Cite another example of information technology companies pushing the boundaries of privacy issues; apologizing, and then pushing again once scandal dies down. As long as the controversy fades, is there anything unethical about such a strategy?

Solution :

Given :

[tex]$H(S) =\frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

Transfer function, [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

[tex]$Y(S) \left(S^2+\frac{10}{3}S+1\right) = (2S-2) \times (S)$[/tex]

[tex]$S^2Y(S) + \frac{10}{3}(SY(S)) + Y(S) = 2(S \times (S)) - 2 \times (S)$[/tex]

Apply Inverse Laplace Transforms,

[tex]$\frac{d^2y(t)}{dt^2} + \frac{10}{3} \frac{dy(t)}{dt} + y(t)=2 \frac{dx(t)}{dt} - 2x(t)$[/tex]

The above equation represents the differential equation of transfer function.

Given : [tex]$x(t)=e^{-t} u(t) \Rightarrow X(S) = \frac{1}{S+1}$[/tex]

We have : [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

[tex]$Y(S) = X(S) \times \frac{6S-6}{3S^2+10 S + 3} = \frac{6S-6}{(S+1)(3S+1)(S+3)}$[/tex]

[tex]$Y(S) = \frac{A}{S+1}+\frac{B}{3S+1} + \frac{C}{S+3}[/tex]

[tex]$A = Lt_{S \to -1} (S+1)Y(S)=\frac{6S-6}{(3S+1)(S+3)} = \frac{-6-6}{(-3+1)(-1+3)} = 3$[/tex]

[tex]$B = Lt_{S \to -1/3} (3S+1)Y(S)=\frac{6S-6}{(S+1)(S+3)} = \frac{-6/3-6}{(1/3+1)(-1/3+3)} = \frac{-9}{2}$[/tex]

[tex]$C = Lt_{S \to -3} (S+3)Y(S)=\frac{6S-6}{(S+1)(3S+1)} = \frac{-18-6}{(-3+1)(-9+1)} = \frac{-3}{2}$[/tex]

So,

[tex]$Y(S) = \frac{3}{S+1} - \frac{9/2}{3S+1} - \frac{3/2}{S+3}$[/tex]

        [tex]$=\frac{3}{S+1} - \frac{3/2}{S+1/3} - \frac{3/2}{S+3}$[/tex]

Applying Inverse Laplace Transform,

[tex]$y(t) = 3e^{-t}u(t)-\frac{3}{2}e^{-t/3}u(t) - \frac{3}{2}e^{-3t} u(t)$[/tex]

       [tex]$=\frac{-3}{2}e^{-\frac{1}{3}t}u(t) + \frac{3}{1}e^{-t}u(t)-\frac{3}{2}e^{-3t} u(t)$[/tex]

where, a = 2

            b = 1

            c= 2

Answer:

first mark me as a brainleast

Answer:

The physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another ( C )

Explanation:

Network physical is simply the process/method of connecting the Network using cables while Logical topology is the general architecture of the communication mechanism in the network for all nodes.

Hence The correct statement is the physical topology addresses how devices are connected, while a logical topology addresses how devices actually communicate to one another

Answer:

D. Turn on the light source and the spectrophotometer.

Explanation:

A spectrophotometer is a machine used to measure the presence of any light-absorbing particle in a solution as well as its concentration. To prepare a spectrophotometer for use, the first step is to turn on the spectrophotometer and allow it to warm up for at least 15 minutes. After this is done, the next step will be to ensure that the samples and blank are ready. Next, an appropriate wavelength is set for the solute being determined. Finally, the absorbance is measured of both the blank and samples.

Answer:

The code is as follows:

<form name = "myForm">

       <div>

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

       </div>

       <div>

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

       </div>

   </form>

Explanation:

This defines the first button

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

This defines the second button

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

The code is self-explanatory, as it follows all the required details in the question

Answer:

jsow

hfhcffnbxhdhdhdhdhdhdddhdhdgdhdhdhdhdhdhhhdhdjsksmalalaksjdhfgrgubfghhhhhhh

Explanation:

j

Answer:

Input power = 465.63 W

current = 1.94 A

Explanation:

we have the following data to answer this question

V = 9130

i = 0.051

the input power = VI

I = 51.0 mA = 0.051

= 9130 * 0.051

= 465.63 watts

the current = 465.63/240

= 1.94A

therefore the input power is 465.63 wwatts

while the current is 1.94A

the input power is the same thing as the output power.

Answer: Plot of  [tex]\log y[/tex] vs [tex]x[/tex] would be linear with a slope  of 4.

Explanation:

Given

Equation is [tex]y=10^{4x}[/tex]

Taking log both sides

[tex]\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x[/tex]

It resembles with linear equation [tex]y=mx+c[/tex]

Here, slope of [tex]\log y[/tex] vs [tex]x[/tex] is 4.

Answer:

1.49 mm

Explanation:

The modulus of elasticity, Y = stress/strain = σ/ε

σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.

ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,

So, Y = σ/ε

Y = F/A ÷ ΔL/L

Y = FL/AΔL

making the change in length ΔL subject of the formula, we have

ΔL = FL/AY

substituting the value of A into the equation, we have

So, ΔL = FL/(πd²/4)Y

ΔL = 4FL/πd²Y

Since Y = 200 GPa = 200 × 10⁹ Pa

Substituting the values of the variables into the equation, we have

ΔL = 4FL/πd²Y

ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]

ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]

ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]

ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N

ΔL = 0.149 × 10⁻² m

ΔL = 1.49 × 10⁻³ m

ΔL = 1.49 mm

The change in length of the circular rod is 1.49 mm

Answer:

1. Drawing grid.

2. Orthogonal.

3. Geometries.

4. CTRL+N.

5. Thirteen (13).

Explanation:

CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.

Some of the features of an AutoCAD software are;

1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.

2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.

3. Tangent is a point where two geometries meet at just a single point.

4. If you want to create a new drawing, simply press CTRL+N for the short cut key.

5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.

Answer:

embroidery scissor

Explanation:

is small, sharp and pointed good for fine work use trimming scallops,clipping threads,and cutting large eyelets.

hope this helps

Answer:

Walls Built Using Bricks and Wood

The brick house is more energy-efficient than the one built with wood.

Explanation:

Because of their high thermal mass, which gives bricks the ability to absorb heat and release it over time, bricks remain more energy-efficient than other building materials, including wood.  In summer, bricks leave your home cool.  In winter, they make it warm.  With these two advantages provided by bricks over other building materials, bricks are the most energy-efficient building material.

Answer:

The container should be made of lead and the liquid should be water.

Explanation:

Since the volume of the container of liquid after expansion is V = V₀(1 + βΔθ) where V = initial volume, β = coefficient of volume expansion, Δθ = temperature change.

So, the volume change V₂ - V₁ where V₁ = volume of liquid and V₂ = volume of container

For steel, V₂ = V₀(1 + β₂Δθ) and V₁ = V₀(1 + β₁Δθ)

So, ΔV = V₀(1 + β₂Δθ) - V₀(1 + β₁Δθ) = V₀[1 + β₂Δθ - 1 - β₁Δθ] = V₀[β₂Δθ - β₁Δθ]

Since we want a minimum value for ΔV and V₀ and Δθ are the same, we need ΔV/V₀Δθ = β₂ - β₁ to be a minimum

where β₂ = coefficient of volume expansion of liquid and β₁ = coefficient of volume expansion of container.

So, trying each combination, with  β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 171 × 10⁻⁶ (C°)⁻¹

With  β₂ = 207  × 10⁻⁶ (C°)⁻¹] and β₁ = 87  × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 120 × 10⁻⁶ (C°)⁻¹

With  β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 36  × 10⁻⁶ (C°)⁻¹

β₂ - β₁ = 1120 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 1084 × 10⁻⁶ (C°)⁻¹

With  β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 ×  10⁻⁶ (C°)⁻¹

β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 1033 × 10⁻⁶ (C°)⁻¹

The combination that gives the lowest value for β₂ - β₁ is β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹

Since  β₁ = 87 × 10⁻⁶ (C°)⁻¹ = coefficient of expansion for lead β₂ = 207 × 10⁻⁶ (C°)⁻¹]  = coefficient of expansion for water, the container should be made of lead and the liquid should be water.

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

Answer:

The right answer is "16.5 hrs".

Explanation:

Given values are:

[tex]x_1=2.6 \ mm[/tex]

[tex]t_1=3 \ hrs[/tex]

[tex]x_2=6.1 \ mm[/tex]

As we know,

⇒ [tex]\frac{x^2}{Dt}=constant[/tex]

or,

⇒ [tex]\frac{x_1^2}{t_1} =\frac{x_2^2}{t_2}[/tex]

⇒ [tex]t_2=(\frac{x_2}{x_1})^2\times t_1[/tex]

By putting the values, we get

       [tex]=(\frac{6.1}{2.6} )^2\times 3[/tex]

       [tex]=5.5\times 3[/tex]

       [tex]=16.5 \ hrs[/tex]      

Answer:

a) -505.229 kJ/Kg

b) -1.724 kJ/kg

Explanation:

T1 = 400°C

P1 = 3 MPa

P2 = 125 kPa

work output   = 530 kJ/kg

surrounding temperature = 20°C = 293 k

A) Calculate heat transfer from Turbine to surroundings

Q = h2 + w - h1

h ( enthalpy )

h1 = 3231.229 kj/kg

enthalpy at P2

h2 = hg = 2676 kj/kg

back to equation 1

Q = 2676 + 50 - 3231.229  = -505.229 kJ/Kg  ( i.e.  heat is lost )

b) Entropy generation

entropy generation = Δs ( surrounding )  + Δs(system)

                                =  - 505.229 / 293   + 0

                                = -1.724 kJ/kg  

Solution :

Characteristic length  = thickness / 2

                                    [tex]$=\frac{0.04}{2}$[/tex]

                                    = 0.02 m

Thermal conductivity for steel is 42.5 W/m.K

[tex]$\text{Biot number} = \frac{\text{convective heat transfer coefficient} \times \text{characteristic length}}{\text{thermal conductivity}}$[/tex]

                  [tex]$=\frac{30 \times 0.02}{42.5}$[/tex]

                  = 0.014

Since the Biot number is less than 0.01, the lumped system analysis is applicable.

[tex]$\frac{T-T_{\infty}}{T_0-T_{\infty}} = e^{-b\times t}$[/tex]

Where,

T = temperature after t time

[tex]$T_{\infty}$[/tex] = surrounding temperature

[tex]$T_0$[/tex] = initial temperature

[tex]$b=\frac{\text{heat transfer coefficient}}{\text{density} \times {\text{specific heat } \times \text{characteristic length }}}$[/tex]

t = time

We calculate B:

[tex]$b=\frac{30}{7833 \times 460 \times 0.02}$[/tex]

  = 0.000416

Thus, [tex]$\frac{100-50}{500-50}=e^{-0.00416 \times t}$[/tex]

t = 5281.78 second

  = 88.02 minutes

Thus the time taken for reaching 100 degree Celsius is 88.02 minutes.

answer

d just too the test

Answer:

what about it?

Explanation:

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

Answer:

b. 828

Explanation:

UDL = 2.6 - 0.3 [5.5] - 0.02 [ 21 ] = 0.53

USR = 0.7 - 0.3 [5.5 / 2 ] - 0.04 [ 21 ] = -0.69

UB = -0.3 [ 1.25 ] - 0.01 = -0.645

Psr = [tex]\frac{e^{-0.53} }{e^{-0.53} + e^{-0.69} + e^{-0.645} }[/tex]

Solving the equation we get 0.184.

Number of people who will take shared ride is:

0.184 * 4500 = 828 approximately.

The following should be done by the team member:

Learn more: brainly.com/question/17429689

Answer:

A. This option is true

B. This option is false

C. This option is true

Explanation:

A. Generally speaking, the greater percentage of cold, the recrystallization grain size would turn out to be smaller. Therefore this true.

B. A higher annealing temperature does not result in smaller recrystallization grain size. Therefore this is false.

C. As the percentage of cold work is greater, the recrystallization temperature would tend to be lower. Therefore this is true.

Answer:

Bachelor's degree

In order to become a licensed architect in the US and the District of Columbia, applicants are required to complete a professional degree in architecture, gain on-the-job experience through a paid internship.

Answer:

Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

Answer:

The answers are in the explanation. The pH is 5.91

Explanation:

The CH3NH2 reacts with HCl as follows:

CH3NH2 + HCl → CH3NH3⁺ + Cl⁻

When 200mL of HCl are added, the moles of CH3NH2 and HCl are reacting completely producing CH3NH3+ and Cl-. That means the species present are:

no H+. All reacted

yes H2O. Because the water is present in the solutions of HCl and CH3NH2

yes Cl-. Is a product of the reaction

Yes CH3NH2. Is consumed in the reaction but comes from the equilibrium of CH3NH3+

yes CH3NH3+. Is the other product of the reaction. MAJOR SPECIES

When 300.00mL of HCl are added, 100mL are in excess:

yes H+. Is in excess: H+ + Cl- = HCl in water. MAJOR SPECIES. Determine the pH of the solution.

yes H2O. Is present because the reactants are diluted.  

yes Cl-. Is a product of reaction and comes from HCl.

Yes CH3NH2. The reactant is over but comes from the equilibrium of CH3NH3+

yes no CH3NH3+. Yes. Is a product and remains despite HCl is in excess.

To find the pH:

At equivalence point the ion that determines pH is CH3NH3+. Its concentration is:

0.100L * (0.200mol/L) = 0.0200 moles / 0.300L = 0.0667M CH3NH3+

The equilibrium of CH3NH3+ is:

Ka = Kw/kb = 1x10-14/4.4x10-4 = 2.273x10-11 = [H+] [CH3NH2] / [CH3NH3+]

As both [H+] [CH3NH2] comes from the same equilibrium:

[H+] =  [CH3NH2] = X

2.273x10-11 = [X] [X] / [0.0667M]

1.5159x10-12 = X²

X = 1.23x10-6M = [H+]

As pH = -log [H+]

pH = 5.91

The pH at the equivalent point for this titration is "5.91".

[tex]CH_3NH_2 = 0.200\ M\\\\ \text{volume} = 100.0\ mL = 0.100\ L\\\\HCl = 0.100\ M\\\\[/tex]

We must now quantify the pH well at the equivalence point.

We know that even at the point of equivalence, moles of acid and moles of the base are equivalent. As such, first, we must calculate the number of moles of the given base.

Calculating the Moles in [tex]CH_3NH_2 = 0.200\ M \times 0.100\ L = 0.0200\ moles[/tex]

Calculating the Moles in [tex]HCl = 0.0200 \ moles[/tex]

Calculating the volume of [tex]HCl[/tex]:

[tex]\to \text{Molarity} = \frac{ \text{moles}}{\text{volume \ (L)}} \\\\\to \text{Volume} = \frac{\text{moles}}{\text{molarity}}\\\\[/tex]

                [tex]= \frac{0.0200 \ moles}{ 0.100\ M}\\\\= 0.200 \ L\\\\= 200 \ mL\\\\[/tex]

Calculating the reaction among the acid and base:

[tex]CH_3NH_2 + HCl \longrightarrow CH_3NH_3^{+} + Cl^-[/tex]

[tex]0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0200[/tex]

Therefore the conjugate acid of the bases exists at the standard solution.

Then we must calculate the new molar mass of [tex]CH_3NH_3^+[/tex].

Total volume[tex]= 100 + 200 = 300\ mL = 0.300\ L[/tex]

[tex][CH_3NH_3^+] = \frac{0.0200\ mole}{ 0.300\ L}= 0.0667\ M[/tex]

Using the ICE table

[tex]CH_3NH_3^+ + H_2O \longrightarrow CH_3NH_2 + H_3O^+[/tex]

[tex]I \ \ \ \ \ \ \ \ \ \ 0.0667 \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ 0\\\\C\ \ \ \ \ \ \ \ -x\ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ +x\\\\E \ \ \ \ \ \ \ \ \ \ \ \ 0.0667-x \ \ \ \ \ \ \ \ \ \ \ \ +x \ \ \ \ \ \ \ \ \ \ \ \+x\\\\\to Ka = \frac{[CH_3NH_2] [H_3O^+] }{[CH_3NH_3^+]}[/tex]

Calculating [tex]K_a[/tex] from [tex]K_b[/tex]

[tex]\to K_a \times K_b = 1\times 10^{-14}\\\\\to K_a = \frac{1\times 10^{-14}}{4.4\times 10^{-4}} = 2.27\times 10^{-11}\\\\[/tex]

                           [tex]= 2.27\times 10^{-11} \\\\= x\times \frac{x}{(0.0667-x)}[/tex]

The x in the 0.0667-x can be ignored since the Ka value is just too small and it also does not follow the five percent criteria.

[tex]\to 2.27 \times 10^{-11} \times 0.0667 = x_2\\\\\to x_2 = 1.515\times 10^{-12}\\\\\to x = 1.23\times 10^{-6}\ M\\\\\to [H_3O^+] = x = 1.23\times 10^{-6}\ M\\\\[/tex]

We have the formula to calculate pH.

[tex]\to pH = - \log [H_3O^+] = - \log 1.23\times 10^{-6}\ M= 5.91[/tex]

The pH at the equivalent point for this titration is "5.91".

Find out more information about the pH here:

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