Q2: Determine the number of 8-inch-high by 8-inch-wide by 16-inch-long concrete blocks required to complete the wall in Figures below. The overhead doors are 10 feet wide by 12 feet high. If lintel blocks are required wherever the #4 horizontal bars are located and above the doors, how many plain blocks and how many lintel blocks are needed for the wall?

Answer:

C. Both technician A and B

Explanation:

If the master cylinder is overfilled it will not allow enough room for the brake fluid to expand due to heat expansion. This blocks the vent port.  If a vent port is not open, brake fluid pressure will increase as brakes heat up.  This will cause the brakes to self apply, cause more heat in the brake fluid and the vehicle will slow down.

There, we can conclude that Both technician A and B are correct.

Answer:

C. Both technician A and B

Explanation:

The event that made both cylinders to be over filled especially the master cylinder and the blocking of the vent port, this will cause the vehicle brake to apply itself after just a little motion of the vehicle.

Therefore both technicians are correct from the information given above.

Hence, we can boldly say the correct answer is C. ie Both technician A and B

Answer:

a) The mass flow rate is 19.71 kg/s

b) The inlet area is 0.41 m²

c) The thrust power is 333.31 kW

d) The propulsive efficiency is 26.7%

Explanation:

Please look at the solution in the attached Word file.

Answer:

∅=Ф  

P = W sin([tex]\alpha[/tex] + Ф)

Explanation:

First, we'll isolate and draw the free-body diagram of the pipe  

Note that since the pipe is moving, the friction force is equal to the product of normal reaction force and the kinetic coefficient of friction  

F = F_max = u_kN  

Also note that the weight makes with the y-axis angle a because the x-axis makes the same angle with the horizontal  

The expression for angle of friction is:

B = tan-1 (u_k)

From here we can express the coefficient of friction as:

u_k = tan(Ф)

Replace u_s by tan(Ф) in the expression for the friction force

F = N tan(Ф)  

diagram is attached

By equating sum of forces in y-direction to zero, we can write the expression for the normal reaction force  

ΣF_y = 0

N — W cos[tex]\alpha[/tex]- P sin Ф= 0

From here we can express N as:

N = W cos[tex]\alpha[/tex] -— P sin Ф

Replace N by the expression above in the expression for friction force F(written in step 1)  

F = (W cos[tex]\alpha[/tex]  — P sin  Ф) tan( Ф)                                 (1)  

Now, we'll equate sum of forces in x-direction to zero  

ΣF_x = 0

-F - W cos[tex]\alpha[/tex]  + P sin  Ф =0

Replace F by expression (1)  

— (W cos[tex]\alpha[/tex]  — P sin Ф) tan(Ф) — W sin[tex]\alpha[/tex]+pcosФ=0

-W cos [tex]\alpha[/tex] tan(Ф) + P sin Ф tan(Ф) — W sin[tex]\alpha[/tex] +pcosФ=0

P(sin Ф tan(Ф) + cosФ) — W(cos [tex]\alpha[/tex] tan(Ф) + sin [tex]\alpha[/tex])

From here we can express the force P needed to pull the pipe as:

P = W(cos[tex]\alpha[/tex]  tan(Ф) + sin[tex]\alpha[/tex])/sinФ*tansФ+cosФ                    (2)

All we have to do now is to simplify the expression (2). We'll start by sin replacing tan(Ф) with sinФ/cosФ

P = W(cos *sinФ/cosФ + sin)/sinФ*sinФ/cosФ+cosФ *cosФ/cosФ

We can multiply the right side of equation by cosФ/cosФ

P = W(cos[tex]\alpha[/tex] *sinФ + sin[tex]\alpha[/tex]cosФ)/sin∅*sinФ+cos∅cosФ *cosФ/cosФ

Finally, we'll replace (cos[tex]\alpha[/tex] *sinФ + sin[tex]\alpha[/tex]cosФ) by sin([tex]\alpha[/tex] + Ф) and (sin∅ sinФ + cos∅ cos Ф) by cos( ∅— Ф)

P wsin([tex]\alpha[/tex] + Ф) /cos(∅ — Ф)                                              (3)  

Since the first derivative of the function is actually tangens of the angle which tangent makes with the x-axis, we'll find it by equating the first derivative by zero(this means that the tangent of the function is horizontal, i.e. that the function is at its maximum or minimum)  

Note that the variable in the expression (3) is 0, since both B and a are known  

dP/d∅ =d/d∅ [sin(Ф+)/cos(∅-Ф) ]

Note that sin(Ф+[tex]\alpha[/tex]) is constant since both Ф and a are known  

dP/d∅ = sin(Ф+[tex]\alpha[/tex]) d/dФ [1/cos(∅-Ф) ]  

Next, we'll apply the reciprocal rule  

= -dP/d∅[cos(∅-Ф)]/cos^2(∅-Ф)*sin(Ф+[tex]\alpha[/tex])

Next, we'll apply the differentiation rule  

=(-sin(∅-Ф))*d/d∅[∅-Ф]*sin(Ф+[tex]\alpha[/tex])/cos^2(∅-Ф)

=(d/d∅[∅]+d/d∅[-∅])*sin(Ф+[tex]\alpha[/tex])sin(∅-Ф)/cos^2(∅-Ф)

dP/d∅ =sin(Ф+[tex]\alpha[/tex])*sin(∅-Ф)/cos^2(∅-Ф)                       (4)

Next step will be to equate the expression (4) to zero, to determine the value of # when the function is minimum  

sin(Ф+[tex]\alpha[/tex])*sin(∅-Ф)/cos^2(∅-Ф) =0  

Note that sin(Ф+[tex]\alpha[/tex]) is constant, so in order for the equation above to be correct, sin(∅-Ф) needs to be equal to zero  

sin(∅-Ф)  = 0

Since sin 0° = sin 180° = 0, two possible solutions for ∅ are:

∅-Ф=0                           Ф=∅  

or  

∅-Ф = 180°                    ∅ = 180° +  Ф

Since the function for P is only good over the range 0 <  ∅ < 90°, since when > 90° the friction force will change its direction, we can conclude that the minimum force P is required to move the pipe at angle:  

∅=Ф  

Finally, replace # by 8 in expression (3) to determine the minimum force P required to move the pipe

P = W sin([tex]\alpha[/tex] + Ф ) / cos ∅ —  ∅)  

P = W sin([tex]\alpha[/tex] + Ф)

Explanation:

a) The total volume equals the sum of the volumes.

500 = x + y

The total octane amount equals the sum of the octane amounts.

89(500) = 87x + 92y

44500 = 87x + 92y

b) desmos.com/calculator/ekegkzllqx

As x increases, y decreases.

c) Use substitution or elimination to solve the system of equations.

44500 = 87x + 92(500−x)

44500 = 87x + 46000 − 92x

5x = 1500

x = 300

y = 200

The required volumes are 300 gallons of 87 gasoline and 200 gallons of 92 gasoline.

Find the answer in the attachment

It is given that :

Let the mean bulk temperature [tex]$=\frac{50+100}{2}$[/tex]

                                                    [tex]$=75^\circ C$[/tex]

From the property table at 1 bar and [tex]$75^\circ C$[/tex],

[tex]$K=0.02917 \ W/\mu K, \ Pr = 0.71055 $[/tex]

Flow is laminar as Re = 4000 for laminar.

Flow Nusselt Number is given by :

[tex]$\overline{Nu} = 0.664 (Re)^{0.5} Pr^{1/3} = \frac{hd}{K}$[/tex]

[tex]$\theta = 4 \times 0.2 \times 0.1 \times (100-50)$[/tex]

  [tex]$=17.32$[/tex]

At 10 bar and [tex]$75^\circ C$[/tex],

[tex]$\rho = 9.999 \ kg/m^3 , \ \mu =20.91 \times 10^{-6}$[/tex]

[tex]$K=30.05 \times 10^{-7} \ W/\mu K, \ Pr = 0.7092, \ C_p=1.019 \ kJ/kg K$[/tex]

[tex]$Re_2 = \frac{9.999 \times 2 \times V}{1 \times 20.9 \times 10^{-6}}$[/tex]

Initial, [tex]$Re_i = \frac{1 \times V}{1 \times 20.82 \times 10^{-6}}$[/tex]

                [tex]$=40000$[/tex]

[tex]$V=40000 \times 0.2 \times 20.82 \times 10^{-6}$[/tex]

[tex]$Re_2 = \frac{9.999 \times 2 \times 40000}{1 \times 20.9 \times 10^{-6}}$[/tex]

[tex]$Re_2=796477.01$[/tex]

Flow is turbulent.

This Nusselt number is given by :

[tex]$Nu=(0.037)(Re)^{0.8}- 8\pi Pr^{1/3}=958.75$[/tex]

[tex]$h=\frac{958.75 \times k}{0.2}$[/tex]

  [tex]$=144.05 \ W /\mu^2C$[/tex]

[tex]$\theta =144.05 \times 0.2 \times 0.1 \times (100.5)$[/tex]

  [tex]$=144.05 \ \omega$[/tex]

Learn More :

https://brainly.in/question/43296102

https://brainly.in/question/48196879

Answer:

Explanation:

The step by step solution is in the attached file.

Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

Answer:

Attached to this solution is a Seventeen pages of code. Cheers!

Explanation:

Answer:

See the explanation for the answer

Explanation:

Check to determine the redesigned beam is satisfactory for cracking:

crack width is controlled by establishing a minimum spacing.

Steps followed to check for cracking in the beams

i) According to aci code 10.6.7 if depth of beam isgreater than 36 in then skin reinforcement has to provide.the skin reinforcement to be provided should be such that it should not be greater than the actual main tension reinforcement.

ii) In second step steel stress is determined:

steel stressfs=Ms/(As*(d-hf/2)) or fs= 0.60fy

where Ms=service load moment

As*(d-hf/2)=area of reinforcement *moment area

iii) S=540/fs-2.5Cc is less than or equal to 12*(36/fs)

here Cc = clear spacing

if S=center to center spacing is with in the limit as specified above then the cracking is with in the control if not then redesign has to done.

In the given problem the data given is :

grade of steel in desigened beam is 75 and in redesigned beam is 60 so the stress in steel is 75*0.6=45ksi and 0.6*60=36ksi respectively

now the spacing is calculated for the two design and redesigned beams

the center to center spacing is given by S=540/fs-2.5Cc

For designed beam

S=540/45-(2.5*2.25)=6.375in which is less than 12*36/fs=12*36/45=9.6in hence it is safe

For redesigned beam

S=540/36-(2.5*2.25)=9.375in and it is less than the maximum spacing which is given by 12*36/fs=12*36/36=12in

Hence, the beam is within the limits and the beam is safe against the cracking.

Modifications to reduce the number of reinforcing bars

The addition of steel does not prevent cracking due to restrained shrinkage but it limits the width of crack by causing the formation of the number of narrow cracks rather than single wide crack.

Larger size bars leads to fewer cracks but wider cracks while smaller size bars leads to number of narrow cracks hence it is advisable to provide number of smaller diameter bars of equal strength of designed bars rather than larger  bars.

Answer:

See explaination for python programming code

Explanation:

Python programming code below

import re

s = "abc" # enter string here

#s = "hello world! HELLOW INDIA how are you? 01234"

# Short version

print filter(lambda c: c.isalpha(), s)

# Faster version for long ASCII strings:

id_tab = "".join(map(chr, xrange(256)))

tostrip = "".join(c for c in id_tab if c.isalpha())

print s.translate(id_tab, tostrip)

# Using regular expressions

s1 = re.sub("[^A-Za-z]", "", s)

s2 = s1.lower()

print s2

import string

values = dict()

for index, letter in enumerate(string.ascii_lowercase):

values[letter] = index + 1

sum = 0

for ch2 in s2:

for ch1 in values:

if(ch2 == ch1):

sum = sum + values[ch1]

print sum

Answer:

See Explaination

Explanation:

if(check1 and check2):

outcome = "BOTH"

elif(check1):

outcome = "ONE"

elif(check2):

outcome = "TWO"

else:

outcome = "NEITHER"

Answer:

Explanation:

Attached is the solution

Answer:

Check the explanation

Explanation:

The process of Alcoholic fermentation involves the converting a single mole of glucose into two moles of carbon dioxide and two moles of ethanol, and in the process producing two moles of ATP. The total chemical formula for alcoholic fermentation is: C6H12O6 → 2 C2H5OH + 2 CO. Sucrose is a dimer of fructose and glucose molecules.

Kindly check the attached image below to see the full step by step explanation to the question above.

The question asks for a function to check if a sentence contains all five vowels at least once, regardless of case sensitivity. A solution involves creating a set of vowels and comparing it to a set of found vowels in the sentence, returning true if all vowels are present.

The question relates to determining whether a given sentence contains all five vowels (a, e, i, o, u) at least once, ignoring case sensitivity. This problem is typically solved using a function that iterates through each character in the sentence, checks if it is a vowel, and then keeps track of whether all vowels have been encountered. The essential steps involve converting the sentence to lowercase (to ignore case sensitivity), then checking for the presence of each vowel. A simple approach is to use a set to keep track of the vowels found, and once the set contains all five vowels, the function can return True. Otherwise, it returns False after checking the entire sentence.

An example implementation could be:

This code creates a set of vowels and then iterates over the sentence, adding each encountered vowel to another set. If, by the end of the sentence, the second set is equal to the set of all vowels, the function returns True; otherwise, it returns False.

Answer:

The expression for the concentration of electrons is P = NA - ND

Explanation:

Please look at the solution in the attached Word file

Answer:

see explaination for all the answers and full working.

Explanation:

deflection=8P*DN/Gd^4

G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2

for outer spring,

deflection=8*3*50^3*5/(70*9^4)=32.66mm

for inner spring

deflection=8*3*30^3*10/(70*5^4)=148.11mm

max stress=k*8*P*C/(3.14*d^2)

for outer spring

c=50/9=5.55

k=(4c-1/4c-4)+.615/c=1.2768

max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2

for inner spring

c=6

k=1.2525

max stress=2.29KN/mm^2

Explanation:

Data

Load = 3kn = 3000N

Modulus of rigidity = 80Gpa= 80000mpa

Outer spring diameter = 50mm

d. = 9mm

N = 5

Inner spring diameter = 30mm

d = 5mm

N = 10

Fo = outer force

Fi = inner force

Ki = stiffness of inner spring

Ko = stiffness of outer spring

Ks = stress factor

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

Answer:

A) the forces are balanced because Jasper weighs the same as Gemma

Explanation:

Answer:

A. The forces are balanced because Jasper weighs the same as Gemma.

Explanation:

Took the test

Answer:

Departure rate = 7.65 vehicle/min

Explanation:

See the attached file for the calculation.

Answer:

Explanation:

Given that, .

Mass of car is

M = 1300kg

Velocity of car

V = 80km/h = 80 × 1000/3600

V = 22.22m/s

Calculate the kinetic energy of the vehicle as follows:

K.E = ½ MV²

K.E = ½ × 1300 × 22.22²

K.E = 320,987.65 J

Given that,

Enthalpy is 45MJ / kg

h = 45MJ / kg

Then, enthalpy is given as.

Enthalpy = Energy / mass

h = E / m

45 × 10^6 = 320,987.65 / m

m = 320,987.65 / 45 × 10^6

m = 7.133 × 10^-3 kg

m = 7.133 mg

Also, given that, density is 680kg/m³

Density is given as

Density = mass / Volume

ρ = m / v

Then, v = m / ρ

v = 7.133 × 10^-3 / 680

v = 1.049 × 10^-5 m³

We know that

1mL = 10^-6 m³

Therefore,

v = 1.049 × 10^-5 m³ × 1mL / 10^-6m³

v = 10.49 mL

Answer:

a.) The mean velocity = 0.0318 m/s

    The  hydrodynamic entry length = 0.636 m

     The  thermal entry length = 0.004 m

(b) The mass flow rate = 0.0051 kg/s

    The hydrodynamic entry length = 0.028 m

     The  thermal entry length = 1.419 m

Explanation:

See the attached files for the calculation.

Answer:

Flow velocity

50.48m/s

Pressure change at probe tip

1236.06Pa

Explanation:

Question is incomplete

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively

solution

In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe

please check attachment for complete solution and step by step explanation

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

[tex]COP_{i} = \frac{250\,K}{300\,K-250\,K}[/tex]

[tex]COP_{i} = 5[/tex]

The real Coefficient of Performance is:

[tex]COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}[/tex]

[tex]COP_{r} = 12.571[/tex]

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

Consider airflow over a plate surface maintained at a temperature of 220°C. The temperature profile of the airflow is given as T(y) = T∞ − (T∞−Ts)exp((−V/α_fluid)y).

The airflow at 1 atm has a free stream velocity and temperature of 0.08 m/s and 20°C, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow

Answer:

A) heat flux on the plate is;q_o = 11737.34 W/m²

B) convection heat transfer coefficient of the airflow is;h = 58.67 W/m².k

Explanation:

The temperature profile of the airflow is given as;

T(y) = T∞ − (T∞−Ts)exp((−V/α_fluid)y)

Let's differentiate with respect to y;

dT/dy = [[(T∞−Ts)V]/α](e^(-vy/α)

Where;

T∞ = 20°C

Ts = 220°C

V = 0.08 m/s

α is thermal diffusivity of air and from the table i attached at a temperature of 220°C, by interpolation it has a value of;

α = 5.33 x 10^(-5) m²/s

Thus, at y =0;

dT/dy = [[(20 − 220)0.08]/(5.33 x 10^(-5))](e^(0))

dT/dy = -300187.62 °C/m

A) Now, heat flux at y = 0 would be given by;

q_o = -k(dT/dy)

Where k is thermal conductivity

from the table attached at 220°C and by interpolation, the thermal conductivity k = 0.0391 W/m.k

Thus,

q_o = -0.0391(-300187.62)

q_o = 11737.34 W/m²

B) the convection heat transfer coefficient of the airflow is gotten from;

q_o = h(Ts - T∞).

Where h is the convection heat transfer coefficient of the airflow

Thus making h the formula, we have;

h = q_o/(Ts - T∞)

h = 11737.34/(220 - 20)

h = 58.67 W/m².k

Answer:

# the function fix_yz is defined

# it takes a string as parameter

def fix_yz(word):

   # new_word is to hold the new corrected string

   new_word = ""

   # loop through the string

   # and check for any instance of y or z.

   # if any instance is found, it is replaced accordingly

   for each_letter in word:

       if each_letter == 'z':

           new_word += 'y'

       elif each_letter == 'Z':

           new_word += 'Y'    

       elif each_letter == 'y':

           new_word += 'z'

       elif each_letter == 'Y':

           new_word += 'Z'        

       else:

           new_word += each_letter

   # the value of new string is returned

   return new_word        

Explanation:

The function is written in Python 3 and it is well commented. An image is attached showing the output of the given example.

The function take a string as input. It then loop through the string and check for any instance of 'y' or 'z'; if any instance is found it is swapped accordingly and then append to the new_word.

The value of bew_word is returned after the loop.

Answer:

See Explanation Below

Explanation:

// Program is written in C++ Programming Language

// Comments are used for explanatory purpose

// Program starts here

#include<iostream>

#include <bits/stdc++.h>

using namespace std;

int main()

{

// Declare 2 string variables to store the secret message and to store the encrypted text

string message, result;

// Prompt user to enter a secret message

cout<<"Enter a secret message: ";

cin>message;

// Convert the input string to char array

int n = message.length();

char char_array[n + 1];

strcpy(char_array, message.c_str());

// Initialise result

result = "";

// Declare an array of all possible alphabets a-z

char possible[26] = { 'a','b','c','d','e','f','g,','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v',w','x','y','z'};

// Generate output string

// Start by getting string position

int count = 0;

while(count<n)

{

// If current character is blank or !

if(char_array[count] = '!' || char_array[count] = ' ')

{

result+=char_array[count];

}

else

{

for(int I = 0; I<26; I++)

{

if(char_array[count] = possible[I])

{

result+=possible[25-I];

}

}

}

count++;

}

// No output required; the program stops here

return 0;

}

// End of program

Answer:

461.65 KJ/Kg

Explanation:

In this question, we are asked to calculate the values of heat transferred in the process.

Please check attachment for complete solution and step by step explanation

Answer:

type of study design is Prospective cohort study

Explanation:

Answer:

[tex]n = 2.36[/tex]

Explanation:

The stress experimented by the circular bar is:

[tex]\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)[/tex]

[tex]\sigma = 10.186\,kpsi[/tex]

The safety factor is:

[tex]n = \frac{24\,kpsi}{10.186\,kpsi}[/tex]

[tex]n = 2.36[/tex]