Ethylene glycol, the ingredient in antifreeze, does not cause health problems because it is a clear liquid

Answer:

a) the friction angle of the sand is 36.87°

b) the deviator stress at failure is 450 kN/m³

Explanation:

Given the data in the question;

For a consolidated drained test

The effective major principle stresses

σ₃ = σ₃' = 80 kN/m²

and

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 80 kN/m² + 240 kN/m²  = 320 kN/m²

now

a) friction angle of the sand

σ₁' = σ₃'tan²( 45° + Ф/2' ) + 2c' tan( 45° + Ф/2 )

for sand; c' = 0

so

σ₁' = σ₃'tan²( 45° + Ф/2' )

we substitute

320 = 80 tan²( 45° + Ф/2' )

Ф' = 2 × [ tan⁻¹ (√[tex]\frac{320}{80}[/tex]) - 45° ]

Ф' = 2 × [ 63.4349° - 45° ]

Ф' = 2 × 18.4349

Ф' = 36.87°

Therefore,  the friction angle of the sand is 36.87°

b)  deviator stress

σ₁' = σ₃'tan²( 45° + Ф/2' )

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )

σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )   3.597

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 150tan²( 45° + 36.87°/2 )

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600 - 150

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex]  = 450 kN/m³

Therefore, the deviator stress at failure is 450 kN/m³

Complete Question:

Throughout your job search, you’ll find that _____(the economy/education/salary) is closely related to the career of your choosing. It’s important to take the time to find out what to expect now, so you can start developing the ____ (task/communication/skill) that you need to excel in that career.

Answer:

Education; skills.

Explanation:

Throughout your job search, you'll find that education is closely related to the career of your choosing. It's important to take the time to find out what to expect now, so you can start developing the skills you need to excel in that career.

This ultimately implies that, all job openings that are being advertised by various organizations usually have a minimum requirement such as academic experience or level i.e the prospective candidate must have attained a level of education. Also, it is very important for undergraduates and potential employees to ensure that they are being proactive, as well as developing the requisite skills needed to excel in their career.

Answer:

both are c

Explanation:

a) Question Completion:

INPUT HOURS OF LABOR

Country   Cookies      Milk

Atlantis       2 hours   1 hour

Neverland  4 hours   1 hour

Answer:

1. Atlantis has the absolute advantage in the production of cookies.

2. No country has the absolute advantage in the production of milk.

Explanation:

Absolute advantage refers to superior production capability.  It is determined when a country, for example, has the ability to produce a particular good or service at lower cost or more efficiently (i.e. with less resources) than the other country.  In the scenario above, Atlantis has an absolute advantage in the production of cookies because it can produce the same quantity of cookies using 2 labor hours that Neverland can produce using 4 labor hours.  But for the production of milk, Atlantis and Neverland share the same comparative advantage less they can use less labor hours to produce milk than they can produce cookies.

Answer: speaks Portuguese

Eu disse a todos a tradução para que possam te ajudar

Explanation: Y’all can help I have no idea

QUESTION 13. Explain the use of quotation marks in the excerpt "I will give each of you

seed. The one who will bring me the most beautiful flower within six months will be chosen but

wife and the future empress of China. ".

QUESTION 14. The palace servant considered the idea of ​​her daughter attending the celeb

organized by the prince of the region, a foolish idea, a madness. This is an OP

Do you agree with this opinion of the character? Justify your answer.

Hey there ..

I think the answer is as they put brake ..

I am not sure .. just check the image also provided above .. ..

If u think this helped u ..plz mark me as brainliest ..

And follow me

Answer:

During starts and stops

Explanation:

"Many of the energy inefficiencies in vehicles occur during starts and stops."

found it in the text.

Answer:

Explanation:

The given equation is :

[tex]\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f[/tex]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

Determine the maximum speed for safe vehicle operation

firstly calculate the stopping sight distance

m = R ( 1 - cos [tex]\frac{28.655*S}{R}[/tex] )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos [tex]\frac{28.65 *s }{678}[/tex]  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + [tex]\frac{u^2}{30(\frac{a}{3.2} )-G1}[/tex]  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + [tex]\frac{u^2}{30[(11.2/32.2)-0 ]}[/tex]

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

Answer:

Hello your question is incomplete attached below is the missing part of the question

answer ;

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Explanation:

Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch

40v and 50mA are in series hence;  Ix = 50mA

also Vcd = 30V

CD is parallel to AB hence; Vcd = Vab = 30 V

Vab = ∝*Ix + 60 v

 30v  = ∝ ( 50mA ) + 60

therefore ∝ = -600

voltage drop in the Vcd branch = 30 V

Voltage drop in the middle branch = 40v - 30v = 10 volts

voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts

Answer:

The fifth deviation is +1

Explanation:

Given

[tex]Deviations = -5, +2, +4,-2[/tex]

Required

Determine the fifth deviation

Represent the fifth deviation with x.

As a theorem, the sum of deviations from mean is always 0.

So, we have:

[tex]-5 +2 +4-2+x = 0[/tex]

[tex]-1+x = 0[/tex]

Add 1 to both sides

[tex]1 -1+x = 0 +1[/tex]

[tex]x = 0 +1[/tex]

[tex]x= +\ 1[/tex]

Hence, the fifth deviation is +1

Answer:

a. molecular interactions.

Explanation:

Conduction is thermal energy transfer by molecular interactions. Therefore, conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Some examples of conductors include metal, steel, aluminum, copper, graphite, etc.

Hence, conduction is thermal energy transfer as a result of the movement of electrons and collision between the molecules of an object.

Answer:

electrical

computer

mechanical

and manufacturing .... I think

The answer is electrical, manufacture, computer, and chemical

Answer:

They have to make sure that the building is safe, appropriate for city conditions, they are very creative, they can build really well, and can be very focused.

Answer:

In Java:

import java.util.Scanner;

import java.lang.Math;

public class MyClass {

   public static void main(String args[]) {

     Scanner input = new Scanner(System.in);

     float x, y, z;

     System.out.print("Enter three numbers: ");

     x = input.nextFloat();      y = input.nextFloat();      z = input.nextFloat();

     System.out.print("Square root of x: ");

     System.out.printf("%.2f",Math.sqrt(x));

     System.out.println();

     System.out.print("Absolute of y - z: ");

     System.out.printf("%.2f",Math.abs(y - z));

     System.out.println();

int fact = 1;

     for(int i = 1;i<=(int)Math.ceil(z);i++){

         fact*=i;

     }

     System.out.print("Factorial of z: "+fact);    

   }

}

Explanation:

This line declares x, y and z as float

     float x, y, z;

This line prompts user for three numbers

     System.out.print("Enter three numbers: ");

The next line get input for x, y and z

     x = input.nextFloat();      y = input.nextFloat();      z = input.nextFloat();

The italicized prints the square root of x rounded to 2 decimal places

     System.out.print("Square root of x: ");

     System.out.printf("%.2f",Math.sqrt(x));

     System.out.println();

The italicized prints the absolute of y - z to 2 decimal places      

     System.out.print("Absolute of y - z: ");

     System.out.printf("%.2f",Math.abs(y - z));

     System.out.println();

This initializes fact to 1

int fact = 1;

The following iteration calculates the factorial of ceil z

     for(int i = 1;i<=(int)Math.ceil(z);i++){

         fact*=i;

     }

     This prints the calculated factorial

     System.out.print("Factorial of z: "+fact);    

   }

}

Answer:

- Vernier thrust is 5.37 kN

- mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Explanation:  

Given that;

For individual thrust chambers (vacuum);

Fc = 73.14 kN ,  Cc = 3279 m/sec

For Overall engine with Vernier (vacuum);

Foa = 297.93 kN = ,  Coa = 3197 m/sec.

- determine the Vernier thrust

Vernier thrust Fv =  Foa - ( 4 × Fc )

Vernier thrust Fv  = 297.93 - ( 4 × 73.14)

Vernier thrust Fv  = 297.93 - 292.56

Vernier thrust Fv  = 5.37 kN

Therefore, Vernier thrust is 5.37 kN

-

Vernier mass flow rate;

we know that

[tex]Co_{a}[/tex] = Fc + Fv  / mc + mv

mv = Foa/Coa - Fc/Cc

we convert kilonewton to kilograms

1 kn = 102 kg

Fc = 73.14 kN = 73.14 × 102 = 7460.28 kg

Foa = 297.93 kN = 297.93 × 102 = 30388.86 kg

we substitute

mv = (30388.86 / 3197) - (( 4 × 7460.28) / 3279)

mv = 9.5054 - 9.1006

mv = 0.4048 kg/s

Therefore, mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Answer:

what does that mean welp I have no idea sorry for answering

Answer:

Technician A Is right

Explanation:

Reinforcements, as the name suggests, are used to enhance the mechanical properties of a plastic. Finely divided silica, carbon black, talc, mica, and calcium carbonate, as well as short fibres of a variety of materials, can be incorporated as particulate fillers.

Answer:

A) attached below

B) Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

Explanation:

attached below is a detailed solution

A) attached below

B) Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A

Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

Answer:

[tex]\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}[/tex]

Explanation:

Given

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,[/tex]

[tex]0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,[/tex]

[tex]0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,[/tex]

[tex]0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.[/tex]

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.[/tex]

[tex]0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,[/tex]

[tex]0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.[/tex]

[tex]0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.[/tex]

[tex]0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.[/tex]

[tex]0.78.[/tex]

The 0.3's is will be plotted as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

The 0.4's is as follows:

[tex]\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}[/tex]

The 0.5's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}[/tex]

The 0.6's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

Lastly, the 0.7's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

The combined steam and leaf plot is:

[tex]\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}[/tex]

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×[tex]\frac{100}{w}[/tex]

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = [tex]\frac{D60}{D10}[/tex]

⇒ Cu = [tex]\frac{0.4}{0.12} = 3.33[/tex]

d.)

Coefficient of Graduation = Cc = [tex]\frac{D30^{2}}{D10 . D60}[/tex]

⇒ Cc = [tex]\frac{0.22^{2}}{(0.4) . (0.12)}[/tex] = 1

Answer:

Tech A is correct.          

Explanation:

A front-wheel-drive pulling to the left can result from several factors. One of them is definitely a faulty break.

A correct diagnosis linking the problem to the brakes is when there is an internal restriction and the pull is constant to one side and gets worse when the brakes are applied.

To confirm this, one would need to lift the vehicle and rotate each wheel by hand to check for excessive friction.

So the restriction may be caused by:

Cheers

Answer:

a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b) length of the specimen is 66.97 mm

Explanation:

Given the data in the question;

a) Determine the maximum stress that can be applied without plastic deformation

when know that; maximum stress σ[tex]_{max}[/tex]  = F / A

where F is the force in the rod ( 39872 N )

A is the cross-sectional area of the rod ( 100 mm² )

so we substitute;

σ[tex]_{max}[/tex]  = 39872 N / 100 mm²

σ[tex]_{max}[/tex]  = 398.72 N/mm²

Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b)  

strain in the members can be calculated using the expression

ε = σ / E

where σ is the stress in the rod

E is the module of elasticity (  110 GPa = 110000 N/mm² )

(Sl-L) / L = σ/E

where Sl-L is the change in length of the member

L is the original length of the specimen

so we substitute

(67.21 - L) / L = 398.72 / 110000

110000( 67.21 - L) = 398.72L

7393100 - 110000L = 398.72L

7393100 = 398.72L+ 110000L

7393100 = 110398.72L

L = 7393100 / 110398.72

L = 66.97 mm

Therefore; length of the specimen is 66.97 mm

Answer:

A Only

Explanation:

Reinforcements, as the name suggests, are used to enhance the mechanical properties of a plastic. Finely divided silica, carbon black, talc, mica, and calcium carbonate, as well as short fibres of a variety of materials, can be incorporated as particulate fillers.

Answer:

The rate of heat supply is 8.901 horse-power.

Explanation:

From Thermodynamics energy efficiency of the electric car ([tex]\eta[/tex]), no unit, is the ratio of translational mechanical power ([tex]\dot E_{out}[/tex]), measured in horse power, to electric power ([tex]\dot E_{in}[/tex]), measured in horse-power. The rate of heat supply ([tex]\dot E_{l}[/tex]), measured in horse-power, by the motor to the engine compartment at full load is difference between electric energy and translational mechanical energy. That is:

[tex]\eta = \frac{\dot E_{out}}{\dot E_{in}}[/tex] (1)

[tex]\dot E_{l} = \dot E_{in}-\dot E_{out}[/tex] (2)

[tex]\dot E_{l} = \left(\frac{1}{\eta}-1\right)\cdot \dot E_{out}[/tex] (3)

If we know that [tex]\eta = 0.91[/tex] and [tex]\dot E_{out} = 90\,hp[/tex], then rate of heat supply is:

[tex]\dot E_{l} = \left(\frac{1}{0.91}-1 \right)\cdot (90\,hp)[/tex]

[tex]\dot E_{l} = 8.901\,hp[/tex]

The rate of heat supply is 8.901 horse-power.

Answer:

Construction, the techniques and industry involved in the assembly and ... Early building materials were perishable, such as leaves, branches, and animal hides. ... The well-developed masonry technology of Mesopotamia was used to build large ... although its precise description is unknown; the concealed faces of stones

Explanation:

Answer:

red RTV is sealant for areas that get hot black RTV is for oily surfaces like oil pans and head covers

Explanation: