is used to determine the shear stress at point P over the section supporting a downward shear force in the -y direction. What is Q

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) =  30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

[tex]R = \dfrac{u^2}{15(e+f_s)}[/tex]

where;

R= radius

[tex]f_s[/tex] = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

[tex]f_s[/tex] = 0.20

So;

[tex]R = \dfrac{30^2}{15(0.08+0.20)}[/tex]

[tex]R = \dfrac{900}{15(0.28)}[/tex]

[tex]R = \dfrac{900}{4.2}[/tex]

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B  in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

[tex]\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%[/tex]

Where:

[tex]V_{gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.

[tex]V_{fe}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

[tex]\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%[/tex]

Where:

[tex]m_{fe}[/tex], [tex]m_{gr}[/tex] - Masses of the ferrite and graphite phases, measured in grams.

[tex]\rho_{fe}, \rho_{gr}[/tex] - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

[tex]\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%[/tex]

[tex]\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%[/tex]

If [tex]\rho_{gr} = 2.3\,\frac{g}{cm^{3}}[/tex], [tex]\rho_{fe} = 7.9\,\frac{g}{cm^{3}}[/tex], [tex]m_{gr} = 3.2\,g[/tex] and [tex]m_{fe} = 96.8\,g[/tex], the volume percentage of graphite is:

[tex]\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%[/tex]

[tex]\%V_{gr} = 10.197\,\%V[/tex]

The volume percentage of graphite is 10.197 per cent.

Following are the solution to the given points:

[tex]\to C_{Gr} = 100\\\\ \to C_{\alpha}= 0[/tex]From [tex]Fe-F_{\frac{e}{3}} c[/tex] diagram.  

[tex]\to W_{\alpha} =\frac{C_{Gr}-C_{o}}{C_{Gr}-C_{\alpha}}[/tex]

           [tex]= \frac{100-3.6}{100-0} \\\\= \frac{100-3.6}{100} \\\\= \frac{96.4}{100} \\\\=0.964[/tex]

Calculating the weight fraction of graphite:  

[tex]\to W_{Gr}=\frac{C_0 - c_d}{C_{Gr} -c_d}[/tex]

            [tex]= \frac{3.6-0}{100-0} \\\\ = \frac{3.6}{100} \\\\= 0.036[/tex]

Calculating the volume percent of graphite:

[tex]\to V_{Gr}=\frac{\frac{W_{Gr}}{P_{Gr}}}{\frac{w_{\alpha}}{P_{\alpha}}+ \frac{W_{Gr}}{P_{Gr}}}[/tex]

           [tex]=\frac{\frac{0.036}{2.3}}{\frac{0.964}{7.9}+\frac{0.036}{2.3}}\\\\=0.11368 \times 100\%\\\\=11.368\%[/tex]

Therefore, the final answer is "0.964, 0.036, and 11.368%"

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Answer:

Option C = internal energy stays the same.

Explanation:

The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.

So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.

Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.

The amount of heat,q = Work,w.

In the concept of free expansion the only thing that changes is the volume.

Answer:

false

Explanation:

the changing of a prisoner sentence or another penalty to another less severe

Answer:

a. [tex]\eta _{th}[/tex] = 77.65%

b. bwr = 6.5%

c. 3538.986 kW

d. -163.169 kJ

Explanation:

a. The given property  are;

P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa

p₄/p₁ = 10

P₂/P₁ = p₄/p₃ = √10

p₂ = 100·√10

[tex]T_{2s}[/tex] = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K

T₂ = T₁ + ([tex]T_{2s}[/tex] - T₁)/[tex]\eta _c[/tex] = 300 + (416.85 - 300)/0.8 = 446.0625 K

p₄ = 10×p₁ = 10×100 = 1000 kPa

p₄/p₃ = √10 =

p₃ = 100·√10

T₃ = 300 K

T₃/[tex]T_{4s}[/tex] = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)

[tex]T_{4s}[/tex] = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K

T₄ = T₃ + ([tex]T_{4s}[/tex] - T₃)/[tex]\eta _c[/tex] = 300 + (215.905- 300)/0.8 = 194.881 K

The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28

T₄ = 446.0625 K

T₆ = 1400 K

[tex]T_{7s}[/tex]/T₆ = (1/√10)^(0.4/1.4)

[tex]T_{7s}[/tex] = 1400×(1/√10)^(0.4/1.4)  = 1007.6 K

T₇ = T₆ - [tex]\eta _t[/tex](T₆ - [tex]T_{7s}[/tex]) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K

T₈ = 1400 K

T₉ = 1086.08 K

T₅ = T₄ + [tex]\epsilon _{regen}[/tex](T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K

[tex]\eta _{th}[/tex] =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))

(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765

[tex]\eta _{th}[/tex] = 77.65%

b. Back work ratio, bwr = [tex]bwr = \dfrac{w_{c,in}}{w_{t,out}}[/tex]

((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))

40.9435/627.84 = 6.5%

c. [tex]w_{net, out} = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)][/tex]

Power developed is given by the relation;

[tex]\dot m \cdot w_{net, out}[/tex]

[tex]\dot m \cdot w_{net, out}[/tex]= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW

d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)

-163.169 kJ

Answer:

The temperature will be greater than 25°C

Explanation:

In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.

mathematically

Change in the internal energy of a system ΔU = ΔQ + ΔW

in an adiabatic process, ΔQ = 0

therefore

ΔU = ΔW

where ΔQ is the change in heat into the system

ΔW is the work done by or done on the system

when work is done on the system, it is conventionally negative, and vice versa.

also W = pΔv

where p is the pressure, and

Δv = change in volume of the system.

In this case, work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C

Answer:

False

Explanation:

Because

The effectiveness (ϵ) of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer.

When scaffolds are now being construct or deconstruct, a competent person must supervise the work and train everybody who'll be assisting, and the further discussion can be defined as follows:

Therefore, the final answer is "Option B".

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Answer and Explanation:

Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:

A                    B                       C (output)

0                    0                        0

0                    1                          1

1                     0                         1

1                     1                          0

The logic circuit is shown below

C = A'B + AB'

If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.

Answer:

effectiveness of the heat exchanger will not be double when the length of the heat exchanger is doubled.

Because effectiveness depends on NTU and not necessarily the length of the heat exchanger

Answer:

α = kt

ω = [tex]\frac{kt^2}{2}[/tex]

θ = [tex]\frac{kt^3}{6}[/tex]  

Explanation:

given data

Initial velocity ω = 0

time t = 10 s

Number of revolutions = 20

solution

we will take here first α = kt     .....................1

and as α = [tex]\frac{d\omega}{dt}[/tex]

so that

[tex]\frac{d\omega}{dt}[/tex] = kt      ..................2

now we will integrate it then we will get

∫dω  = [tex]\int_{0}^{t} kt\ dt[/tex]   .......................3

so

ω = [tex]\frac{kt^2}{2}[/tex]

and

ω = [tex]\frac{d\theta}{dt}[/tex]     ..............4

so that

[tex]\frac{d\theta}{dt}[/tex]  = [tex]\frac{kt^2}{2}[/tex]

now we will integrate it then we will get

∫dθ = [tex]\int_{0}^{t}\frac{kt^2}{2} \ dt[/tex]      ...............5

solve it and we get

θ = [tex]\frac{kt^3}{6}[/tex]  

Answer:

Please check explanation for answer

Explanation:

Here, we are concerned with stating the advantages and disadvantages  of using a 6 tube passes instead of a 2 tube passes of the same diameter:

Advantages

* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface

* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too

            Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.

Disadvantages

* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.

* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of  manufacturing costs

Answer:

Option A - fail/ not fail

Explanation:

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______

Answer:

Explanation:

For a job of applications engineer which require excellent technical skills, commitment  to working , ability to deal well and confidently with customers a willingness to travel and very intelligent and well-balanced personality.

The ten questions you should ask Maria to determine if she is qualified for the job are :

Answer:

The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.

Explanation:

This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.

Therefore this is the formula for Half wave rectifier

Vrms = Vm/2 and Vdc

= Vm/π:

Where,

Vrms = rms value of input

Vdc = Average value of input

Vm = peak value of output

Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.

Answer:

a. 0.4544 N

b. [tex]5.112 \times 10^{-5 M}[/tex]

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

[tex]H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O[/tex]

[tex]NaOH\ Mass = Normality \times equivalent\ weight \times\ volume[/tex]

[tex]= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL[/tex]

= 0.27264 g

[tex]NaOH\ mass = \frac{mass}{molecular\ weight}[/tex]

[tex]= \frac{0.27264\ g}{40g/mol}[/tex]

= 0.006816 mol

Now

Moles of [tex]H_2SO_4[/tex] needed  is

[tex]= \frac{0.006816}{2}[/tex]

= 0.003408 mol

[tex]Mass\ of\ H_2SO_4 = moles \times molecular\ weight[/tex]

[tex]= 0.003408\ mol \times 98g/mol[/tex]

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

[tex]= \frac{acid\ mass}{equivalent\ weight \times volume}[/tex]

[tex]= \frac{0.333984 g}{49 \times 0.015}[/tex]

= 0.4544 N

b. And, the acid solution molarity is

[tex]= \frac{moles}{Volume}[/tex]

[tex]= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}[/tex]

= 0.00005112

=[tex]5.112 \times 10^{-5 M}[/tex]

We simply applied the above formulas

The volume of the 0.3200 M, NaOH required to neutralize the H₂SO₄, is

21.30 mL, which gives the following acid solution approximate values;

1) Normality of the acid solution is 0.4544 N

2) The molarity of the acid is 0.2272

Molarity of the NaOH = 0.3200 M

Volume of NaOH required = 21.30 mL

1) The normality of the acid solution is found as follows;

The chemical reaction is presented as follows;

H₂SO₄(aq) + 2NaOH (aq) → Na₂SO₄ (aq) + H₂O

Number of moles of NaOH in the reaction is found as follows;

[tex]n = \dfrac{21.30}{1,000} \times 0.3200 \, M = \mathbf{0.006816 \, M}[/tex]

Therefore;

The number of moles of H₂SO₄ = 0.006816 M ÷ 2 = 0.003408 M

[tex]Normality = \mathbf{ \dfrac{Mass \ of \, Acid \ in \ reaction}{Equivalent \ mass \times Volume \ of \ soltute}}[/tex]

Which gives;

[tex]Normality = \dfrac{ 98 \times 0.003408 }{49 \times 0.015} = \mathbf{0.4544}[/tex]

2) The molarity is found as follows;

[tex]Molarity = \dfrac{0.003408 \, moles}{0.015 \, L} = \mathbf{0.2272 \, M}[/tex]

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Answer:

A. Yes

B. Yes

Explanation:

We want to evaluate the validity of the given assertions.

1. The first statement is true

The sine rule stipulates that the ratio of a side and the sine of the angle facing the side is a constant for all sides of the triangle.

Hence, to use it, it’s either we have two sides and an angle and we are tasked with calculating the value of the non given side

Or

We have two angles and a side and we want to calculate the value of the side provided we have the angle facing this side in question.

For notation purposes;

We can express the it for a triangle having three sides a, b, c and angles A,B, C with each lower case letter being the side that faces its corresponding big letter angles

a/Sin A = b/Sin B = c/Sin C

2. The cosine rule looks like the Pythagoras’s theorem in notation but has a subtraction extension that multiplies two times the product of the other two sides and the cosine of the angle facing the side we want to calculate

So let’s say we want to calculate the side a in a triangle of sides a, b , c and we have the angle facing the side A

That would be;

a^2 = b^2 + c^2 -2bcCosA

So yes, the cosine rule can be used for the scenario above

Answer:

The final volumetric flow rate will be "76.4 m³/s".

Explanation:

The given values are:

[tex]\dot{m_{1}}=10 \ kg/s[/tex]

[tex]\dot{m_{2}}=20 \ Kg/s[/tex]

[tex]T_{1}=293 \ K[/tex]

[tex]T_{2}=313 \ K[/tex]

[tex]P_{1}=P_{2}=P_{3}=10 \ bar[/tex]

As we know,

⇒  [tex]E_{in}=E_{out}[/tex]

[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]

[tex]e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}[/tex]

[tex]\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}[/tex]

⇒  [tex]\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}[/tex]

         [tex]=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}[/tex]

On substituting the values, we get

         [tex]=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}[/tex]

         [tex]=76.4 \ m^3/s[/tex]

Answer:

Working volttage

Explanation:

SMT electrolytic capacitors are marked with working voltage. The value of these capacitors is measured in micro farads. It is a surface mount capacitor which is used for high volume manufacturers. They are small lead less and are widely used. They are placed on modern circuit boards.

Answer:

Explanation:

La vaca

El pato

Answer:

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

Given the following values

Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

Preload stress

αi = 0.75Sp = 0.75 * 120 = 90 kpsi

stiffness constant

C = [tex]\frac{Kb}{Kb + Km}[/tex]  = [tex]\frac{3}{3+2}[/tex] = 0.2

A) yielding factor of safety

nP = [tex]\frac{sPAt}{Cp + Fi}[/tex] = [tex]\frac{120* 0.1419}{0.2*14.17 + 12.771}[/tex]

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

[tex]nL = \frac{SpAt - Fi}{CP}[/tex] = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50