Which of the following is NOT part of a car's drive train?
A axle
B rotor
C differential
D transmission

The amplitude of the sinusoid is 28384 *x soít cos.

Amplitude is defined as the greatest deviation from equilibrium of a point on a vibrating body or wave in terms of displacement or distance traveled. In most cases, amplitude is calculated by looking at a wave graph and determining the height of the wave from rest. The strength or intensity of the wave is gauged by its amplitude.

Sinusoid is defined as a signal with sine wave characteristics. In the liver, spleen, and bone marrow, sinusoids and irregular tubules transport blood in place of venules and capillaries. The sine or cosine functions from trigonometry form the foundation of sinusoidal signals, which are periodic functions.

Thus the amplitude of the sinusoid is 28384 *x soít cos.

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Answer:

a) 34 mm

b) 39 mm

c) 93.16%

Explanation:

power transmitted P = 28 kW 28000 W

angular speed N = 440 rpm

angular speed in rad/s Ω = 2\(\pi\)N/60

Ω = (2 x 3.142 x 440)/60 = 46.08 rad/s

allowable shear stress τ = 80 MPa = 80 x \(10^{6}\) Pa

torque T = P/Ω = 28000/46.08 = 607.64 N-m

a)  for the minimum diameter of a solid shaft, we use the equation

τ\(d^{3}\)= \(\frac{16T}{ \pi}\)

80 x \(10^{6}\) x \(d^{3}\) =  \(\frac{16*607.64}{3.142}\) = 3094.28

\(d^{3}\) = 3094.28/(80 x \(10^{6}\)) = 0.0000386785

d = \(\sqrt[3]{0.0000386785}\) ≅ 0.034 m = 34 mm

b) For a hollow shaft with outside diameter D = 40 mm = 0.04 m

we use the equation,

T = \(\frac{16}{\pi }\) x τ x \(\frac{D^{4} - d^{4}}{D^{4} }\)

where d is the internal diameter of the pipe

607.64 =  \(\frac{16}{3.142}\) x 80 x \(10^{6}\) x  \(\frac{0.04^{4} - d^{4}}{0.04^{4} }\)

3.82 x \(10^{-12}\) = \(0.04^{4} - d^{4}\)

\(d^{4}\) = \(\sqrt[4]{2.56*10^{-6} }\)

d = 0.039 m = 39 mm

c) we assume weight is proportional to cross-sectional area

for solid shaft,

area = \(\pi r^{2}\)

r = diameter/2 = 34/2 = 17 mm

area = 3.142 x \(17^{2}\) = 907.92 mm^2

for hollow shaft, radius is also gotten as before

external area =  \(\pi r^{2}\) = 3.142 x \(20^{2}\) = 1256.64 mm^2

internal diameter =  \(\pi r^{2}\)  = 3.142 x \(19.5^{2}\) = 1194.59 mm^2

true area of hollow shaft = external area minus internal area

area = 1256.64 - 1194.59 = 62.05 mm^2

material weight saved is proportional to 907.92 - 62.05 = 845.87 mm^2

percentage weight saved is proportional to  845.87/907.92 x 100%

= 93.15%

The American society of civil engineering give us infrastructure in 2013 A grade of D

The overall score for America's infrastructure in all 16 categories is D+ in the 2013 Report Card, which is slightly higher than the D in ASCE's 2009 Report Card.

Six infrastructure sectors profited through increased private sector investment, focused initiatives in municipalities and states to perform renovations or repairs, or a one-time boost in government money.

Importantly, it is the first time the grades have increased since the American Society of Civil Engineers initially rated the quality of America's network in 1998. Nevertheless, a D+ rating remains unacceptable.

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TLDs play an important role in organizing and identifying websites on the internet, and understanding the different types of TLDs can help individuals and organizations choose the best domain name for their needs.

A top-level domain (TLD) is the last segment of a domain name, and it can be either generic or specific. Generic TLDs are used by a wide range of organizations and individuals, while specific TLDs are restricted to certain types of organizations or geographic regions. Examples of generic TLDs include .com, .org, and .net, while examples of specific TLDs include .edu, .gov, and .uk. Therefore, the answer to your question is "both generic TLD and specific TLD."

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Answer:

\(\triangle h=4.935m\)

Explanation:

From the question we are told that:

Liquid density \(\rho=800\)

Diameter of pipe \(d=4cm \approx 0.004m\)

Diameter of venture \(d=10cm \approx 0.010m\)

Specific weight of mercury P_mg \(133 kN/m^3\)

Flow rate \(r=0.05 m^3/s\)

Area A:

            \(A_1=\frac{\pi}{4}0.1^2\\A_1=0.00785m^2\\A_2=\frac{\pi}{4}0.04^2\\A_2=0.001256m^2\\\)

Generally the Bernoulli's equation is mathematically given by

\(\frac{P_1}{\rho_1g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}\\\)

Where

\(V_1=\frac{r}{A_1} \\\\ &V_1=\frac{r}{A_2}\)

Therefore

\(P_1-P_2=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})\)

Generally the equation for pressure difference b/w manometer fluid is given as

\(P_1-P_2=(p_mg-pg)\triangle h\)

Therefore

\((p_mg-pg)\triangle h=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})\)

\(\triangle h=\frac{\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})}{(p_mg-pg)}\)

\(\triangle h=\frac{\frac{(800)(0.05)^2}{2}(\frac{(0.1)^2-(0.4)^2}{(0.1)^2(0.04)^2})}{(1.33*10^3-800*9.81)}\)

\(\triangle h=4.935m\)

Therefore elevation change is mathematically given by

\(\triangle h=4.935m\)

Answer:

true

Explanation:

Answer:

I think it is false!

Explanation:

Answer: I think it's true

Explanation:

Because if you were part of a play, you would have a part but if you work on props, you don't have a part onstage.

Answer:

Hence the magnitude of the pure moment m will be \(2\times 10^5.\)

Explanation:

Now,

We know that,

\(\frac{M}{I} = \frac{E}{R}\)

Thus, \(M =\frac{EI}{R}\)

Here

R = 10000 mm  

\(I=\frac{1}{12}\times 120\times 10^{3}\\= 10^{4} mm^{4}\)

\(E=2\times 10^{5}n/mm^{2}\\\\E=2\times 10^{5}n/mm^{2}\\\\M={(2\times 10^{5}\times 10^{4})/{10000}}\\\\M=2\times 10^{5}\)

Hence, the magnitude of the pure moment m will be \(2\times 10^5.\)

Answer:

The HTML ul tag is used for the unordered list. There can be 4 types of bulleted list: To represent different ordered lists, there are 4 types of attributes in <ul> tag. This is the default style. In this style, the list items are marked with bullets.

Answer:

Ordered

\({ \tt{ < ol > .... < / ol > }}\)

Unordered

\({ \tt{ < ul > .... < /ul > }}\)

Answer:

Following are the solution to this question:

Explanation:

\(Y=Y_{NS}+Y_{ew}\)

   \(=\frac{q_{NS}}{S_{NS}}+\frac{q_{ew}}{S_{ew}}\\\\=\frac{600}{1800}+\frac{400}{1800}\\\\=\frac{6}{18}+\frac{4}{18}\\\\=\frac{3}{9}+\frac{2}{9}\\\\=\frac{1}{3}+\frac{2}{9}\\\\=\frac{3+2}{9}\\\\=\frac{5}{9}\\\\=0.55555555556\\\\=0.57\)

calculating length \((C_D)=\frac{1.5 \ L +5}{1-y}\\\\\)

                                       \(=\frac{1.5 \times (1.2) +5}{1-0.57}\\\\=\frac{1.8 +5}{0.43}\\\\=\frac{6.8}{0.43}\\\\= 15.8 \ sec\)

critical value:

\(\to V_{C_1}= 600 \ \frac{veh}{hr}\\\\\to V_{C_2}= 400 \ \frac{veh}{hr}\\\\\)

Total critical value:

\(V_C= V_{C_1}+ V_{C_2}\\\\\)

    \(= 600 + 400\\\\= 1 000 \ \frac{veh}{hr}\\\\\)

Active green time\(= 15.8-1.2=14.6\)

\(g_1=\frac{600}{1000}\times 14.6= 8.76 \ sec\\\\g_2=\frac{400}{1000}\times 14.6= 5.84 \ sec\)

Native green time:

\(G_1= 8.76 -1.2+4 =11.56 \sec\\\\ G_2= 5.84 -1.2+4 =8.64 \sec\)

The function decimalToRep that returns the representation of an integer in a give base is given as follows:

def repToDecimal(str,base):

result = 0

#initializing a dictionary

dict={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'A':10,'B':11,'C':12,'D':13,'E':14,'F':15}

for i in str:

n = dict[i]

result = base * result + n

return result

#calling the method and displaying the result

print(repToDecimal('10',10))

print(repToDecimal('10',8))

print(repToDecimal('10',2))

print(repToDecimal('10',16))

It is to be noted that a function is essentially a "chunk" of code that you may reuse instead of writing it out several times.

Programmers can use functions to break down or breakdown an issue into smaller segments, each of which performs a specific purpose.

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Answer:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}\)

Explanation:

Given

\(0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,\)

\(0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,\)

\(0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,\)

\(0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.\)

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

\(0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.\)

\(0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\)

\(0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.\)

\(0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.\)

\(0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.\)

\(0.78.\)

The 0.3's is will be plotted as thus:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}\)

The 0.4's is as follows:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}\)

The 0.5's is as follows:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}\)

The 0.6's is as thus:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}\)

Lastly, the 0.7's is as thus:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}\)

The combined steam and leaf plot is:

\(\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}\)

A contractor has a job that should be completed in 100 days. At present, he has 80 men on the job and it is estimated that they will finish the work in 130 days. Of the 80 men, 50 are each paid ₱120.00 a day, 25 at ₱180.00 a day, and 5 at ₱250.00 a day. For each day beyond the original 100 days, a contractor has to pay ₱500.00.

liquidated damages.(a) How many more men should the contractor add so that he would complete the work on time?In the first case, we see that the contractor already has 80 men and they are working for 130 days to complete the job. So, we can use the following formula to determine the additional number of workers required to finish the work in 100 days.

b) If of the additional men, 2 are paid ₱180.00 a day, and the rest at ₱120.00 a day, would the contractor save money by employing more men and not paying the fine Let’s assume that the contractor adds 440 workers, of which 2 are paid ₱180.00 a day and the rest are paid ₱120.00 a day.

The total cost of the new workers is, therefore, ₱9600.00 + ₱4500.00 + ₱49800.00 = ₱63,900.00.The cost of liquidated damages would be calculated as follows:  $$LD = (130-100) \cdot 500 = ₱15,000.00$$.

Therefore, the contractor would save money if he employs more men and not pays the fine. The contractor’s savings would be:$$Savings = LD - Additional cost$$$$= 15000.00 - 63900.00 $$$$= -48900.00$$

Thus, we can see that the contractor would save ₱48,900.00 by employing more men and not paying the fine.

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Answer:

Voltage Regulator

Technician A is correct.

Explanation:

Technician B is not correct.  The voltage regulator is not installed between the output terminal of the alternator and the positive terminal of the battery as claimed by Technician B.  Technician A's opinion that the voltage regulator controls the strength of the rotor's magnetic field is correct.  The computer can also be used to control the output of the alternator by controlling the field current.

OSHA 300 Logs and OSHA 300A SummariesWorkers have the right to review OSHA 300 Logs and OSHA 300A Summaries.

These logs and summaries are part of the OSHA recordkeeping requirements and provide information about work-related injuries and illnesses in the workplace. The OSHA 300 Log is a detailed record of all reportable injuries and illnesses, while the OSHA 300A Summary is a summary of the annual totals from the OSHA 300 Log. Workers have the right to access and review these records to ensure transparency and awareness of workplace safety issues. However, it's important to note that specific rights may vary depending on the jurisdiction and applicable regulations.

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Answer:

A. 336.3 ft.

B. 279.83 ft.

Explanation:

 The formula for Stopping Site Distance SSD is given s follows;

\(SSD=\dfrac{2 \times R \times \theta \times \pi}{180}\)

\(R = \dfrac{M}{1 - cos\theta}\)

R = 5730/D = 5730/10 = 573 ft

\(\theta = \dfrac{SSD \times D}{200}\)

D = Degree of curvature

θ = (SSD×D)/20 = (28.65×SSD)/573

Where:

\(\dfrac{M}{R} = 1 - cos \left(\dfrac{28.65 \cdot SSD}{R} \right)\)

D = 10°

M is the distance of the obstruction from the highway = 24.5 ft

\(\dfrac{24.5}{573} = 1 - cos \left(\dfrac{28.65 \cdot SSD}{573} \right)\)

\(cos \left(\dfrac{28.65 \cdot SSD}{573} \right) = 1 - \dfrac{24.5}{573} =\dfrac{1097}{1146}\)

\(cos^{-1} \left(\dfrac{1097}{1146}\right) = \dfrac{28.65 \cdot S}{573} =16.815\)

573×16.815= 28.65×SSD

SSD = 9635.13/28.65 = 336.3 ft

B. When M = 17, we have

\(\dfrac{17}{573} = 1 - cos \left(\dfrac{28.65 \cdot SSD}{573} \right)\)

\(cos \left(\dfrac{28.65 \cdot SSD}{573} \right) = 1 - \dfrac{17}{573} = \dfrac{556}{573}\)

\(cos^{-1} \left( \dfrac{556}{573}\right) = \dfrac{28.65 \cdot SSD}{573} = 13.99 \approx 14\)

SSD = 573×14/28.65 = 279.83 ft.

An 8-bit ring counter is required to be designed, where its states are 0xFE, OXFD, 0x7F. The requirement is to use only two 74XX series ICs and no other components.

If the ring counter starts in an invalid state, it must be self-correcting. This is an interesting problem to be solved. Ring counters are also known as circular counters or shift registers. The counters move from one state to another by shifting the data in the counter. The given sequence is 0xFE, OXFD, 0x7F.

These are the hexadecimal equivalent values of 1111 1110, 1111 1101, and 0111 1111, respectively. These values are the previous states of the counter when it shifts to the next state. To start the counter, any state value can be used. But it must be ensured that it is a valid state. That is the state value must be one of the given sequence values,

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Answer:

c. The tension in the cord connected to the truck is greater than 1200 lb

e. The normal force between A and B is 1200 lb.

Explanation:

The correct question should be

A 1000 lb boulder B is resting on a 200 lb platform A when truck C accidentally accelerates to the right (truck in reverse). Which of the following statements are true (select two answers)?

A free body diagram is shown below.

The normal force between the the boulder and the platform will be the sum of the force, i.e 1000 lb + 200 lb = 1200 lb

For the combination of the bodies to accelerate upwards, then the tension must be greater than the normal force, i.e T > 1200 lb

Answer:

Space Rovers

Explanation:

The engineering advancement of the invention space rovers has changed space exploration tremendously. The first space rover ever to go out onto another planet was made by NASA and put on mars to learn more about the red planet and see if there where any plausible signs of past or maybe even present life. This mission was incredibly successful and for the first time in space exploration history, there was a spacecraft that you could move around that could with do the conditions of the planet for weeks, months, sometimes years at a time, moving around taking pictures and constantly collecting data. This was revolutionary for scientists, this way they could get copious amounts of information just on rover and not having to send multiple crafts only to die in the first few hours but sending back a relative amount of information. With rovers you can have them collect samples and bring the back to earth, not something  that could’ve been done without a human, until rovers. They have completely changed and improved collecting data from other planets and have been one of the most revolutionary inventions when it comes to space exploration.

This is what I put for the answer, hope it helps in some way.

To achieve 100% statement coverage, additional tests are needed to cover different combinations of drink preferences (milk and sugar). For 100% decision coverage, tests should cover both the selection of drink type and the decisions related to adding milk and sugar.

a. Control Flow Diagram:

Start

|

V

Choose Drink Type (Hot or Cold)

|

V

IF Hot Drink

|   |

|   V

|   Ask for Milk Preference

|   |

|   V

|   IF Milk Required

|   |   |

|   |   V

|   |   Add Milk

|   |   |

|   |   V

|   |   Ask for Sugar Preference

|   |   |

|   |   V

|   |   IF Sugar Required

|   |   |   |

|   |   |   V

|   |   |   Add Sugar

|   |   |   |

|   |   |   V

|   |   V

|   V

|   Dispense Hot Drink

|

V

ELSE (Cold Drink)

   |

   V

   Dispense Cold Drink

|

V

End

b. Given the tests:

Test 1: Cold drink

Test 2: Hot drink with milk and sugar

Statement Coverage achieved: The statement coverage achieved would be 10 out of 15 statements (66.7%).

Decision Coverage achieved: The decision coverage achieved would be 2 out of 3 decisions (66.7%).

c. Additional tests for 100% statement coverage:

Test 3: Hot drink without milk and sugar

Test 4: Hot drink with milk only

Test 5: Hot drink with sugar only

Test 6: Hot drink without milk and without sugar

Additional tests for 100% decision coverage:

Test 7: Cold drink

Test 8: Hot drink with milk and sugar

Test 9: Hot drink without milk and sugar

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In the two scenario above concerning port fuel-injection and sequential fuel injectors, only Technician b is correct.

A vehicle often has one fuel injector per cylinder and if one has a four-cylinder car, it would also have four fuel injectors.

Note that Fuel-pressure regulators that are found on a port fuel-injected systems is one that often work with injector pressures that is said to be of 30 to 55 PSI.

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The exhibit provided does not contain enough information to determine the location from where the syslog messages have been retrieved.

However, based on the context of the question, it is likely that the syslog messages have been retrieved from a syslog server or client.

Syslog is a standard protocol used to send system messages, such as error messages, warning messages, and informational messages, to a centralized logging server or client. This allows network administrators to collect and monitor system events from different network devices in one central location, making it easier to troubleshoot and diagnose network issues. Therefore, it is common for syslog messages to be retrieved from a syslog server or client. However, the specific location of the syslog messages cannot be determined from the exhibit alone.

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To determine whether the project specifications have been met, the contractor must perform a Proctor test and a sand-cone test.

Projects are tasks, activities, or operations that are designed to accomplish a specific goal within a given period of time.

The Proctor test measures the compaction of soil and the sand-cone test measures the volume of soil that can be filled in a given area.
The Proctor test data provided shows that the mass of compacted soil + mold ranges from 240.29 kg to 263.45 kg. The mass of the can + dry soil ranges from 19.81 kg to 20.99 kg. The mass of the compaction mold ranges from 2.031 kg to 2.031 kg. This data indicates that the contractor has achieved a relative compaction of 90%.
The sand-cone test data provided shows that the initial mass of the sand-cone apparatus was 5.912 kg, the final mass of the sand-cone apparatus was 2.378 kg and the mass of soil recovered from the hole was 2.883 kg. The moisture content of the soil from the hole was 7% and the density of the sand was 1300 kg/m3. This data indicates that the contractor has achieved a volume of 1.114x10^3 m3, which is within the project specifications.
Based on the data provided, it can be concluded that the contractor has met the project specifications and has achieved the required relative compaction of 90%. If the contractor has not met the project specifications, then a course of action should be taken to address the issue. This could include additional tests to verify the compaction and volume of soil, adjusting the compaction process, or additional backfilling and compaction.

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The probability that one telephone call passes through the switchboard in three minutes is 1.49%.

Using the Poisson probability formula, we can calculate the probability of exactly one call passing through the switchboard in 3 minutes as follows:

P(X = 1) = (e^(-6) * 6^1) / 1!

Where

6 = 2 per minute * 3 minute

X is the number of calls passing through the switchboard in 3 minutes.

So, we have

P(X = 1) = (e^(-6) * 6^1) / 1!

= (0.00248 * 6) / 1

= 0.0149

Therefore, the probability that one telephone call passes through is approximately 0.0149, or about 1.49%.

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Answer:

Carbon Monoxide.

The Boolean OR operation is also known as: A. Boolean sum

The Boolean OR operation is also known as the Boolean sum. This  Boolean OR operation returns a value of true if at least one of its inputs is true, and false if both inputs are false. According to Boolean Logic, all operations are true or false, i.e., yes or no. All relationships between operations can be represented using logical operators such as AND, OR, or NOT. The branch of mathematics known as Boolean algebra deals with operations on logical values with binary variables.

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The size conductor that should be used to supply each motor is 12 AWG and the size conductor  that should be used for the feeder conductor is 1/0AWG.

a. Using NEC table 430.250

Running connect for 150HP  connected to 460V = 21 A

Conductor size=25% current to 125% current (Increase)

Increase current=21×125/100

Increase current=26.25 A

Nearest standard conductor size using NEC table 310.15(B)(16)=25 A

Since the motor contain NEMA design code the conductor will be choosing from 75° column which means that the size conductor that should be used to supply each motor is 12AWG copper conductors.

b. Feeder conductor

Using NEC section 430.24  plus ampere rating of other motors

Running current

Using NEC Table 430.250

Running current=26.25+21+21+21+21+21

Running current=131.25 A

The value that is greater than 131.25 A  under  75° column of NEC Table 310 (B)(16) is 150 A which means that the size conductor that should be used for the feeder conductor is 1/0 AWG conductors.

Therefore the size conductor that should be used to supply each motor is 12 AWG and the size conductor  that should be used for the feeder conductor is 1/0AWG.

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