ago
Electrical protective devices are designed to automatically
lutik1710 [3]

Answer:

can i have your number please

5 0
2 years ago
Read 2 more answers
Which regulations are related to guard rail height and dimensions and uniformity of stairs?
galina1969 [7]

Answer:

C.

structural safety

Explanation:

Guards protecting floor surfaces must be 36 inches in height, while guards for stairs must be 34 inches in height measured vertically from the tread nosing. A guard may also serve as the required handrail (34 to 38 inches high) provided the top rail meets the requirements for grip size.

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3 years ago
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. A two-dimensional fluid motion is represented by a point vortex of strength Γ set at distance c from an infinite straight soli
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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.

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8 0
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

8 0
3 years ago
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