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I am Lyosha [343]
3 years ago
12

A plane surface 25 cm wide has its temperature maintained at 80°C. Atmospheric air at 25°C ows parallel to the surface

Engineering
1 answer:
Gre4nikov [31]3 years ago
3 0

Answer:

See the detailed answer in attached file.

Explanation :

Download docx
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An electric winch operates on 140 volts​ [V] and draws 6 amperes​ [A] of current. The winch has an efficiency of 64​%. The winch
Elis [28]

Answer:

Explanation:

efficiency = (energy output / energy input)

energy input = ivt where i is current = 6 A, v is volts = 140 and t time = 11 seconds

energy input = 6 × 140 × 11 = 9240 J

energy output = 0.64 × 9240 = 5913.6 J

510 pound-force = 2268.593 N

at height h, potential energy = mgh = 5913.6 J

2268.593 × h = 5913.6

h = 5913.6 / ( 2268.593) = 2.61 m

3 0
3 years ago
An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% fil
den301095 [7]

Answer:

Explanation:

From the given information:

Addition of all the materials = 65+ 36+ 5 +6 = 112  which is higher than 100 percentage; SO we need to find;

The actual percentage of each material which can be determined as follows:

Percentage of the coarse aggregate will be = 65 × 112/100

= 72.80%

Percentage of the Fine aggregate will be = 36 × 112/100

= 40.32%

Percentage of the filler  will be = 5 × 112/100

= 5.6%

Percentage of the   asphalt binder will be = 6 × 112/100

= 6.72 %

So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:

Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)

Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)

Gt = 100/ (50.722)

Gt =1.972

Also;Given that the bulk density = 143.9 lb/ft³

LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft

Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43

=2.305

The percentage of air  void  = (Gt -Gm )× 100/ Gt

= (1.972 - 2.305) ×  100/ 1.972

= -16.89%

The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)

= 617.647/54.42

= 11.35%

Thus; the percentage voids in mineral aggregate =  -16.89% + 11.35%

the percentage voids in mineral aggregate = -5.45%

The percent voids filled with asphalt. = 100 × 11.35/-5.45

The percent voids filled with asphalt = - 208.26 %

7 0
3 years ago
The current through a particular circuit
timofeeve [1]

Answer:

I think a circular particular circuit element is given by it equals 5  200.8 e blah blah blah blah

Explanation:

3 0
3 years ago
For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists:
lana [24]

Answer:

a) The planar defect that exists is twin boundary defect.

b) The planar defect that exists is the stacking fault.

Explanation:      

I am using bold and underline instead of a vertical line.

a. A B C A B <u>C</u><u> </u>B A C B A

In this stacking sequence, the planar defect that occurs is twin boundary defect because the stacking sequence at one side of the bold and underlined part of the sequence is the mirror image or reflection of the stacking sequence on the other side. This shows twinning. Hence it is the twin boundary inter facial defect.

b. A B C A <u>B C  B C</u> A B C

In this stacking sequence the planar defect that occurs is which occurs is stacking fault defect. This underlined region is HCP like sequence. Here BC is the extra plane hence resulting in the stacking fault defect. The fcc stacking sequence with no defects should be A B C A B C A B C A B C. So in the above stacking sequence we can see that A is missing in the sequence. Instead BC is the defect or extra plane. So this disordering of the sequence results in stacking fault defect.

6 0
4 years ago
In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t
gladu [14]

Answer:

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}

Where nCx is the combinatorial number

nCx=\frac{n!}{x!(n-x)!}

Let's calculate f(x) :

f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064

f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288

f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432

f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216

For the cumulative distribution function that we are looking for :

P(X\leq x)=F(x)

F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1

F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1

The cumulative distribution function for X is :

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

5 0
3 years ago
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