If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respectively, determine the resultant moment produced by all the weights about point A.

Answer:

A.) 7.13 degree north east

B.) 806.23 km/h

C.) 7.44 hours

D.) 0.06 hours

Explanation:

Assume Washington is due east of San Francisco and Francisco with Washington in the positive x-direction

Also, the cross wind is blowing from north to south of 100 km/hr in y coordinate direction.

A.) Using Cartesian system of coordinates, the direction of flight for the plane to actually arrive in Washington can be calculated by using the formula

Tan Ø = y/x

Substitute y = 100 km/h and x = 800km/h

Tan Ø = 100/800

Tan Ø = 0.125

Ø = Tan^-1(0. 125)

Ø = 7.13 degrees north east.

Therefore, the direction of flight for the plane to actually arrive in Washington is 7.13 degree north east

B.) The speed in the San Francisco to Washington direction can be achieved by using pythagorean theorem

Speed = sqrt ( 800^2 + 100^2)

Speed = sqrt (650000)

Speed = 806.23 km/h

C.) Let us use the speed formula

Speed = distance / time

Substitute the speed and distance into the formula

806.23 = 6000/ time

Make Time the subject of formula

Time = 6000/806.23

Time = 7.44 hours

D.) If there is no cross wind,

Time = 6000/800

Time = 7.5 hour

Time difference = 7.5 - 7.44

Time difference = 0.06 hours

The given question is incomplete. The complete question is as follows.

Steam is contained in a closed rigid container with a volume of 1 m3. Initially, the pressure and temperature of the steam are 10 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine

(a) the temperature at which condensation first occurs, in [tex]^{o}C[/tex],

(b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.

(c) What is the volume, in [tex]m^{3}[/tex], occupied by saturated liquid at the final state?

Explanation:

Using the property tables

  [tex]T_{1} = 500^{o}C[/tex],    [tex]P_{1}[/tex] = 10 bar

  [tex]v_{1} = 0.354 m^{3}/kg[/tex]

(a) During the process, specific volume remains constant.

  [tex]v_{g} = v_{1} = 0.354 m^{3}/kg[/tex]

  T = [tex](150 - 160)^{o}C[/tex]

Using inter-polation we get,

      T = [tex]154.71^{o}C[/tex]

The temperature at which condensation first occurs is [tex]154.71^{o}C[/tex].

(b) When the system will reach at state 3 according to the table at 0.5 bar then

  [tex]v_{f} = 1.030 \times 10^{-3} m^{3}/kg[/tex]

  [tex]v_{g} = 3.24 m^{3} kg[/tex]

Let us assume "x" be the gravity if stream

   [tex]v_{1} = v_{f} + x_{3}(v_{g} - v_{f})[/tex]

   [tex]x_{3} = \frac{v_{1} - v_{f}}{v_{g} - v_{f}}[/tex]

               = [tex]\frac{0.3540 - 0.00103}{3.240 - 0.00103}[/tex]

               = 0.109

At state 3, the fraction of total mass condensed is as follows.

  [tex](1 - x_{5})[/tex] = 1 -  0.109

                = 0.891

The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.891.

(c) Hence, total mass of the system is calculated as follows.

     m = [tex]\frac{v}{v_{1}}[/tex]

         = [tex]\frac{1}{0.354}[/tex]

         = 2.825 kg

Therefore, at final state the total volume occupied by saturated liquid is as follows.

     [tex]v_{ws} = m \times v_{f}[/tex]

                 = [tex]2.825 \times 0.00103[/tex]

                 = [tex]2.9 \times 10^{-3} m^{3}[/tex]

The volume occupied by saturated liquid at the final state is [tex]2.9 \times 10^{-3} m^{3}[/tex].

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

[tex] V = A*t [/tex]

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]    

Now, using the density we can find the mass:

[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]

Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):

[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

Answer:

Explanation:

The value of a will be zero as it is provided that the particle is on the x-axis.

Calculate the velocity of particles along x-axis.

[tex]{\bf{V}} = 2{x^2}t{\bf{\hat i}} + [4y(t - 1) + 2{x^2}t]{\bf{\hat j}}{\rm{ m/s}}[/tex]

Substitute 0 for y.

[tex]\begin{array}{c}\\{\bf{V}} = 2{x^2}t{\bf{\hat i}} + \left( {4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t} \right){\bf{\hat j}}{\rm{ m/s}}\\\\ = 2{x^2}t{\bf{\hat i}} + 2{x^2}t{\bf{\hat j}}{\rm{ m/s}}\\\end{array}[/tex]

Here,

[tex]A = 2{x^2}t \ \ and\ \ B = 2{x^2}t[/tex]

Calculate the magnitude of vector V .

[tex].\left| {\bf{V}} \right| = \sqrt {{A^2} + {B^2}}[/tex]

Substitute

[tex]2{x^2}t \ \ for\ A\ and\ 2{x^2}t \ \ for \ B.[/tex]

[tex]\begin{array}{c}\\\left| {\bf{V}} \right| = \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\\\ = \left( {2\sqrt 2 } \right){x^2}t\\\end{array}[/tex]

The velocity of the fluid particles on the x-axis is [tex]\left( {2\sqrt 2 } \right){x^2}t{\rm{ m/s}}[/tex]

Calculate the direction of flow.

[tex]\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} )[/tex]

Here, θ is the flow from positive x-axis in a counterclockwise direction.

Substitute [tex]2{x^2}t[/tex] as A and [tex]2{x^2}t[/tex] as B.

[tex]\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2{x^2}t}}{{2{x^2}t}}} \right)\\\\ = {\tan ^{ - 1}}\left( 1 \right)\\\\ = 45^\circ \\\end{array}[/tex]

The direction of flow is [tex]45^\circ[/tex] from the positive x-axis.

Answer:

The function is [tex]-\sin x[/tex] and the constant of integration is [tex]C = - 1[/tex].

Explanation:

The resultant expression is equal to the sum of a constant multiplied by the integral of a given function and an integration constant. That is:

[tex]a = k\cdot \int\limits {f(x)} \, dx + C[/tex]

Where:

[tex]k[/tex] - Constant, dimensionless.

[tex]C[/tex] - Integration constant, dimensionless.

By comparing terms, [tex]k = 2[/tex], [tex]C = -1[/tex] and [tex]\int {f(x)} \, dx = \cos x[/tex]. Then, [tex]f(x)[/tex] is determined by deriving the cosine function:

[tex]f(x) = \frac{d}{dx} (\cos x)[/tex]

[tex]f(x) = -\sin x[/tex]

The function is [tex]-\sin x[/tex] and the constant of integration is [tex]C = - 1[/tex].

Answer:

In consolidated AB Co 300 shares.

Explanation:

Consolidation is a process in which two different organizations are united. In this question A Co and B Co are consolidated and a new Co names AB Co is formed. The shares of both the companies will be combined and their total share capital will be increased.

Answer:T = 167.3 N

Explanation:

Given that the

Weight mg = 110 N

The mass m of the ball will be

m = 110/9.8 = 11.22 kg

As the direction of the ball’s velocity is changing, the force responsible for this is centripetal force F. And

F = mV^2/r

Where

V = 5.0 m/s

r = L = 4.9 m

m = 11.22

Substitute all these parameters into the formula

F = (11.22 × 5^2)/4.9

F = 280.6/4.9

F = 57.27 N

Tension T = F + mg

Substitute F and mg into the formula

T = 57.27 + 110

T = 167.3 N

Therefore, the tension in the rope at that point is 167.3 N

Answer:

The conversion in the real reactor is = 88%

Explanation:

conversion = 98% = 0.98

process rate = 0.03 m^3/s

length of reactor = 3 m

cross sectional area of reactor = 25 dm^2

pulse tracer test results on the reactor :

mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2

note:  space time (t) =

t = [tex]\frac{A*L}{Vo}[/tex]   Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor

therefore (t) = [tex]\frac{25*3*10^{-2} }{0.03}[/tex] = 25 s

since the reaction is in first order

X = 1 - [tex]e^{-kt}[/tex]

[tex]e^{-kt}[/tex] = 1 - X

kt = In [tex]\frac{1}{1-X}[/tex]

k = In [tex]\frac{1}{1-X}[/tex] / t  

X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then  

K = 0.156 [tex]s^{-1}[/tex]

Calculating Da for a closed vessel

; Da = tk

      = 25 * 0.156 = 3.9

calculate Peclet number Per using this equation

0.65 = [tex]\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})[/tex]

therefore

[tex]\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0[/tex]

solving the Non-linear equation above( Per = 1.5 )

Attached is the Remaining part of the solution

Answer:

219kJ

Explanation:

The work done (W) on a gas in an isothermal process is given by;

W = -P₁V₁ ln[tex]\frac{V_{2}}{V_1}[/tex]      -----------------(i)

Where;

P₁ = initial pressure of the gas

V₁ = initial volume of the gas

V₂ = final volume of the gas

From the question;

P₁ = 600kPa = 6 x 10⁵Pa

V₁ = 0.45m³

V₂ = 0.2m³

Substitute these values into equation (i) as follows;

W = -6 x 10⁵ x 0.45 x ln [tex]\frac{0.2}{0.45}[/tex]

W = -6 x 10⁵ x 0.45 x ln (0.444)

W = -6 x 10⁵ x 0.45 x -0.811

W = 2.19 x 10⁵

W = 219 x 10³

W = 219kJ

Therefore, the work done on the gas during the compression process is 219kJ

Answer:

See explanation

Explanation:

Solution:-

- Three students measure the volume of a liquid sample which is 6.321 L.

- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:

                                                 Students

                      Trial          A            B               C

                         1            6.35        6.31          6.38

                        2            6.32        6.31          6.32

                        3            6.33        6.32         6.36

                        4            6.36        6.35         6.36

- We will define the two terms stated in the question " precision " and "accuracy"

- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.

- The mean measurement taken by each student would be as follows:

                       [tex]E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\[/tex]

- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:

                        [tex]Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\[/tex]

- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:

                   Var ( A )          <          Var ( B )        <    Var ( C )

                   most precise                                      Least precise

- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .

                       [tex]P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\[/tex]

- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:

                   P ( B )         <       P ( A )         <      P ( C )

            most accurate                                least accurate

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

       P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

      h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

       s1 = s-P1 = 0.6492 KJ/kg.K

       v1 = v-P1 = 0.001010 m^3 / kg

       P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

       s2 = s1 = 0.6492 KJ/kg.K   .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

                           [tex]w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}[/tex]

- From the following relation we can determine ( h2 ) as follows:

                          h2 = h1 + wp

                          h2 = 191.81 + 8.0699

                          h2 = 199.88 KJ/kg

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

              P3 = 8 MPa

              T3 = ?  ( assume fluid exist in the saturated vapor phase )

              h3 = hg-P3 = 2758.7 KJ/kg

              s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

                          [tex]q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}[/tex]

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

             P4 = 10 KPa

             s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

- Compute the quality of the mixture at condenser inlet by the following relation:

                           [tex]x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947[/tex]

- Determine the isentropic ( h4s ) at this state as follows:

                          [tex]h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}[/tex]        

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

                         [tex]h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\[/tex]

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

                        [tex]w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}[/tex]

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

                       [tex]W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}[/tex]

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

                        [tex]n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31[/tex]

Answer: The thermal efficiency of the cycle is 0.31

Answer:

a) 2.45 KN

b) 0.0375 m

Explanation:

[tex](a) \quad \sigma_{v}=400 \times 10^{6} \mathrm{Pa} \quad A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(5)^{2}=19.635 \mathrm{mm}^{2}=19.635 \times 10^{-6} \mathrm{m}^{2}[/tex]

[tex]P_{U}=\sigma_{U} A=\left(400 \times 10^{6}\right)\left(19.635 \times 10^{-6}\right)=7854 \mathrm{N}[/tex]

[tex]P_{\text {al }}=\frac{P_{U}}{F S}=\frac{7854}{3.2}=2454 \mathrm{N}[/tex]

(b) [tex]\quad \delta=\frac{P L}{A E}=\frac{(2454)(60)}{\left(19.635 \times 10^{-6}\right)\left(200 \times 10^{9}\right)}=37.5 \times 10^{-3} \mathrm{m}[/tex]

Question:Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline . Which of them is correct ?

Answer: Technician  B is   correct

Explanation: Two types of engines exist , the two stroke (example, used in chainsaws)  is a type of engine that uses two strokes--a compression stroke and a return stroke to produce power in a crankshaft combustion cycle and the  four stroke engines(eg lawnmowers) which  uses four strokes,  2-strokes during  compression and exhaustion accompanied by 2 return strokes for each of the initial process to produce power in a combustion cycle.

While a 2 stroke system engine, requires mixing of oil and fuel to the crankshaft before  forcing  the mixture into the cylinder and do not require a pressurized system.  The 4 stroke system uses a splash and pressurized system where oil is not mixed with gasoline but drawn from the sump and  directed to  the main moving  parts of  crankshaft through its channels.

We can therefore say that Technician A is wrong while Technician B is  correct

Answer:

I have attached the diagram for this question below. Consult it for better understanding.

Find the cross sectional area AB:

A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m

Forces is given by:

F = 2.75 × 10³ N

Horizontal Stress can be found by:

σ (x) = F/A

σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m

σ (x) = 143.23 × 10⁶ Pa

Horizontal Strain can be found by:

ε (x) = σ (x)/ E

ε (x) = 143.23 × 10⁶ / 200 × 10⁹

ε (x) = 716.15 × 10⁻⁶

Find Vertical Strain:

ε (y) = -v · ε (y)

ε (y) = -(0.3)(716.15 × 10⁻⁶)

ε (y) = -214.84 × 10⁻⁶

For L = 0.05m

Change (x) = L · ε (x)

Change (x) = 35.808 × 10⁻⁶m

For W = 0.012m

Change (y) = W · ε (y)

Change (y) = -2.5781 × 10⁻⁶m

For t= 0.0016m

Change (z) = t · ε (z)

where

ε (z) = ε (y) ,so

Change (z) = t · ε (y)

Change (z) = -343.74 × 10⁻⁹m

A = A(final) - A(initial)

A = -8.25 × 10⁻⁹m²

(Consult second picture given below for understanding how to calculate area)

The resulting change in the 50-mm gauge length; the width of portion AB of the test coupon; the thickness of portion AB; the cross- sectional area of portion AB are respectively; Δx = 35.808 × 10⁻⁶ m; Δy = -2.5781 × 10⁻⁶m; Δ_z = -343.74 × 10⁻⁹m; A = -8.25 × 10⁻⁹m²

Let us first find the cross sectional area of AB from the image attached;

A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m

We are given;

Tensile Load; F = 2.75 kN = 2.75 × 10³ N

Horizontal Stress is calculated from the formula;

σₓ = F/A

σₓ = (2.75 × 10³)/(19.2 × 10⁻⁶)m

σₓ = 143.23 × 10⁶ Pa

Horizontal Strain is calculated from;

εₓ = σₓ/E

We are given E = 200 GPa = 200 × 10⁹ Pa

Thus;

εₓ = (143.23 × 10⁶)/(200 × 10⁹)

εₓ = 716.15 × 10⁻⁶

Formula for Vertical Strain is;

ε_y = -ν * εₓ

We are given ν = 0.30. Thus;

ε_y  = -(0.3) * (716.15 × 10⁻⁶)

ε_y  = -214.84 × 10⁻⁶

A) We are given;

Gauge Length; L = 0.05m

Change in gauge length is gotten from;

Δx = L * εₓ

Δx = 0.05 × 716.15 × 10⁻⁶

Δx = 35.808 × 10⁻⁶ m

B) From the attached diagram, the width is;

W = 0.012m

Change in width is;

Δy = W * ε_y

Δy = 0.012 * -214.84 × 10⁻⁶

Δy = -2.5781 × 10⁻⁶m

C) We are given;

Thickness of plate; t = 1.6 mm = 0.0016m

Change in thickness;

Δ_z = t * ε_z

where;

ε_z = ε_y

Thus;

Δ_z = t * ε_y

Δ_z = 0.0016 * -214.84 × 10⁻⁶

Δ_z = -343.74 × 10⁻⁹m

D) The change in cross sectional area is gotten from;

ΔA = A_final - A_initial

From calculating the areas, we have;

A = -8.25 × 10⁻⁹ m²

Read more about stress and strain in steel plates at; https://brainly.com/question/1591712

Answer:1 common law

Explanation:

It also says that if Cathy wasn’t wearing a hardhat hat she is responsible for her own injury, more than one answer is correct.

In this case, if Cathy's employer completes compliance and general duty requirements then the organization may not be held liable and again, the law can generally hold Cathy responsible for the injuries as she was not wearing the proper kits for such work.

According to OSHA, Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.

You are entitled to a secure workplace. To stop workers from being murdered or suffering other types of harm at work, the Occupational Safety and Health Act of 1970 (OSH Act) was passed. According to the legislation, companies are required to give their workers safe working environments.

Therefore, more than one answer is correct.

Learn more about OSHA, here:

https://brainly.com/question/13127795

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Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

Mathematical modeling aids in technological design by simulating how proposed system might behave. The correct option is 2.

Mathematical modelling describes a real world problem in mathematical terms or in the form of equations. This makes an engineer to discover new features about the problem and designer to alter his design for better function and output.

Mathematical models allow engineers and designers to understand how the proposed model and actual prototype will be produced.

Thus, the correct option is 2.

Learn more about mathematical modelling

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Answer:

height ≈ 60.60 m

Explanation:

The surveyor is trying to find the height of the hill . He takes a sight on the top of the hill and finds the angle of elevation is 40°. The distance from the hill where he measured the angle of elevation of 40° is not known.

Now he moves 150 m on level ground directly away from the hill and take a second sight from this point and measures the angle of elevation as 22°. This illustration forms a right angle triangle. The opposite side of the triangle is the height of the hill. The adjacent side of the triangle which is 150 m is the distance on level ground directly away from the hill.

Using tangential ratio,

tan 22° = opposite/adjacent

tan 22° = h/150

h = 150 × tan 22°

h = 150 × 0.40402622583

h = 60.6039338753

height ≈ 60.60 m

Answer:

The power that fluid supplies to the turbine is 1752.825 kilowatts.

Explanation:

A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:

[tex]-\dot Q_{loss} - \dot W_{out} + \dot m \cdot \left[(h_{in}-h_{out}) + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right] = 0[/tex]

Output power is cleared:

[tex]\dot W_{out} = -\dot Q_{loss} + \dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right][/tex]

If [tex]\dot Q_{loss} = 0.3\,kW[/tex], [tex]\dot m = 0.85\,\frac{kg}{s}[/tex], [tex]h_{in} = 2800\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 1550\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 45\,\frac{m}{s}[/tex] and [tex]v_{out} = 20\,\frac{m}{s}[/tex], then:

[tex]\dot W_{out} = -0.3\,kW + \left(0.85\,\frac{kg}{s} \right)\cdot \left\{\left(2800\,\frac{kJ}{kg}-1550\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(45\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]

[tex]\dot W_{out} = 1752.825\,kW[/tex]

The power that fluid supplies to the turbine is 1752.825 kilowatts.

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

Answer:

Electricity cost = $9.36

Explanation:

Given:

Electric power = 400 W = 0.4 KW

Unit cost of electricity = $0.13/kWh

Overall time = 1/4 (30 days) (24 hours) = 180 hours

Find:

Electricity cost

Computation:

Electricity cost = Electric power  x Unit cost of electricity x Overall time

Electricity cost = 0.4 x $0.13 x 180

Electricity cost = $9.36

Given:

Electric power = 400 W = 0.4 KW

Over all Time  = 30(1/4) = 7.5 days

Unit cost of electricity = $0.13/kWh

Find:

Electricity cost.

Computation:

Electricity cost = Electric power x Unit cost of electricity x Over all Time

Electricity cost = 0.4 x 0.13 x 7.5

Electricity cost = $

Answer:

Exit temperature = 32°C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

Answer:

D) Lower efficiency and lower effectiveness.

Explanation:

Given;

Two finned surfaces with long fins which are identical,

with difference in the convection heat transfer coefficient,

The first finned surface has a higher convection heat transfer coefficient, but gives the same heat rate as the second, which will make it (first finned surface) to have lower efficiency and lower effectiveness than the second finned surface.

Therefore, the correct option is "(D) Lower efficiency and lower effectiveness"