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alexandr1967 [171]
3 years ago
9

The correct area in sq. Inches and sq. Feet is: Select one: a. 966.76 sq. Inches and 8.056 sq. Feet b. 96.676 sq. Inches and 8.0

56 sq. Feet c. 96.676 sq. Inches and 0.671 sq. Feet
Engineering
1 answer:
kogti [31]3 years ago
8 0

Answer:

c. 96.676 sq. Inches and 0.671 sq. Feet

Explanation:

From the list of the given option, we are told to chose the correct area in sq. inches that correspond to sq. Feet.

If we recall from the knowledge of our conversion  table that,

1 sq feet = 144 sq inches

Then, let's confirm if the option were true.

a.  966.76 sq. Inches and 8.056 sq. Feet

Assuming

if 1 sq feet = 8.056

in sq inches, we have ( 8.056 × 144 ) sq inches

= 1160.064 sq. inches

So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 1 is wrong

b. 96.676 sq. Inches and 8.056 sq. Feet

if 1 sq feet = 8.056

in sq inches, we have ( 8.056 × 144 ) sq inches

= 1160.064 sq. inches

So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 2 is wrong/

c. 96.676 sq. Inches and 0.671 sq. Feet

if 1 sq feet = 0.671

in sq inches, we have ( 0.671 × 144 ) sq inches

=  96.624 sq. Inches which is closely equal to 96.676 sq. Inches

Therefore, this is the correct answer as it proves that 96.676 sq. Inches = 0.671 sq. Feet

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Yakvenalex [24]

Answer:

Activation\ Energy=2.5\times 10^{-19}\ J

Explanation:

Using the expression shown below as:

N_v=N\times e^{-\frac {Q_v}{k\times T}

Where,

N_v is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

{Q_v} is the activation energy

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Given that:

N_v=2.3\times 10^{13}

N = 10 moles

1 mole = 6.023\times 10^{23}

So,

N = 10\times 6.023\times 10^{23}=6.023\times 10^{24}

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (425 + 273.15) K = 698.15 K  

T = 698.15 K

Applying the values as:

2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

Q_v=2.5\times 10^{-19}\ J

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Answer:

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Explanation:

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This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:

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