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3 years ago

Which option can result from a lack of stable social and professional bonds?

Engineering

2 answers:

PilotLPTM [1.2K]3 years ago

6 0

B.) An unstructured, chaotic life

Natasha_Volkova [10]3 years ago

6 0

Answer:

It B

Explanation:

;)))))

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(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor

serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm = 0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

A) would the 1040 steel be a possible candidate for this application

Yield strength of 1040 steel < stress ( in order to be a possible candidate )

stress = p / A0 = ( 18000 ) / ( $\frac{\pi }{4}$ ) * 0.0075^2

= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )

Therefore 1040 steel is not a possible candidate for this application

B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm

Area1 = ( $\frac{\pi }{4}$ ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%

6 0

3 years ago

John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa

IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= $35000\times (\frac{F}{P}, 4 \ percent,5 )$

= $35000\times 1.216 7$

= $42584.50$

Now,

time (t) will be:

= $5\times 12$

= $60 \ monthly \ payments$

So, monthly payment will be:

= $42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )$

= $42584.50\times 0.0193$

= $821.88$

6 0

3 years ago

Consider two different versions of algorithm for finding gcd of two numbers (as given below), Estimate how many times faster it

juin [17]

Answer:

Explanation:

Step 1:

a) The formula for compute greatest advisor is

gcd(m,n) = gcd (n,m mod n)

the gcd(31415,14142) by applying Euclid's algorithm is

gcd(31,415,14,142) =gcd(14,142,3,131)

=gcd=(3,131, 1,618)

=gcd(1,618, 1,513)

=gcd(1,513, 105)

=gcd(105, 43)

=gcd(43, 19)

=gcd(19, 5)

=gcd(5, 4)

=gcd(4, 1)

=gcd(1, 0)

STEP 2

b) The number of comparison of given input with the algorithm based on checking consecutive integers and Euclid's algorithm is

The number of division using Euclid's algorithm =10 from part (a)

The consecutive integer checking algorithm:

The number of iterations =14,142 and 1 or 2 division of iteration.

14,142 ∠= number of division∠ = 2*14,142

Euclid's algorithm is faster by at least 14,142/10 =1400 times

At most 2*14,142/10 =2800 times.

5 0

3 years ago

The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si

alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

A) Classifying the soil according to USCS system

( using 2nd image attached below )

description of sand :

The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0

3 years ago

A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r

allochka39001 [22]

Answer: 37.4K

Explanation:

See attachment

7 0

3 years ago