1. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle.

i. True
ii. False

2. The cosine rule is used when we are given either a) three sides or b) two sides and the included angle.

i. True
ii. False

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

Answer:

Explanation:

La vaca

El pato

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:

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:

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 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:

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:

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:

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:

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]

Learn more about the normality and the molarity of a solution here:

https://brainly.com/question/6532653

https://brainly.com/question/14112872

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 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:

false

Explanation:

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

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%"

Learn more Graphite:

brainly.com/question/4770832