The efficiency of a steam power plant can beincreased by bleeding off some of the steam thatwould normally enter the turbine and usingit topreheat the water entering the boiler. In this process,liquid water at 50oC and 1000 kPa is mixed withsuperheated steam at 200oC and 1000 kPa. If the plantoperators want to produce a saturated liquid at 1000kPa, what ratio of mass flow rates of water andsuperheatedsteam are required

Answer:

Attached below is the derived next state equations

Explanation:

Attached below is the derived next state equations

used for the solution of the given problem.

Answer:

True, That is correct. Soil removed from an excavation site is indeed called spoil.

Spoil definition: The waste material (such as soil) brought up during the course of an excavation.

Hope I helped! Sorry if not. Have a wonderful week and follow me for more help! Remember your worth and love yourself! Adíos! ;D

Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.

Thus,  The debris that is dug up during an excavation, such as soil. Depending on the location and the type of excavation, the composition of the spoil material can change.

It could consist of dirt, gravel, boulders, debris, or other substances that were removed or moved during the excavation process.

Typically, the spoil is put aside and used for a variety of tasks, including backfilling, grading, or disposal, as necessary for the project and permitted by local laws.

Thus, Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.

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Answer:

Die of intoxication by water first

Explanation:

We assume that the weight of the man is 154.35 pounds which is 70 kg

LD50 water = 90g per kg

Maximum concentration = 90x70

= 6300grams

Convert grams to liters

6300/100

= 6.3 litres

From here we get amount of kool aid

6.3 x 1.97x10^-4

= 1.24x10^-3

= 1.24grams

1.24 grams is below 420 kool aid is lower than LD50 with about 6 grams for 1 kg (6x70kg = 420). So 420 is lethal dose. But 1.24 is less than this so the man has to die of water intoxication first.

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

Determine the maximum gage pressure in the tank

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

Answer:

1.933 KN-M

Explanation:

Determine the largest permissible bending moment when the composite bar is bent  horizontally

Given data :

modulus of elasticity of steel = 200 GPa

modulus of elasticity of aluminum = 75 GPa

Allowable stress for steel = 220 MPa

Allowable stress for Aluminum = 100 MPa

a = 10 mm

First step

determine moment of resistance when steel reaches its max permissible stress

next : determine moment of resistance when Aluminum reaches its max permissible stress

Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M

attached below is a detailed solution

Answer:

t2 = 256 hours

Explanation:

Given data:

Carbon concentration ( C ) at 1.2mm from surface, C = 0.38 wt%

Duration( t ) of heat treatment for 0.38wt% at 1.2mm =  9-hr

Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm

Assuming same concentration of 0.38wt%  we will apply Fick's second law for constant surface concentration

attached below is the remaining part of the solution

x1 = 1.2 mm

x2 = 6.4 mm

t1 = 9-hr

t2 = ?

t2 = [tex](\frac{6.4}{1.2} )^{2} * 9[/tex]  = 256 hours

Hi sorry I need points I'm New

Answer: My favorite color is Red favorite letter C favorite fruit watermelon.

Answer:

0.08704 W

Explanation:

converting the mm to m (1000mm = 1m)

cross-sectional area of the fins, Ac = (0.003) (0.0004) = 0.0000012‬m^2

The wetted perimeter of the cross-section, P = 2 (0.003 + 0.0004) = 0.0068‬m

Thickness of solid in direction of heat flow, B^2 = (heat transient coefficient, h) (The wetted perimeter of the cross-section, P) ÷ (Thermal conductivity, k) (cross-sectional area of the fins, Ac)

B^2 = (8 W/m2K)(0.0068‬m) ÷ (175 W/mK)(0.0000012‬m^2)

=259.0476m^-2

B= square root of the result

B = 16.09m^-1

we now look for:

The Coordinate, x = B, multiplied by Length, L

x = (16.09m^-1) (0.04m) = 0.6436‬

 finding the side area of a fin =  P multiplied by Length, L

= 0.0068‬m X 0.04m = 0.000272m^2

Neglecting inefficiency, assuming the fins are all 100% efficient, the power they would dissipate =

h, Heat-transfer coefficient (PL) (temperature of at the base - temperature at the ambient air)

= (8) (0.000272m^2)(340 K- 300k)

= 0.08704 W

Where are your answer options?

Answer:

Implement a hazard communication program

Explanation: i took the quiz

Answer:

When a switch is in the "off" position the circuit is open. Electric charges cannot flow when a switch is in the off position.

Explanation:

A switch that can open or close an electric circuit can be used to stop the flow of current.

A switch can be defined as an electrical component (device) that is typically designed and developed for interrupting the flow of current or electrons in an electric circuit.

This ultimately implies that, a switch that can open or close an electric circuit can be used to stop the flow of current.

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Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

[tex]P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}[/tex]

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

[tex]\eta_t[/tex] = The efficiency of the pump = 78%

[tex]\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}[/tex]

[tex]Q_{max}[/tex] = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump [tex]Q_{max}[/tex] ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

Answer:

El =50 degre

Explanation:

Yan ang answer

Answer:

Explanation:

Given that:

[tex]y = \int^t_og'(t-s) f(s) ds \ \text{is solution to } \ my"ky= f(t)[/tex]

where;

[tex]g'(0) = \dfrac{1}{m}[/tex]     and [tex]mg"+kg = 0[/tex]

[tex]\text{Using Leibniz Formula to prove the above equation:}[/tex]

[tex]\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt[/tex]

So, [tex]y = \int ^t_0 g' (t-s) f(s) \ ds[/tex]

[tex]\text{By differentiation with respect to t;}[/tex]

[tex]y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\ y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds[/tex]

[tex]y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)[/tex]

[tex]Since \ \ mg" (t) +kg (t) = 0 \\ \\ \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\ put \ t \ =0 \ we \ get;\\g" (0) = - \dfrac{k}{m } g(0) \\ \\ g"(0) = 0 \ \ \ \ ( because \ g(0) =0) \\ \\[/tex]

[tex]Now \ differentiating \ equation (111) \ with \ respect \ to \ t \\ \\ g"'(t) = -\dfrac{k}{m}g(t) \\ \\ replacing \ it \ into \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\ y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\ my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}[/tex]

Answer:

Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.

Explanation:

Answer: hello attached below is the properly written chain reaction  to your question

answer :

d[NO] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]

d[N] / dt = [tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]

Explanation:

write out expressions for  d[NO] / dt and d[N] / dt

Given :

properly written chain reaction ( attached below)

Expression for d[NO] / dt can be written as

[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH][/tex]

Expression for d[N] / dt can be written as

[tex]k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH][/tex]

Answer:

C Electrical Connections

Explanation:

In reading says . However, electrical

connections aren’t shown in construction drawings.

Answer:

Explanation:

The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.

From the given information:

The elastic modulus is:

[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]

Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.

The shear yield strength by using von Mises criteria is estimated as;

[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]

The shear yield strength by using Tresca criteria is:

[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]

Answer:

a

Explanation:

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

[tex]\theta[/tex] = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑[tex]f_{y}[/tex] = 0, ∑M = 0

If ∑[tex]M_{A}[/tex] = 0

⇒[tex]F_{BC}[/tex]×0.9 - P × 0.6 = 0

⇒[tex]F_{BC}[/tex]×3 - P × 2 = 0

⇒[tex]F_{BC}[/tex] = [tex]\frac{2P}{3}[/tex]

If ∑[tex]M_{B}[/tex] = 0

⇒[tex]F_{AD}[/tex]×0.9 = P × 0.3

⇒[tex]F_{AD}[/tex] ×3 = P

⇒[tex]F_{AD}[/tex] = [tex]\frac{P}{3}[/tex]

Now,

Area, A = [tex]\frac{\pi }{4} X (0.012)^{2}[/tex] = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , [tex]\delta[/tex] = [tex]\frac{P l}{A E}[/tex]

Now,

[tex]\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 9.1626 × 10⁻⁹ P

[tex]\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 1.83253 × 10⁻⁸ P

Given that,

[tex]\theta[/tex] = 0.015°

⇒[tex]\theta[/tex] = 2.618 × 10⁻⁴ rad

So,

[tex]\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}[/tex]

⇒2.618 × 10⁻⁴ = (  1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

The magnitude of Force P is; P = 25715.15 N

If we draw a free body diagram of the rigid beam system, then for beam AB we can take moments in the following manner;

Taking moments about point A, we have;

(F_bc * 0.9) - P(0.6) = 0

F_bc = ²/₃P

Taking moments about B gives;

P(0.3) - F_ad * 0.9 = 0

F_ad = ¹/₃P

Normal stress for BC is;

σ_bc = F_bc/A_bc

σ_bc = (²/₃P)/(π * 0.006²)

σ_bc = (²/₃P)/(1.131 × 10⁻⁴) N/m²

σ_ad = (¹/₃P)/(π * 0.006²)

σ_ad = (¹/₃P)/(1.131 × 10⁻⁴) N/m²

We know that;

Elongation is; ΔL = PL/AE = (P/A) * (L/E)

Where E for 304 stainless steel is 193 GPa = 193 × 10⁹ Pa

Thus;

ΔL_bc =  (²/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))

ΔL_bc = 1.83253P × 10⁻⁸

Likewise;

ΔL_ad = (¹/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))

ΔL_ad = 9.1626P × 10⁻⁹ m

Converting the beam tilt angle from degrees to radians gives;

θ = 0.015° = 0.00026179939 rads

Using small angle analysis, we can say that;

θ = (ΔL_bc - ΔL_ad)/36

θ = P((1.83253P × 10⁻⁸) - (9.1626P × 10⁻⁹))/36

Solving gives P = 25715.15 N

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Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Explanation:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.

Explanation:

hope it helps <33

Answer:

I would say Firewall security. Malware is pretty corrupting, but the firewall I think is good for protection if you have a good one.

Explanation:

Answer:

it is indeed C

Explanation:

Answer:

c

Explanation:

Answer:

3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O

Explanation:

It is all about unit conversions.

The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.

160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm

By the way, now we can determine the volume of MOS, in order to work with density.

1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³

But density is mg/m³, so we convert mm³ to m³

1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³

Now, we apply density to determine the mass of MOS

Density = mass /volume → Density . volume = mass

1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg

To make more easier the calculate, we convert mg to g.

3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g

To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol

We need to know moles of Si and O₂ in the MOS

Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles

1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:

5.86×10⁻¹⁸ mol of SiO₂ may have:

(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si

(5.86×10⁻¹⁸ .2) /1 =  1.17×10⁻¹⁷ moles of O₂

Let's count the atoms (1 mol of anything contain NA particles)

5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si

1.17×10⁻¹⁷ mol of O₂  . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O

The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O

We are given;

Area of MOS device; A = 1 mm²

Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm

Formula for volume is;

V = Area * thickness

V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³

Converting volume to m³ gives;

V = 1.6 × 10⁻¹³ m³

Now, to get the mass of MOS, we will use the formula;

Mass = Density * volume

we are given density = 2.20mg/m³

Thus;

Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³

Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g

From periodic table, the molar mass of SiO₂ = 60.08 g/mol

Then;

Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol

Number of moles of MOS = 5.86 × 10⁻¹⁸ moles

Now, 1 mol of SiO₂  is formed from 1 mol of Si and 2 mol of O. Thus;

5.86×10⁻¹⁸ mol of SiO₂ will have;

5.86×10⁻¹⁸ moles of Si

and (5.86×10⁻¹⁸ .2) =  11.72 × 10⁻¹⁸ moles of O₂

From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;

5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si

11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O

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