Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. Ex = 6 − 5x, (0, 1) The equation ex = 6 − 5x is equivalent to the equation f(x) = ex − 6 + 5x = 0. F(x) is continuous on the interval [0

Answer:

BE = 559lb T

BC = 300lb T

Explanation:

Base on the figure attached, let us first find the load on joint B, C and D

FC = 5/8 * 800 = 500lb

FD = 13/28 * 800 = 150lb

FB is also equal to 150lb

The angle alpha α at joint D will be:

Tan α = 6/12

α = tan ^-1 (0.5)

α = 26.6

by using resolution of forces, resolve the force at that point to horizontal and vertical component.

Horizontal component

Fd + DE sin α = 0

DE sinα  = -Fd

DE = -Fd/sinα

DE = -150/26.6

DE = - 335.4lb

Vertical component

-CD - DEcosα = 0

CD = -DEcosα

CD = -(-335.4)cos26.6

CD = 300lb

So let us do the same at joint C and joint E

At joint C

Horizontal component

Fc + CE = 0

CE = -Fc

CE = -500lb

Vertical component

CD - BC = 0

BC = CD

BC = 300lb T

At joint E

The horizontal component will be:

-CE - DEsinα - BEsinα + EFsinα = 0

-(-500) - (-335.4)sin26.6 - BEsin26.6 + EFsin26.6 + EFsin 26.6

650 - BEsin26.6 + EFsin26.6 = 0

Make BE the subject of the formular

BEsin26.6 = 650 + EFsin26.6

BE = (650 - EFsin26.6)/ sin26.6

BE = 1453.2 + EF ................ (1)

The vertical component

DEcosα - BEcosα - EFcosα = 0

substitute all the parameters

- 335.4cos26.6 - BEcos26.6 - EFcos26.6 = 0

Substitute BE in the equation 1 into the equation above

-300 - (1453.2 + EF)cos26.6 - EFcos26.6 = 0

-1599.7 - 1.789EF = 0

EF = -1599.7 / 1.789

EF = -894.2lb

Substitute EF in equation 1

BE = 1453.2 - 894.2

BE = 559lb T

Therefore the forces in members BE and BC

BE = 559lb T

BC = 300lb T

Explanation:

10000=5000(1.12^x)

2=1.12^x

(log_1.12)(2)=x

x= about 6.1163

10000=5000(1.2^x)

2=1.2^x

(log_1.2)(2)=x

x= about 3.8019

compare them by saying like 20% will be 6.12/3.8 times faster

Answer:

a) the rate of heat transfer from the pipe to the air is 23.866 watts

b) YES, the rate of heat transfer changes to 3518.61 watt

Explanation:

Given that:

steam is saturated at 17.90 bar.

the pipe is stainless steel and has an outside diameter of 6.75 cm

length = 34.7 m

Air flows over the pipe at 7.6 m/s

Bulk fluid temperature of 27°C

we know that

hD/k = 0.028 (Re)^0.8 (Pr)^0.33

Outside diameter of pipe = 6.75 cm

length of the pipe = 34.7 m

velocity of air = 7.6 m/s

Cp of air = 1.005 kJ/Kgk

viscosity of air = 1.81 × 10⁻⁵ kg/(m.sec)

thermal conductivity of air = 2.624 × 10⁻⁵ kw/m.k

so as

hD/k = 0.028 (Re)^0.8 (Pr)^0.33

hD/k = 0.028 (Dvp / u)^0.8 (Cpu / k)^0.33

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([0.0675 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 0.0414 w/m².k

a)

Now to find the rate of heat transfer Q

Q = hAΔT

Q = 0.0414 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 23.866 watts

therefore the rate of heat transfer from the pipe to the air is 23.866 watts

b)

Now the flow direction changes to parallel flow, then

(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([34.7 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))

h = 6.1036 w/m².k

so from the steam table, saturated steam at 17.70 bar, temperature of steam will be 105.383°C

so to find the rate of heat transfer Q

Q = hAΔT

Q = 6.1036 × (2π × 0.03375 × 34.7) × (105.383 - 27)

Q = 3518.61 watt

Therefore the rate of heat transfer changes to 3518.61 watt

Answer:

science is hard but human cloning this is legendary but still its up to the government but me i would allow this because its something that science can upgrade to

Explanation:

Answer:

The outer diameter of the ball is 6.2138 cm

Explanation:

The formula to apply is ;

Heat loss ,

[tex]Q/t=kA*\frac{( T_1-T_2)}{d}[/tex]

where ;

Q/t=total heat loss from the ball = 1600 w

k=coefficient of heat transmission through the ball= 2.75 W/m.˚C

A=area in m² of the ball with the coefficient of heat transmission

T₁=Hot temperature

T₂=Cold temperatures

d=thickness of the ball

Area of spherical ball using internal diameter, 3cm= 0.03 m will be

Radius = half the diameter = 0.03/2 = 0.015

Area = 4 *π*r²

Area = 4*π*0.015² = 0.002827 m²

Apply the formula for heat loss to get the thickness as:

1600 = {2.75 * 0.002827 *(250-30 ) }/d

1600 =1.711/d

1600d = 1.711

d=1.711/1600 = 0.001069 m

d= 0.1069

Using internal radius and the thickness to get outer radius as;

3 + 0.1069 = 3.1069 cm

Outer diameter will be twice the outer radius

2*3.1069 = 6.2138 cm

Answer:

The answer is "828.75"

Explanation:

Please find the correct question:

For W21x93 BEAM,

[tex]Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3[/tex]

For A992 STREL,

[tex]F_y= 50\ ks[/tex]

Check for complete section:

[tex]\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}[/tex]

Design the strength of beam =[tex]\phi_b Z_x F_y\\\\[/tex]

                                                [tex]=0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\[/tex]

The color of white light bends the most when it is refracted by a prism is green after the violet.

The shorter the wavelength of the mild, the greater its miles refracted. As a result, crimson mild is refracted the least, and violet mild is refracted the most - inflicting the colored mild to unfold out to shape a spectrum.

The hues of white mild withinside the order of lowering wavelength are crimson orange-yellow inexperienced blue indigo violet. Fro the listing we are given withinside the alternatives inexperienced has the shortest wavelength than others. Therefore inexperienced will bend the most.

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1. When the interviewer asked if you have any questions at the end of the interview don't say no. You should always say yes your interviewer is expecting you to ask a few good questions before ending the interview.

2. Always thank the interviewer for their time and effort to interview you. This would look very good for you and its a nice way to help wrap up the interview.

Answer:

Draw up the contract for construction management.

Answer:

Work with construction management to solve specific problems as they arise.

Ensure compliance with safety standards.

Explanation:

The final specific internal energy : 190 kJ/kg

The laws of thermodynamics 1 state that: energy can be changed but cannot be destroyed or created  

The equation is:  

[tex]\tt E_{in}-E_{out}=\Delta E~system\\\\\Delta E=\Delta U+\Delta KE+\Delta PE\\\\\Delta U=m(U_2-U_1)\\\\Q-W=\Delta U+\Delta KE+\Delta PE[/tex]  

Energy owned by the system is expressed as internal energy (U)  

This internal energy can change if it absorbs heat Q (U> 0), or releases heat (U <0). Or the internal energy can change if the system does work or accepts work (W)  

The sign rules for heat and work are set as follows:  

• The system receives heat, Q +  

• The system releases heat, Q -  

• The system does work, W -  

• the system accepts work, W +  

A closed system of mass 20 kg⇒m=20 kg

Heat transfer of 1000 kJ from the system to the surroundings⇒Q=-1000 kJ

The work done on the system is 200 kJ⇒W=+200 kJ

The initial specific internal energy of the system is 250 kJ /kg⇒U₁ = 250 kj/kg

Neglect changes in kinetic and potential energy⇒ΔKE+ΔPE=0, so

Q-W = ΔU

Input in equation

[tex]\tt -1000-200=20(U_2-250)\\\\-1200=20U_2-5000\\\\3800=20U_2\\\\U_2=190~kJ/kg[/tex]

Answer:

cubes

Explanation:

Design for software engineering does not need special care for the placement of dimensions. Software engineering generally deals with arranging lines of coding.

In order to achieve a software's goal by using lower-level, machine-understandable algorithms and a set of parameters based on project requirements or environmental constraints, software design is the process of outlining solutions, planning, and specifying a method of implementation based on the needs of the user.

Software engineering is the application of engineering principles, which are often employed in the design, development, testing, deployment, and management of physical systems to the design.

Therefore, the main focus of software engineering is organizing coded lines that are kept in a device's or computer's memory rather than in actual physical space.

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Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

Answer: a(s) is of degree n and b(s) is of degree n-1.

a)Characteristic equation in form suitable for Evans's root-locus method is given by a(s) + b(s)K = 0, where a(s) and b(s) are polynomials of s with real coefficients. Now, given equation can be represented as: s + (1/τ) = 0 => s + (1/τ) = 0s + (1/τ)K = 0=> K = -τs/τ + 0Thus, L(s) = s, a(s) = 1, b(s) = 1 and K = -τ.

b)The characteristic equation is given by:s^2 + cs + c + 1 = 0For the root-locus method, we have to write the characteristic equation in the form a(s) + b(s)K = 0. Since the degree of a(s) is 2, we select K such that the degree of b(s) is also 2, and so b(s) will be monic.s^2 + cs + c + 1 = 0=> (s + c/2)^2 + 1/4 - c^2/4 + c = 0=> s^2 + (2c)s + (1 + c - c^2/4) = 0Now, taking a(s) = s^2 + (2c)s + (1 + c - c^2/4) and b(s) = 1 with K = -1/c^2, we have:a(s) + b(s)K = s^2 + (2c)s + (1 + c - c^2/4) - 1/c^2 = 0

c) The characteristic equation is given by:(s + c)3 + A(Ts + 1) = 0

i. versus parameter A:s = -c is a repeated root of multiplicity 3 when A = 0For s = -c, the characteristic equation becomes:-A(Tc + 1) = 0If A = 0, the characteristic equation will be (s + c)^3 = 0 and will have a repeated root at s = -c with a multiplicity of 3.

ii. versus parameter T:s = -c is a repeated root of multiplicity 3 when T = 0.For s = -c, the characteristic equation becomes:3c^2s + A = 0If A = 0, the characteristic equation will be (s + c)^3 = 0 and will have a repeated root at s = -c with a multiplicity of 3.

iii. versus the parameter c:We cannot draw the root locus of (s + c)3 + A(Ts + 1) = 0 with respect to the parameter c because there is no c term in the characteristic equation; it only contains c cubed.

d) The characteristic equation is given by:1 + (kp + k1/s + kDs/Ts + 1)G(s) = 0Assuming G(s) = Ac(s)/d(s), where c(s) and d(s) are monic polynomials with the degree of d(s) greater than that of c(s), the characteristic equation becomes:

d(s) + kp d(s) + k1 d(s)/s + kDs c(s)/T = 0.

Thus, a(s) = d(s) + k1 d(s)/s and b(s) = kDc(s)/T + kp d(s) with K = -1.

Therefore, a(s) is of degree n and b(s) is of degree n-1.

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Answer: the cancer risk for an adult male drinking the water for 10 years is 2.88 × 10⁻⁶

Explanation:

firstly, we find the time t required to travel for the contaminant to the well;

Given that, contamination flowing rate = 0.5 ft/day

Distance of well from the site = 1 mile =  5280 ft

so t = 5280 / 0.5 = 10560 days

k is given as 1.94 x 10⁻⁴ 1/day

next we find the Pollutant concentration Ct in the well

Ct = C₀ × e^-( 1.94 x 10⁻⁴ × 10560)

Ct = 0.3 x e^-(kt)  

Ct= 0.0386 mg/L

next we determine the chronic daily intake, CDI

CDI =  (C x CR x EF x ED) / (BW x AT)

where C is average concentration of the contaminant(0.0368mg/L), CR is contact rate (2L/day), EF is exposure frequency (350days/Year), ED is exposure duration (10 years), BW is average body weight (70kg).

now we substitute  

CDI = (0.0368 x 2 x 350 x 10) / ((70x 365) x 70)

= 257.7 / 1788500

= 0.000144 mg/Kg.day

CDI = 1.44 x 10⁻⁴ mg/kg.day  

Finally we calculate the cancer risk, R

Slope factor SF is given as 0.02 Kg.day/mg

Risk, R = I x SF

= 1.44 x 10⁻⁴ mg/kg.day  x 0.02Kg.day/mg    

R = 2.88 × 10⁻⁶

therefore the cancer risk for an adult male drinking the water for 10 years is 2.88 × 10⁻⁶

Answer:

Anne is a mechanical engineer.

Explanation:

There are many different types of engineers (mechanical, industrial, electrical, chemical and civil) around the world. When a job involves designing machines, it falls under mechanical engineering. Such job requires a person to be as creative as possible, so he/she can come up with an innovative design that will be reasonable when used.

Designing machines includes spacecrafts. Therefore, Anne is a mechanical engineer working for NASA.

Answer:

1) titration

2) titrand

3) equivalence point

4) titrant

5) Burette

6) Indicator

Explanation:

The process in which a known volume of a standard solution is added to another solution so that the standard solution can react with the solution of unknown concentration such that its concentration is determined  can be referred to as titration.

The solution which is added to another solution  is called the titrant.  The titrand is the solution of unknown concentration

A burette is a glassware used to slowly add a known volume of the titrant to the titrand.

The indicator used signals the point when the reaction is complete by a color change. At this point, a stoichiometric amount of titrant has been added to the titrand. This is also referred to as the equivalence point.

Answer:

The answer is "Option D".

Explanation:

[tex]\to \epsilon_T= 0\\\\\to R+R_1=500 \ lb \\\\\to \epsilon \ M_{pin} > 0\\\\\to -R \times 20 +(500 \times 5) =0\\\\\to 2500 =20R\\\\\to 20R= 2500\\\\\to R= \frac{2500}{20}\\\\\to R= \frac{250}{2}\\\\\to R= 125 \ lb \\\\\to R_1 =500-125\\[/tex]

         [tex]=375 \ lb \\[/tex]

Answer:

is it multiple choice or....

Answer:

both

Explanation:

Tech A says that most electric fuel pumps are now mounted inside of the fuel tank. Tech B says that fuel flows through the center of the electric pump and is used to cool the pump. Who is correct? quizlet said so  

In this case, both Technicians are right.

In conclusion, both Technicians are right.

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Answer: aerofoil type blades

Explanation: they are more difficult to make but offer better performance and higher rotational speeds making them ideal for electrical energy generation.

Answer:

1) titration

2) titrand

3) equivalence point

4) titrant

5) Burette

6) Indicator

Explanation:

The process of adding a known volume a standard solution to another solution to react with it in order to determine the concentration of the unknown solution is known as titration.

The solution to which another solution of known concentration is added is called the titrand while the solution of known concentration is called the titrant.

A burette is a glassware used to slowly add a known volume of the titrant to the titrand. An indicator shows the point when the reaction is complete by a color change. This is the point when the required amount of one solution has been added to the second solution. It is also called the equivalence point.

I. The process of slowly adding a solution to react with another solution and determine the concentration of one of the solutions based on the reaction between them: Titration.

II. A solution of an unknown concentration that has another solution slowly added to it: Analyte.

III. When the required amount of one solution has been added to the second solution to complete the reaction: Endpoint.

IV. A solution of known concentration that is slowly added to a solution of unknown concentration: Titrant.

V. A glassware that allows a solution to be precisely and slowly added to another solution: Burette.

VI. A reagent added to the analyte solution that changes color when the reaction is complete: Indicator.

Titration is also referred to as titrimetry or volumetric analysis and it can be defined as a chemical process in which a solution of known concentration is slowly added to a solution of unknown concentration, in order to determine the concentration or quantity of the analyte (solution of an unknown concentration)

For example, titration can be used to determine the concentration of a basic solution by titrating it with an acid solution of known concentration that is required to neutralize the basic solution.

In Chemistry, the following terms and apparatus are used for the titration of solutions:

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This question is incomplete, the complete question is;

Determine the greatest pressure drop allowed over the 10-m-long pipe caused by viscous friction, so the flow remains laminar.

The 125-mm-diameter smooth pipe is transporting SAE 10W-30 oil with ρ=920 kg/m3 and µ=0.2 N.s/m2 .

Answer: the greatest pressure drop allowed is 14247 Pascals

Explanation:

LAMINAR FLOW INF PIPE

greatest pressure drop

(P1 - P2) = pressure drop

(P1 - P2) = 32uvl / D²

so (P1 - P2) ∝ V

greatest pressure drop is only at very high velocity

now Re (Reynold Number = svd / u

Re ∝ V

so velocity is high when Re is high

Re = 2000 ( Highest for LAMINAR)

Re = 2000 = svd / u

now in the question we were given that

s = 920 kg/m³, u = 0.2 NS/m², d = 125mm = 0.125m, l = 10m

so we substitute;

2000 = (920 × v × 0.125) / 0.2

v = 3.48 m/s

now pressure drop = (P1 - P2) = 32UVL / d²

we substitute

(P1 - P2)  = (32(0.2) (3.48)(10)) / (0.125)²

(P1 - P2) = 14247 Pascals

therefore the greatest pressure drop allowed is 14247 Pascals

Answer:

Explanation:

From the given information;

The first objective is to derive an expression for the execution time in cycles

For an execution time T for the non-pipelined processor;

[tex]T = \dfrac{N \times S}{R}[/tex]

where

N = instruction count

S = average number of clock cycles to fetch & execute an instruction

R = clock rate

From the average number of clock

[tex]S = 1 + \delta _{branch\_penality }[/tex]

where;

[tex]\delta _{branch\_penality }= {Dynamic \ count \times Delay \ Slots}[/tex]

Similarly, we are given that the dynamic count is 20%

∴

The execution time required for all the delayed slots that is filled with NOP instructions can be estimated as:

[tex]\delta _{branch\_penality }=0.20 \times 1.00[/tex]

[tex]\delta _{branch\_penality }=0.20[/tex]

Therefore;

S = 1 + 0.20

S = 1.2

Recall that:

For an execution time T for the non-pipelined processor;

[tex]T = \dfrac{N \times S}{R}[/tex]

[tex]T = \dfrac{N \times 1.2}{R}[/tex]

[tex]T = \dfrac{ 1.2 \ N}{R}[/tex]

To derive another expression that reflects the execution time of 70% delay; we have:

[tex]\delta_{branch\_penality} = 0.20 \times ( 1-0.7)[/tex]

[tex]\delta_{branch\_penality} = 0.20 \times ( 0.3)[/tex]

[tex]\delta_{branch\_penality} = 0.06[/tex]

S = 1 + 0.06

S = 1.06

For an execution time T for the non-pipelined processor;

[tex]T = \dfrac{N \times S}{R}[/tex]

[tex]T = \dfrac{N \times 1.06}{R}[/tex]

[tex]T = \dfrac{ 1.06 \ N}{R}[/tex]

Finally, the compiler's contribution to the increase in speed up percentage for the above two cases is:

[tex]=\begin {pmatrix}\dfrac{T_{all \ delay \ slots }}{T_{70\% \ delay \ slots } } -1 \end {pmatrix} \times 100[/tex]

[tex]= \begin {pmatrix}\dfrac{1.2 }{1.06 } -1 \end {pmatrix} \times 100[/tex]

= (1.1320 - 1) × 100

= 0.1320 × 100

= 13.20%

Therefore, the compiler's contribution to increasing performance as expressed as speed up percentage is 13.20%

Answer:

attached below is the solution

Explanation:

hello  attached below is the sequence of push and pop operations on stack memory

For The collatz method below

public void collatz(int n) {

if (n == 1) return;

else if (n % 2 == 0)

collatz(n / 2);

else collatz(3*n + 1); }

Answer:

a

Explanation:

just took the quiz

Answer:

1) ∀x [ Tiger(x) → Fast(x) ]

2) эx [ Tiger (x) ∧ Fierce (x) ∧ Dangerous(x) ]

3) ∀x [ Prime(x) → Odd(x) ]

4) ∀x [ prime (x) ∧  ~Two(x) → Odd (x) ]

5) ∀x [ Fruits(x) → ( yellow(x) ∨ Red(x) ]

6) ∀xэy [ I(x) → greater (y, x) ]

Explanation:

Translating the statements into first Order Logic and their negations

1)   All tigers are fast. Domain: animals.

∀x [ Tiger(x) → Fast(x) ]

2)  Some tigers are fierce and dangerous. Domain: animals

эx [ Tiger (x) ∧ Fierce (x) ∧ Dangerous(x) ]

3) Every prime number is odd. Domain: positive integers

∀x [ Prime(x) → Odd(x) ]

4)  All prime numbers except two are odd. Domain: positive integers

∀x [ prime (x) ∧  ~Two(x) → Odd (x) ]

5) All fruits are either yellow or red. Domain: produce.

∀x [ Fruits(x) → ( yellow(x) ∨ Red(x) ]

6) For every integer number, there exist a bigger integer. Domain: integers.

∀xэy [ I(x) → greater (y, x) ]

Answer:

Answer:

1) ∀x [ Tiger(x) → Fast(x) ]

2) эx [ Tiger (x) ∧ Fierce (x) ∧ Dangerous(x) ]

3) ∀x [ Prime(x) → Odd(x) ]

4) ∀x [ prime (x) ∧  ~Two(x) → Odd (x) ]

5) ∀x [ Fruits(x) → ( yellow(x) ∨ Red(x) ]

6) ∀xэy [ I(x) → greater (y, x) ]

Explanation:

Translating the statements into first Order Logic and their negations

1)   All tigers are fast. Domain: animals.

∀x [ Tiger(x) → Fast(x) ]

2)  Some tigers are fierce and dangerous. Domain: animals

эx [ Tiger (x) ∧ Fierce (x) ∧ Dangerous(x) ]

3) Every prime number is odd. Domain: positive integers

∀x [ Prime(x) → Odd(x) ]

4)  All prime numbers except two are odd. Domain: positive integers

∀x [ prime (x) ∧  ~Two(x) → Odd (x) ]

5) All fruits are either yellow or red. Domain: produce.

∀x [ Fruits(x) → ( yellow(x) ∨ Red(x) ]

6) For every integer number, there exist a bigger integer. Domain: integers.

∀xэy [ I(x) → greater (y, x) ]

Explanation:

Answer:

The wind turbine has little impact on the environment compared to conventional power plants. Meanwhile the Wind turbine can affect the wildlife meaning birds are being harmed by the blades.

Explanation: Hope that helps:)

Answer:

C₁₀ = 6.3 KN

Explanation:

The catalog rating of a bearing can be found by using the following formula:

C₁₀ = F [Ln/L₀n₀]^1/3

where,

C₁₀ = Catalog Rating = ?

F = Design Load = 2.75 KN

L = Design Life = 1800 rev/min

n = No. of Hours Desired = 10000 h

L₀ = Rating Life = 500 rev/min

n₀ = No. of Hours Rated = 3000 h

Therefore,

C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3

C₁₀ = (2.75 KN)(2.289)

C₁₀ = 6.3 KN