Let us assume X is the input dataset, and Y is the output variable. Which of the following statements iscorrect? import pandas as pd import numpy as npfrom sklearn. linear_model importLinearRegressionmodel = LinearRegression ( )model. fit(X, Y)model. coef. • A. The model must be fit to access themodel coefficients B. The model coefficients can beaccessed before fitting the model C. It is not necessary to create a LinearRegression() instance to fit themodelD. None of the options

Answer:

Parasite drag and induced drag.

Answer:

peak value of the rectified voltage across the load is 12.7667 V

Explanation:

given data

load operates = 1 K

solution

we ger here peak value of the rectified voltage across the load that is express as

Vo = V (secondary )  - 2 Vd   ....................1

we get here input voltage of transfer that is

V (primary) = √2 × vrms   ..................2

V (primary) = √2 × 120

V (primary) = 169.70   =   170 V

and

V (secondary ) = V (primary) × \(\frac{1}{12}\)       ................3

V (secondary ) = 170× \(\frac{1}{12}\)     = 14.167 V

so

put value in eq 1 we get

Vo = 14.167  - 2 (0.7)

Vo = 12.7667 V

Answer:

b. abc

Explanation:

The phase rotation and its sequence is order in which voltage is reached to their respective sources. The wave forms of a polyphase flow through AC source to the panel. The three phase system has two possible sequence which can be a-b-c or c-b-a. The correct answer is therefore b.

Solution :

Taken moment about point B;

\($\sum M_B = 0$\)

\($\left[(w)(4)(2)-N_A \sin 30^\circ (3)(\sin 30^\circ)-N_A \cos 30^\circ(3 \cos 30^\circ +4) \right] =0$\)

\($N_A = 1.2376w$\)

\($\sum F_x = 0$\)

\($N_A \sin 30^\circ - B_x = 0$\)

\($B_x = 1.2376w \times \sin 30^\circ$\)

\($B_x = 0.6188w$\)

\($\sum F_y = 0$\)

\($B_y+N_A \cos^\circ - (w \times 4) = 0$\)

\($B_y+(1.2376w) \cos 30^\circ - (w \times 4)=0 $\)

\($B_y = 2.9282w$\)

Now calculating the resultant force at B,

\($F_B= \sqrt{B_x^2+B_y^2}$\)

\($F_B= \sqrt{(0.6188w)^2+(2.9282w)^2}$\)

     = 2.9929w

For no failure,

\($N_A<4 \ kN $\)

2.9929 w < 4 kN

\($w < 3.232 \ kN/m$\)

And,

\($F_B < 8 \ kN$\)

2.9929 w < 8 kN

w < 2.673 kN/m

For the safe operation, w = 2.673 kN/m  

Answer:

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

  \(X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915\times10^{-3}\approx j4.99984\,\Omega\)

The capacitive reactance is ...

  \(X_C=\dfrac{1}{j\omega C}=\dfrac{-j}{100\pi\cdot 318.31\times10^{-6}F}\approx -j10.00000\,\Omega\)

Branch 1

The impedance of branch 1 is ...

  Z1 = 8 +j4.99984 Ω

so the current is ...

  I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

  Z2 = 5 -j10 Ω

so the current is ...

  I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

  It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

  It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

__

The phasor diagram of the currents is attached.

_____

Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

Answer:

12 mins

Explanation:

The summation of the forces in vertical direction

= Fb - Fd - w = 0 ∴ Fd = Fb - w  ----- ( 1 )

Fb ( buoyant force ) = Pair * g * Vballoon ------- ( 2 )

Pair = air density , Vballoon = volume of balloon

Vballoon = \(\frac{\pi D^3}{6}\) ,  where D = 4  ∴  Vballoon = 33.51 ft^3

g = 32.2 ft/s^2

From property tables

Pair = 2.33 * 10^-3 slug/ft^3

μ ( dynamic viscosity ) = 3.8 * 10^-7 slug/ft.s

Insert values into equation 2

Fb = ( 2.33 * 10^-3 ) * ( 32.2 ) *( 33.51 )  = 2.514 Ib

∴ Fd = 2.514 - 2.5 = 0.014 Ib ( equation 1 )

Assuming that flow is Laminar  and  RE < 1

Re = (Pair * vd) / μair   -------- ( 3 )

where:  Pair = 2.33 * 10^-3 slug/ft^3 ,  vd = ( 987 * 4 ) ft^2/s ,   μair = 3.8 * 10^-7 slug/ft.s

Insert values into equation 3

Re = 2.4 * 10^7 (  this means that the assumption above is wrong )

Hence we will use drag force law

Assume Cd = 0.5

Express  Fd using the relation below

Fd = 1/2* Cd * Pair * AV^2

therefore V = 1.39 ft/s

Recalculate  Reynold's number using  v = 1.39 ft/s

Re = 34091

from the figure Cd ≈ 0.5 at Re = 34091  

Finally calculate the rise time ( time taken to reach an altitude of 1000 ft )

t = h/v

  = 1000 / 1.39 = 719 seconds ≈ 12 mins

False. Secondary clearance may be required on any size cutter depending on the specific machining operation and the material being machined.

Secondary clearance is not only required on large cutters. It is an essential feature for various cutting tools, regardless of their size, to reduce friction and heat generation during the cutting process. This clearance ensures smooth operation and extends the tool's life. A cutting tool or cutter is typically a hardened metal tool used in machining to cut, shape, and remove material from a workpiece using machining tools and abrasive tools through shear deformation. The majority of these instruments are made just for metals. models are Texture scissors, kitchen shears, spring stacked scissors, pruning shears, paper trimmers, make blades, string trimmers, and turning shaper. A cutting device is a sharp instrument mounted in a machine apparatus and utilized for cutting materials. Machines and processing machines utilize various sorts of cutting instruments. Because of their hardness, diamonds can be used in cutting tools to cut other hard materials.

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False. Depending on the precise machining procedure and the material being machined, secondary clearance may be necessary for any size cutter.

Not only large cutters need secondary clearance. Reduced friction and heat generation during the cutting process is a crucial attribute for all cutting tools, regardless of size. The tool's life is increased and its operation is ensured by this clearing. When cutting, shaping, or removing material from a workpiece using machining tools and abrasive tools through shear deformation, a cutting tool or cutter is often a hardened metal instrument. Most of these instruments were created specifically for metals. Texture shears, culinary shears, spring-stacked shears, pruning shears, paper shears, create blades, string shears, and turning shaper are examples of models.

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The displacement of end D with respect to end A is \(-0.488 \times 10^{-3}\;m\)

Given the following data:

To determine the displacement of end D with respect to end A:

First of all, we would do a sectional cut of the circular rods at point A and then determine the sum of forces acting on it:

\(\sum F=0\\\\ 10000+N_1=0\\\\N_1 = -10000\;N\)

Mathematically, the displacement of a circular rod is given by this formula:

\(d = \frac{NL}{AE}\)

Where:

Substituting the given parameters into the formula, we have;

\(d_1 = \frac{-10000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_1 = \frac{-4350}{0.0003142 \times 68.9 \times 10^9}\\\\d_1 =-0.201 \times 10^{-3}\;meter\)

For section cut BC:

\(\sum F=0\\\\ 10000-20000+N_2=0\\\\N_2 = 10000\;N\)

\(radius = outer\;radius^2 - inner\;radius^2\\\\radius = 0.04^2 -0.03^2=0.0007\;m\)

\(d_2 = \frac{10000 \times 0.435}{\frac{3.142 \times 0.0007^2}{4} \times 68.9 \times 10^9} \\\\d_2 = \frac{4350}{0.00054985 \times 68.9 \times 10^9}\\\\d_2 =0.115 \times 10^{-3}\;meter\)

For section cut D:

\(\sum F=0\\\\ -20000-N_3=0\\\\N_3 =- 20000\;N\)

\(d_3 = \frac{-20000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_3 = \frac{-8700}{0.0003142 \times 68.9 \times 10^9}\\\\d_3 =-0.402 \times 10^{-3}\;meter\)

For the total displacement:

The total displacement is equal to the displacement of end D with respect to end A.

\(d_T = d_1 +d_2+d_3\\\\d_T = -0.201 \times 10^{-3} + 0.115 \times 10^{-3}-0.402 \times 10^{-3}\\\\d_T = -0.488 \times 10^{-3}\;m\)

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Answer: Application.

Explanation:

The question on wether to contine the use of cadavers in the lab for test is being centered around its application. Cadaver which is same as a corpse or dead body is used in crash site during automobil test in lab, some of this cadavers are been disrespected with their applications in the automobile industries because many didn’t consent to be used in those experiments or test.

The dependability of a measurement is also indicated by error bars based on confidence intervals (C.I. ), which are a component of inferential statistics. These error bars show a range that would, 95% of the time, reflect the population mean.

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Answer:

First, we need to calculate the volume of concrete needed to line the canal. The volume of a cylinder is given by the formula:V = πr^2hWhere V is the volume, r is the radius of the cylinder, and h is the height (length) of the cylinder.Since we are lining a canal, the cross-sectional shape of the canal is a rectangle, not a circle. However, we can still use the formula for the volume of a cylinder by treating the width of the canal as the radius and the height of the canal as the length.In this case, the width of the canal is w and the height is h, so the volume of the concrete needed to line the canal is:V = πw^2hWe are given that the length of the canal is 227 ft, so h = 227 ft. We are not given the width of the canal, so we cannot use the formula to calculate the volume of concrete needed.Next, we need to calculate the volume of concrete needed with the 12% allowance for waste and overexcavation. The volume of concrete needed with the allowance is the original volume plus 12% of the original volume.The volume of concrete needed with the allowance can be calculated using the following formula:V' = V + (V * 12%)Where V' is the volume of concrete needed with the allowance and V is the original volume.Since we do not know the original volume, we cannot use this formula to calculate the volume of concrete needed.To solve this problem, we need to know the width of the canal in order to calculate the volume of concrete needed. Without this information, it is not possible to determine the volume of concrete that must be delivered.

Explanation:

Answer:

999999999999999999999

The uncompressed values from given data is s = 10 p(s) = 1/16

compressed file                                                                  uncompressed file

l=-(0.5 log 2 0.5)...= h bit/sample                                         2 bits sample

100,000 bits                                                                                     x bits

x bits ÷ 100000 bits = 2 bits/sample ÷ h bits sample

A data stream is the transmission of a series of coherent signals that have been digitally encoded to carry information. The sent symbols are often organized into a number of packets.

Data streaming is now commonplace. Any communication sent via the Internet is done so as a data stream. When speaking on a phone, the sound is transmitted as a data stream. Depending on the data format selected, Data Stream comprises several sorts of data. When determining the time of an event, attributes like the Timestamp attribute are helpful. An algorithmically encoded ID called "Subject ID" has been taken out of a cookie.

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When the process is in control but does not meet specification, it is referred to as a special cause error.

In statistical process control, a process is considered to be in control when it operates within the defined limits and shows only random variations. However, when a process is in control but does not meet the desired specifications, it indicates the presence of a special cause error.

Special cause errors are attributed to specific factors or events that cause the process to deviate from the expected outcome. These errors are typically unpredictable and require investigation and corrective action to bring the process back within the desired specifications.

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Answer:

D) All of the above.

The two ways through which a computer model is likely to be used by an engineer in order to help refine a design are as follows:

Thus, the correct options for this question are A and D.

A Computer model may be defined as a type of computer program that significantly runs on a computer that typically develops a model, or simulation, of a real-world feature, phenomenon, or any other event.

According to the context of this question, an engineer would try to perform the ways in order to support the refining of the design through the help of calculating the possible costs of building a design and the run simulations to test a problem with the design.

Therefore, the correct options for this question are A and D.

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As a general overview, the term "Weld Center Vertices" typically refers to a feature or option that is used in 3D modeling or computer-aided design (CAD) software.

This feature is usually used in conjunction with a "Fill Hole" mode, which is a tool that is used to fill in holes or gaps in 3D models.

When the "Weld Center Vertices" option is enabled in a "Fill Hole" mode, the software will attempt to connect the vertices or points around the hole by creating a new surface or face that is centered on the vertices. This can be useful for creating a more uniform and seamless 3D model, particularly when dealing with complex shapes or irregular surfaces.

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Therefore, the final pressure of the gas is 1.52 MPa (megaPascals).

We can use the ideal gas law to solve this problem:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to find the initial number of moles of gas:

n1 = (P1 V1)/(R T1)

= (300 x 10^3 Pa) x (0.3 m^3)/(8.31 J/(mol K) x (20 + 273) K)

= 4.97 mol

where R = 8.31 J/(mol K) is the gas constant.

Next, we can use the fact that the process is isothermal (i.e., at constant temperature) to find the final volume of the gas after it is compressed to a pressure of 800 kN/m^2:

P1 V1 = P2 V2

V2 = (P1 V1)/P2

= (300 x 10^3 Pa) x (0.3 m^3)/(800 x 10^3 Pa)

= 0.1125 m^3

Now we can use the fact that the process is adiabatic (i.e., no heat is exchanged with the surroundings) and that the initial and final volumes are the same to find the final pressure of the gas:

P1 V1^γ = P2 V2^γ

where γ is the adiabatic index (a property of the gas), which depends on the specific gas. For simplicity, we will assume that γ = 1.4, which is a reasonable value for diatomic gases such as nitrogen and oxygen.

P2 = P1 (V1/V2)^γ

= (300 x 10^3 Pa) x (0.3 m^3/0.1125 m^3)^1.4

= 1.52 x 10^6 Pa

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Answer:

2000-7000 * 6% = - 0,000126

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

   Bearing life:\(L_1 = L , L_2 = 2L\)

   Catalog rating: \(C_1 = C , C_2 = ? ,\)

From given equation bearing life equation,

\(F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)\)

we Dividing eqn (2) with (1)

\(\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C\)

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

\(R_1 = 0.9 , R_2 = 0.99\)

Now calculating life adjustment factor for both value of reliability from Weibull parametres

\(a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}\)

\(= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968\)

Similarly

\(a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215\)

Now calculating bearing life for each value

\(L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L\)

Now using given ball bearing life equation and dividing each other similar to previous problem

\(\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C\)

Catalog rating increased by factor of 0.61

At amateur operating frequencies, the FTDX10 is designed for 50 Ohm resistive impedance.

Impedance is the sum of resistance and reactance. It was defined as anything that can obstructs the flow of electrons within the electrical circuit. As a result, it influences current generation in the electrical circuit. It could be found in all the possible circuit components and across all possible electrical circuits. Impedance is represented mathematically by the letter Z and has the unit ohm. It's a mix of resistance and reactance.

Z stands for impedance, which is an expression of the resistance to alternating and/or direct electric current that an electronic component, circuit, or system offers. Resistance and reactance are two independent scalar (one-dimensional) phenomena that make up the vector (two-dimensional) quantity known as impedance.

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The dependency on hunting decreased from Stone Age to the Bronze Age is because in Bronze Age, people learned to grow crops and keeping domestic livestock, so there was no  reason for hunting.

Stone Age is the age when humans started making tools of stones and used them in hunting and cutting. It was the prehistoric period.

Bronze Age is after the Stone Age where humans know about metals started to use bronze. It is the period between 3300 BC to 1200 BC.

Thus, the dependency on hunting decreased from Stone Age to the Bronze Age is because in Bronze Age, people learned to grow crops and keeping domestic livestock, so there was no  reason for hunting.

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Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

\(\tau_{ave} = \frac{T}{2tA_{m}}\)

\(A_m\) = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, \(A_m\) = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

\(\tau_{ave} = \frac{T}{2tA_{m}}\)

Plugging in then values, gives;

\(\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa\)

The shear stress will be 80,000,000 Pa or 80 MPa.

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

   class Program

   {

       static void Main(string[] args)

       {

           int[] Number = new int[10];  // Declare array of 10 elements

           //Accept Input

           int counter = 0;

           while(counter<10)

           {

               Console.WriteLine("Enter Number " + (counter + 1)+": ");

               string var = Console.ReadLine();

               Number[counter] = Convert.ToInt32(var);

               counter++;                  

           }

           //Initialize largest to first element of the array

           int Largest = Number[0];

           //Determine Largest

           for(int i=0;i<10;i++)

           {

               if(Largest < Number[i])

               {

                   Largest = Number[i];

               }

           }

           //Print Largest

           Console.WriteLine("The largest input is "+ Largest);

           Console.ReadLine();

       }

   }

}

Answer:

Plzzzz answer fast......

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

the speed Vc of the current and its direction measured clockwise from the north is 0.71 m/s and 231.02° respectively

Explanation:

Given the data in the question and as illustrated in the diagram below;

The absolute velocity of the ship Vs is 6 Knots due east

so we convert to meter per seconds

Vs = 6 Knots × \(\frac{0.51444 m/s}{1 Knots}\) = 3.0866 m/s

Next we determine the relative velocity of the ship Vs/c

Vs/c = AB / t

given that distance between A to B = 10 nautical miles which requires 2 hours

so we substitute

Vs/c = 10 nautical miles / 2 hrs

Vs/c = [10 nautical miles × \(\frac{1852 m}{1 nautical-miles}\) ] / [ 2 hrs × \(\frac{3600s}{1hr}\) ]

Vs/c = 18520 / 7200

Vs/c = 2.572 m/s

Now, from the second diagram below, { showing the relative velocity polygon }

Now, using COSINE RULE, we calculate the velocity current.

Vc = √( V²s + V²s/c - 2VsSs/ccos10 )

we substitute

Vc = √( (3.0866)² + (2.572)² - (2 × 3.0866 × 2.572 × cos10 ) )

Vc = √( (3.0866)² + (2.572)² - (2 × 3.0866 × 2.572 × 0.9848 ) )

Vc = √( 9.527099 + 6.615184 - 15.6361 )

Vc = √0.506183

Vc = 0.71 m/s

Next, we use the SINE RULE to calculate the direction;

Vc/sin10 = Vs/c / sinθ

we substitute

0.71 / sin10 = 2.572 / sinθ

0.71 / 0.173648 = 2.572 / sinθ

4.0887 = 2.572 / sinθ

sinθ  = 2.572 / 4.0887

sinθ = 0.62905

θ = sin⁻¹( 0.62905 )

θ = 38.98°

So, angle measured clock-wise will be;

θ = 270° - 38.98°

θ = 231.02°

Therefore, the speed Vc of the current and its direction measured clockwise from the north is 0.71 m/s and 231.02° respectively

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

Article 430 that is found in  National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

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Answer:

B. Larger than that of the larger-diameter cylinder

Explanation:

By looking at the equation for the Nusselt number for a cylinder in cross flow, Nu = hd/k, and assuming both cylinders are made of the same material, you can see that the heat transfer coefficient h will have to be much larger for the smaller cylinder. You can verify this by using random numbers for each variable (shown below).

Verification:

ds = 1

db = 10

k = 12

Nu = 10

h small = ?

h big = ?

Nu = hd/k

10 = h small * ds / 12

120 = 1 * h small

h small = 120

Nu = hd/k

10 = h big*db / 12

120 = 10 * h big

h big = 12

This shows that the heat transfer coefficient for the smaller diameter, ds, must be bigger than the heat transfer coefficient for the larger diameter, db.

For two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same. The average heat transfer coefficient for the smaller-diameter cylinder is " Smaller than that of the larger-diameter cylinder." (Option C)

The given information states that for two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same.

The Nusselt number is a dimensionless number that relates the convective heat transfer rate to the conductive heat transfer rate.

Since the Nusselt number is the same for both cylinders, and the heat transfer coefficient is directly related to the Nusselt number, the smaller-diameter cylinder will have a smaller average heat transfer coefficient compared to the larger-diameter cylinder.

Thus option C is the right answer.

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https://brainly.com/question/16055406

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Answer:

Top Final year projects for civil engineering students

• Geographic Information System using Q-GIS. ...

• Structural and Foundation Analysis. ...

• Construction Project Management & Building Information Modeling. ...

• Tall Building Design. ...

• Seismic Design using SAP2000 & ETABS.

Explanation:

Hope it's help