Q#3:(A)Supose we extend the circular flow mode to add imports and export copy the circular flow digram onto a sheet paper and then add a foreign country as athird agent.Draw a through sketch of the flows of imports exports and the payment for each on your digrams?

Answer:

The meole fraction of acetone in the outlet liquid is [tex]x_1 = 0.0072[/tex]

Explanation:

1.

The schematic diagram to represent this process is shown in the diagram attached below:

2.

the mole fraction of acetone in the outlet liquid is determined as follows:

solute from Basis Gas flow rate [tex]G_s = 10(1-0.012) =9.88 kmol/hr[/tex]

Let the entering mole be :[tex]y_1 = 1.2[/tex] % = 0.012

[tex]y_1 =(\dfrac{y_1}{1-y_1})[/tex]

[tex]y_1 =(\dfrac{0.012}{1-0.012})[/tex]

[tex]y_1 =0.012[/tex]

Let the outlet gas concentration be [tex]y_2[/tex] = 0.1% = 0.001

[tex]y_2 = 0.001[/tex]

Thus; the mole fraction of acetone in the outlet liquid is:

[tex]G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1[/tex]

[tex]9.88(0.012-0.001)=15*x_1[/tex]

[tex]9.88(0.011) = 15x_1[/tex]

[tex]x_1 = \dfrac{0.10868}{15}[/tex]

[tex]x_1 = 0.0072[/tex]

The mole fraction of acetone in the outlet liquid is [tex]x_1 = 0.0072[/tex]

Answer:

Elena es una líder de tipo autoritario.

Explanation:

El liderazgo autoritario representa el control individual por parte del líder de la toma de decisiones y el proceso de elección y planificación en una determinada organización.

En este liderazgo no se promueve la participación efectiva del equipo en los proyectos, solo el líder toma todas las decisiones necesarias y generalmente oprime a sus subordinados. Así, genera muchas veces situaciones de tensión y agotamiento entre los empleados y demás subordinados.

Answer:

"Use your gloves correctly" and "Inspect your gloves for any damage before wearing."

Explanation:

If you don't use your gloves correctly you could risk your hands being infected/burned. If your gloves are damaged they will provide minimum protection but there is still a potential safety hazard.

Answer:

Use your gloves correctly

Inspect your gloves for any damage before wearing

Answer:

Explanation:

Find the temperature at exit of compressor

[tex]T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k[/tex]

Find the work done by the compressor

[tex]\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg[/tex]

Find the actual workdone by the compressor

[tex]\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg[/tex]

Find the temperature at exit of the turbine

[tex]T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k[/tex]

Find the actual workdone by the turbine

[tex]1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg[/tex]

Find the temperature of the regeneration

[tex]\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k[/tex]

Find the heat supplied

[tex]Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg[/tex]

Find the thermal efficiency

[tex]n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4[/tex]

60.4%

Find the mass flow rate

[tex]m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42[/tex]

Find the actual workdone by the compressor

[tex]\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg[/tex]

Find the actual workdone by the turbine

[tex]\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg[/tex]

Find the temperature of the compressor exit

[tex]\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k[/tex]

Find the temperature at the turbine exit

[tex]4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k[/tex]

Find the temperature of regeneration

[tex]\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k[/tex]

Answer:

a) 60.4%;  18.42 kg/s

b) 37.8% ;    35.4 kg/s

Explanation:

a) at an isentropic efficiency of 100%.

Let's first find the exit temperature of the compressor T2, using the formula:

[tex](r_p) ^k^-^1^/^k = \frac{T_2}{T_1}[/tex]

Solving for T2, we have:

[tex] T_2 = 300 * (8)^1^.^6^6^7^-^1^/^1^.^6^6^7 = 689.3 K [/tex]

Let's now find the work dine by the compressor.

[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]

The actual work done by the compressor =

[tex] W_c = 1 * 2020.4 = 2020.4 KJ/kg [/tex]

Let's find the temperature at the exit of the turbine, T4

[tex](r_p) ^k^-^1^/^k = \frac{T_3}{T_4}[/tex]

Solving for T4, we have:

[tex]T_4 = \frac{1800}{(8)^1^.^6^6^7^-^1^/^1^.^6^6^7} = 783.3 K[/tex]

Let's find the work done by the turbine.

[tex]\frac{W_t}{m} = c_p(T_3 - T_4)[/tex]

[tex]\frac{W_t}{m} = 5.19(1800 - 783.3) = 5276.6 KJ/kg[/tex]

The actual work done by the turbine:

= 1 * 5276.6 = 5276.6 KJ/kg

Let's find the regeneration temperature, using the formula:

[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]

Substituting figures, we have:

[tex] 0.75 = \frac{T_r - 689.3}{783.3 - 689.3} [/tex]

[tex] T_r = [0.75(783.3 - 689.3)] + 689.3 = 759.8 [/tex]

Let's calculate the heat supplied.

[tex]Q = c_p(T_3 - T_r)[/tex]

[tex] Q = 5.19(1800 - 759.8) [/tex]

Q = 5388.2 kJ/kg

For thermal efficiency, we have:

[tex] n = \frac{W_t - W_c}{Q} [/tex]

Substituting figures, we have:

[tex] n = \frac{5276.6 - 2020.4}{5388.2} = 0.604 [/tex]

0.604 * 100 = 60.4%

For mass flow rate:

Let's use the formula:

[tex] m = \frac{W_n_e_t}{P} [/tex]

Wnet = 60MW = 60*1000

[tex] m = \frac{60*10^3}{5276.6 - 2020.4} = 18.42 [/tex]

b) at an isentropic efficiency of 80%.

Let's now find the work done by the compressor.

[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]

The actual work done by the compressor =

[tex] W_c = \frac{2020.4}{0.8}= 2525.5 KJ/kg [/tex]

Let's find the work done by the turbine.

[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]

[tex] \frac{W_t}{m} = 5.19(1800 - 787.5) = 5276.6 KJ/kg[/tex]

The actual work done by the turbine:

= 0.8 * 5276.6 = 4221.2 KJ/kg

Let's find the exit temperature of the compressor T2, using the formula:

[tex]\frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] 2525.5 = 5.19(T_2 - 300) [/tex]

Solving for T2, we have:

[tex] T_2 = \frac{2525.5 + 300}{5.19} = 787.5 [/tex]

Let's find the temperature at the exit of the turbine, T4

[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]

[tex] 4221.2 = 5.19(1800 - T_4) [/tex]

Solving for T4 we have:

[tex] T_4 = 958 K[/tex]

Let's find the regeneration temperature, using the formula:

[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]

Substituting figures, we have:

[tex] 0.75 = \frac{T_r - 787.5}{985 - 787.5} [/tex]

[tex] T_r = [0.75(958 - 787.5)] + 787.5 = 935.5 K [/tex]

Let's calculate the heat supplied.

[tex]Q = c_p(T_3 - T_r)[/tex]

[tex] Q = 5.19(1800 - 935.5) [/tex]

Q =  4486.2 kJ/kg

For thermal efficiency, we have:

[tex] n = \frac{W_t - W_c}{Q} [/tex]

Substituting figures, we have:

[tex] n = \frac{4221.2 - 2525.2}{4486.2} = 0.378 [/tex]

0.378 * 100 = 37.8%

For mass flow rate:

Let's use the formula:

[tex] m = \frac{W_n_e_t}{P} [/tex]

Wnet = 60MW = 60*1000

[tex] m = \frac{60*10^3}{4221.2 - 2525.2} = 35.4 kg/s [/tex]

Answer:

A counter which counts from 0 to 255 with seven segment display

Timer Mode Control (TMOD)

Explanation:

The image attached that is supposed to be attached to the question is shown in the first file below.

Answer:

t = 2.19 seconds

Explanation:

The free body diagram showing the center of mass A and B is attached in the second diagram below.

NOTE : that from the second diagram; Mass A and B do not have any acceleration

Taking the moment about wheel A:

[tex]\sum M_A = I_A \alpha _A[/tex]

[tex]-f(r_A) = I_A \alpha _A ----- (1)[/tex]

The equilibrium forces in the y-direction is 0

i.e

[tex]F_y = 0[/tex]

So;

[tex]N +T sin 30^0 -W_A = 0 ----- (2)[/tex]

The equilibrium forces in the x-direction is as follows:

[tex]\sum F_x = 0[/tex]

[tex]Tcos 30^0 + f= 0 -----(3)[/tex]

The kinetic friction f can be expressed as :

[tex]f = \mu _k N[/tex]

From above equation (2) and equation (3);

[tex]N + [\dfrac{-f}{cos 30^0}]sin 30^0 -150 =0[/tex]

[tex]N - \mu _k N \ tan 30^0 -150 =0[/tex]

[tex]N = \dfrac{150}{1-0.3 \ tan 30^0}[/tex]

N = 181.423 lb

Similarly; from equation(1)

[tex]\alpha_A = - \dfrac{f(r_A)}{I_A}[/tex]

[tex]\alpha _A = \dfrac{-\mu_k N(r_A)}{I_A}[/tex]

[tex]\alpha _A = \dfrac{-0.3*181.423*1.25}{\frac{150}{32.2}*I^2}[/tex]

[tex]\alpha _A =-14.6045 \ rad/s^2[/tex]

However; from the kinematics ; as moments are constant ; so is the angular acceleration is constant )

Thus;

[tex]\omega _A - \omega_o^A = \alpha_A t[/tex]

[tex]\omega _A = \omega_o^A + \alpha_A t[/tex]

[tex]\omega _A = 100 -14.6045 \ t ---- (4)[/tex]

Let's take a look at wheel B now;

Taking the moment about wheel B from the equation of motion:

[tex]\sum M_B = I_B \alpha _B[/tex]

[tex]f(r_B) = I_B \alpha _B[/tex]

[tex]\mu_k N (r_B) = I_B \alpha_B[/tex]

[tex]\mu_k N (r_B) = \dfrac{W_B}{g}* k^2_B \alpha_B[/tex]

[tex]\alpha_B = \dfrac{0.3*181.423*1}{\frac{100}{32.2}*0.75^2}[/tex]

[tex]\alpha = 31.1563 \ rad/s^2[/tex]

Again; from the kinematics; as the moments are constant which lead to the angular accleration;

[tex]\omega _B = \omega _o^B + \alpha _B \ t[/tex]

[tex]\omega _B =0 + 31.156 \ t-----(5)[/tex]

From equation 4 and 5 which attain the same angular velocity; we have;

[tex]\omega^A = \omega^B[/tex]

100 - 14.6045 t = 31.1563 t

100 = 31.1563 t + 14.6045 t

100 = 45.761 t

t = 100/45.761

t = 2.19 seconds

Answer:

(a) Qh(dot)=600 kW, Qc(dot)=400 kW  is an irreversible process.

(b) Qh(dot)=600 kW, Qc(dot)=0 kW  is an impossible process.

(c) Qh(dot)=600 kW, Qc(dot)=200 kW  is a reversible process.

Explanation:

T(hot) = 1200k, T(cold) = 400

efficiency n = (Th - Tc ) / Tc

n = (1200 - 400) / 1200 = 0.667 (this will be the comparison base)

(a)

Qh = 600 kW, Qc = 400 kW

n = (Qh - Qc) / Qh ⇒ (600 - 400) / 600

n = 0.33

0.33 is less than efficiency value from temperature 0.67

∴ it is irreversible process

(b)

Qh = 600 kW, Qc = 0

n = (Qh - Qc) / Qh ⇒ (600 - 0) / 600 = 1

efficiency in any power cycle can never be equal to one.

∴ it is an impossible process.

(c)

Qh = 600 kW, Qc = 200 kW

n = (Qh - Qc) / Qh = (600 - 200) / 600

n = 0.67 (it is equal to efficiency value from temperature)

∴ it is a reversible process

Answer:

A one -two pages menu was written with questions directed to the CSP which is stated below in the explanation section

Explanation:

Solution

If CSP has no team or limited staff, you will need to ask the following questions to understand how the CSP is set up:

To access evidence in the cloud :

Answer:

The solution is presented in explanation

Explanation:

This problem can be solved in following steps:

1) In the first round the peasant will take the goat to the other side.

2) Now, the peasant will come back alone.

3) The peasant will now take the wolf with him to other side.

4) The peasant will return with the goat to riverbank.

5) Now, he will take cabbage to the other side of the river, where the wolf is already present.

6) Peasant will leave cabbage and wolf on other side and come back to riverbank alone. Since, wolf does not eat cabbage.

7) Now, finally the peasant will take goat to the other side of river.

In this way, all three of them shall be transported to the other side of the river without eating each other.

Answer:

option A. Inferring

Explanation:

inferring/ inference as reading strategy simply is the process by which one uses what he/she knows to make a guess about what you don't know or reading between the lines. Readers in making  inferences uses  clues found inside text along with their own  views or experiences to help them figure out what is not directly said,thereby causing a  personal and memorable text. for one to draw an inference from the passage via reading, Identify if its an Inference Question.inferring involves Trusting the Passage or what you are seeing,  then you start Hunting for Clues thereafter you Narrow Down the Choices. and then come to a conclusion or Practice.

Answer:

263.69 W.

(None of the option).

Explanation:

So, from the question above we are given the following parameters or data or information which is going to allow us to solve this question and they are;

(1). diameter of 25 cm.

(2). Initial temperature of 300°C.

(3).temperature drops to 200°C = final temperature.

Step one: Calculate the Reynolds number.

Reynolds number = 3 × 0.25/1.562 × 10^-5 = 48015.365.

Step two: Calculate average heat transfer coefficient.

The average heat transfer coefficient = k/D { 2 + (0.4Re^1/2 + 0.06Re^2/3} px^0.4 × (u/uz)^1/4.

The average heat transfer coefficient = 0.10204 × [ 2 + (87.65 + 79.26) (0.8719) × 0.8909.

average heat transfer coefficient = 0.20204 ( 2 + 129.652).

average heat transfer coefficient = 13.43/m^2.k.

Step three: The rate of heat transfer loss due to convection = (average heat transfer coefficient ) × πD^2 × ( T1 - T2).

The rate of heat transfer loss due to convection= 13.43 × π(0.25)^2 × (300 - 200).

=>The rate of heat transfer loss due to convection = 263.69 W.

Answer:

  No

Explanation:

We expect current to be proportional to the number of identical bulbs. The total resistance is the ratio of voltage to current, so will be inversely proportional to the number of bulbs.

The current readings look wrong in that the first bulb caused the current to be 1 A, but each additional bulb increased it by 2 A. If that is what happened, the bulbs were not identical. That may be OK, but we expect the point of the experiment is to let you see the result described above.

In any event, the total resistance is not calculated properly. It should be the result of dividing voltage (15 V) by current.

Answer:

No, it is not right.  

Explanation:

Your table is not consistent with bulbs of the same resistance.

Current comes from a measurement, but resistance comes from a calculation.

I presume that the measured currents are correct.

Ohm's Law states that the current flowing in a circuit is directly proportional to the voltage.

We usually write it as

V/I = R

1. One bulb in circuit

[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{1 A}}= \mathbf{15 \, \Omega}[/tex]

2. Two bulbs

[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{3 A}} = \mathbf{5 \, \Omega}[/tex]

3. Three bulbs

[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{5 A}} = \mathbf{3 \, \Omega}[/tex]

4. Four bulbs

[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{7 A}} = \mathbf{2.1 \, \Omega}[/tex]

Answer:

b. 1655.7 KJ/kg ( net work produced )

c. 2324.86 KJ/kg

d. 0.25539 --- 25.5%

e. 3980.63 KJ/kg

Explanation:

Given:-

Condenser exit parameters:

 P1 = 50 KPa  , saturated liquid

Boiler exit / Turbine exit parameters:

P3 = 2 MPa

T3 = 1100°C

Solution:-

- Adiabatic and reversible processes for pump and turbine are to be applied

- Assume changes in elevation heads within the turbomachinery to be negligible.

- Assume steady state conditions for fluid flow and the use of property tables will be employed.

Isentropic compression of water in pump:

 Pump inlet conditions :                       Pump exit to Boiler pressure:

  P1 = 50 KPa, sat liquid                         P2 = P3 = 2 MPa

  h1 = 340.54 KJ/kg                               s2 = s1 = 1.0912 KJ/kg.K

  s1 =  1.0912 KJ/kg.K                              h2 = 908.47 KJ/kg

- Apply energy balance for the pump and determine the work input ( Win ) required by the pump:

                     Win = h2 - h1

                     Win = 908.47 - 340.54

                     Win = 567.93 KJ/kg

Isentropic expansion of steam in turbine:

 Turbine inlet conditions :                  Turbine exit to condenser pressure:

  P3 = 2MPa, T3 = 1100°C                      P4 = P1 = 50 kPa

  h3 = 4889.1 KJ/kg                        s4 = s3 = 8.7842 KJ/kg.K  .. superheated

  s3 =  8.7842 KJ/kg.K                   h4 = hg = 2665.4 KJ/kg

- Apply energy balance for the turbine and determine the work output ( Wout ) produced by the turbine:

                     Wout = h3 - h4

                     Wout = 4889.1 - 2665.4

                     Wout = 2223.7 KJ/kg

- The net work-output obtained from the cycle ( W-net ) is governed by the isentropic processes of pump and turbine.

                  W_net = Wout - Win

                  W_net = 2223.7 -  567.93

                  W_net = 1655.77 KJ/kg   ... Answer

- The fraction of work generated by turbine is used to operate the pump. The a portion of Wout is used to drive the motor of the pump. The pump draws ( Win ) amount of work from pump. The ratio of work extracted from turbine ( n ) would be:

                 n = Win / Wout

                 n = 567.93 / 2223.7

                 n = 0.25539  ... Answer ( 25.5 % ) of work is used by pump

- The amount of heat loss in the condenser ( consider reversible process ). Apply heat balance for the condenser, using turbine exit and condenser exit conditions:

                Ql = h4 - h1

                Ql = 2665.4 - 340.54

                Ql = 2324.86 KJ/kg ... Answer

- The amount of heat gained by pressurized water in boiler ( consider reversible process ). Apply heat balance for the boiler, using pump exit and boiler exit conditions:

                Qh = h3 - h2

                Qh = 4889.1 - 908.47

                Qh = 3980.63 KJ/kg ... Answer

Answer:

The thickness of the walls of each hollow lump of "iron ore" is 2.2 cm

Explanation:

Here we have that the density of solid gold = 19.3 g/cm³

Density of real iron ore = 5.15 g/cm³

Diameter of sphere of gold = 4 cm

Therefore, volume of sphere = 4/3·π·r³ = 4/3×π×2³ = 33.5 cm³

Mass of equivalent iron = Density of iron × Volume of iron = 5.15 × 33.5

Mass of equivalent iron = 172.6 cm³

∴ Mass of gold per lump = Mass of equivalent iron = 172.6 cm³

Volume of gold per lump = Mass of gold per lump/(Density of the gold)

Volume of gold per lump = 172.6/19.3 = 8.94 cm³

Since the gold is formed into hollow spheres, we have;

Let the radius of the hollow sphere = a

Therefore;

Total volume of the hollow gold sphere = Volume of gold per lump - void sphere of radius, a

Therefore;

[tex]33.5 = 8.94 - \frac{4}{3} \times \pi \times a^3[/tex]

[tex]\frac{4}{3} \times \pi \times a^3 = 33.5 - 8.94[/tex]

[tex]a^3 = \frac{24.6}{\frac{3}{4} \pi } = 5.9[/tex]

a = ∛5.9 = 1.8

The thickness of the walls of each hollow lump of "iron ore" = r - a = 4 - 1.8 = 2.2 cm.

Answer:

Ts = 311.86 K = 38.86°C

Explanation:

The convection heat transfer coefficient for vertical orientation of the board is given by the formula:

[tex]h = 1.42(\frac{T_{s} - T_{f} }{L})^{0.25}[/tex]

where,

h = heat transfer coefficient

[tex]T_{s}[/tex] = surface temperature

[tex]T_{f}[/tex] = Temperature of fluid (air) = 30°C + 273 = 303 K

L = Characteristic Length = 50 cm = 0.5 m

Since the heat transfer through convection is given as:

[tex]Q_{conv} = hA_{s}(T_{s} - T_{f})[/tex]

using value of h, we get:

[tex]Q_{conv} = 1.42(\frac{T_{s} - T_{f} }{L})^{0.25} A_{s} (T_{s} - T_{f} )[/tex]

[tex]Q_{conv} = 1.42 A_{s} \frac{(T_{s} - T_{f} )^{1.25} }{L^{0.25} }[/tex]

where,

[tex]A_{s}[/tex] = Surface Area = (0.5 m)(0.5 m) = 0.25 m²

Now, the radiation heat transfer is given by:

[tex]Q_{rad} =[/tex] εσ[tex]A_{s} [(T_{s})^{4} - (T_{surr})^{4}][/tex]

where,

ε = emissivity of surface = 0.7

σ = Stefan Boltzman Constant = 5.67 x 10⁻⁸ W/m².k⁴

[tex]T_{surr}[/tex] = Temperature of surroundings = 25°C +273 = 298 k

Now, the total heat transfer rate will be:

[tex]Q_{total} = Q_{conv} + Q_{rad}[/tex]

using values:

[tex]Q_{total} =[/tex] [tex]1.42 A_{s} \frac{(T_{s} - T_{f} )^{1.25} }{L^{0.25} } +[/tex] εσ[tex]A_{s} [(T_{s})^{4} - (T_{surr})^{4}][/tex]

we know that the total heat transfer from the board can be found out by:

[tex]Q_{total} = (0.18 W) (121) = 21.78 W[/tex]

using values in the equation:

21.78 = (1.42)(0.25)[tex](T_{s} - 303)^{1.25}/0.5^{0.25}[/tex] + (0.7)(5.67 x 10⁻⁸)(0.25)[tex][(T_{s})^{4} - 298^{4}][/tex]

21.78 = (0.4222)[tex](T_{s} - 303)^{1.25}[/tex] + 9.922 x 10⁻⁹[tex](T_{s} )^{4}[/tex] - 78.25

100.03 = (0.4222)[tex](T_{s} - 303)^{1.25}[/tex]+ 9.922 x 10⁻⁹[tex](T_{s} )^{4}[/tex]

Solving this equation numerically by Newton - Raphson Method (Here, any numerical method or an equation solver can be used), we get the value of Ts to be:

Ts = 311.86 K = 38.86°C

The film temperature is the average of surface temperature and surrounding temperature. Therefore,

Film Temperature = (25°C + 38.86°C)/2 = 31.93°C

Since, this is very close to 30°C.

Hence, the assumption is good.

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

I inferred you meant emerging technologies we see today.

Explanation:

1. 3D Printing

A three dimensional printing allows a digital model to be printed (constructed) into a physical object.

A box, an industrial design and many more could be printed within minutes.

2. AI voice recognition devices

Another trend in the tech world is the rise in artificial intelligence been used in voice recognition devices that not only recognize one's voice but also take commands from the user.

This technology allows one to listen to the internet, such as listening to music online.

The tech world is ever changing with new technology beyond one's consumption on the increase daily.

Answer:

61.6 hours will be needed to reduce the diameter of solid spherical ball from 2 cm to 0.50 cm.

Explanation:

Find the given attachments

Oxygen combines with nitrogen in the air to form NO at about 2500 celsius.

The enormous energy of lightning breaks nitrogen molecules and enables their atoms to combine with oxygen in the air forming nitrogen oxides. These dissolve in rain, forming nitrates, that are carried to the earth.

At these high temperatures, nitrogen and oxygen from the air combine to produce nitrogen monoxide. One nitrogen molecule (N2) reacts with one oxygen molecule (O2) to make two nitrogen monoxide molecules (NO).

Learn more about NO here:

https://brainly.com/question/27548777

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Answer:

Explanation:

a) Present worth of the system:

First step :

Calculation of bank installment:

We are given:

·nitial costs = $32,000

Borrow amount = ½ of purchase price

Payment of borrow amount = EOY 2 to EOY 4 (3 Equal installments)

Bank loan interest = 8.5% = 8.5/100 = 0.085

Assume the installment amount is F. They will be paid at end of year 2 to end of year 4. Their present value must be equal to borrow amount.

Present value of cost to be incurred in future can be calculated by below formula:

F P (1 + i)

F= Future cost

i = Rate of interest

n = time (in years)

Therefore,

F F $32,000 2 F (1 +0.085)2 (1 +0.085) 3 (1 +0.085) + +

.: 2.35394F = 16,000 or F = 16,000 2.35394 - $6,797.11

Step 2: Present worth of the system:

Given Data:

Initial costs = $32,000

Expected life = 7 years

Salvage value = $5,000

O&M Costs = $2,000 per year

MARR = 12.5%

= 12.5/100

= 0.125

Present value of uniform recurring payments is given by below formula:

P=A (1 + i) - 1 i(1+i)n 72

Where,

P = Present Value

A = Recurring payments per annum

i = rate of interest

n = time (in years)

Hence present value of O&M costs,

P1 = -2,000 x (1 + 0.125) 7-1 0.125 x (1 + 0.125) 7 -$8,984.60

Present worth of the system calculated in below table:

Description

F ($)

MARR (i)

per year

n (years)

P ($)

Initial investment (1/2 of purchase price)

-16,000.00

0.125

0

-16,000.00

Bank installment EOY2

-6,797.11

0.125

2

-5,370.56

Bank installment EOY3

-6,797.11

0.125

3

-4,773.83

Bank installment EOY4

-6,797.11

0.125

4

-4,243.40

Salvage value

5,000.00

0.125

7

2,192.31

O&M Costs

-8,984.60

0.125

0

-8,984.60

The current worth of the new system

-37,180.07

Part b) Decision rule of judgment:

Assuming current value of costs is lower than current value of benefit, an alternative is known to be economic to use based on current worth analysis.

Part c) Decision for the water filtration system:

Given Data:

· Annual savings from filtration system (A) = $13,000 per year

· Expected life (n) = 7 years

· MARR = 12.5%

= 12.5/100

= 0.125

Present value of benefits = 13,000 X (1 + 0.125)7 - 1 0.125 x (1 + 0.125) 7 $58,399.91

Since current value of costs is less than current value of benefit, this is an worthwhile system and has to be be purchased.

Answer:

non-functional requirement,

Yes they can.

The application loading time is determined by testing system under various scenarios

Explanation:

non-functional requirement are requirements needed to justify application behavior.

functional requirements are requirements needed to justify what the application will do.

The loading time can be stated with some accuracy level after testing the system.

Answer:

a) The operation of a heat pump involves the extraction of energy in the form of heat Q₁ from a cold source

b) The modifications required to convert a plant operating on an ideal Carnot cycle to a plant operating on a Rankine cycle involves

i) Complete condensation of the vapor at the condenser to saturated liquid for pumping to the boiler

ii) Heating of the pumped, pressurized water to the boiler pressure

Explanation:

a) 1 - 2. Wet vapor enters compressor where it undergoes isentropic compression to state 2 by work W₁₂  

2 - 3. The vapor enters the condenser at state 2 where it undergoes isobaric and isothermal condensation to a liquid with the evolution of heat  Q₂

3 - 4. The condensed liquid is expanded isentropically with the work done equal to W₃₋₄

4 - 1. At the state 4, with reduced pressure from the previous expansion, the liquid makes its way to the evaporator where it absorbs heat, Q₁, from the body to be cooled.

b. i) Complete condensation of the vapor at the condenser to saturated liquid for pumping to the boiler

Here the condensation process is modified from partial condensation to complete condensation at the same temperature which reduces the size of the pump required to pump the liquid water as opposed to pumping steam plus liquid

ii) Heating of the pumped, pressurized water to the boiler pressure

The pumped water at state 4 will be required to be heated to saturated water temperature equivalent to the boiler pressure, hence heat will need to be added at state.

Sketches of the schematic of a Basic Rankine cycle is attached  

Answer:1. Driving under the influence of any drug that makes you drive unsafely is:

a. Permitted if it is prescribed by a doctor

b. Against the law

c. Permitted if it is a diet pill or cold medicine

2. Which fires can you put out with water:

a. Tire fires

b. Gasoline fires

c. Electrical fires

3. How far should a driver look ahead of the vehicle while driving:

a. 9-12 seconds

b. 12-15 seconds

c. 18-21 seconds

4. To prevent shifting, there should be at least one tie-down for ever ____feet of cargo: a. 10

b. 15 c. 18

5. Which of these statements about downshifting is true:

a. When you downshift for a curve, you should do so before you enter the curve

b. When you downshift for a hill, you should do so after you start down the hill

c. When you downshift for a curve, you should do so after you enter the curve

6. How do you test hydraulic brakes for a leak:

a. Move the vehicle slowly and see if it stops when the brake is applied

b. With the vehicle stopped, pump the pedal three time, apply pressure then hold

For five seconds and see it the pedal moves.

c. Step on the brake pedal and accelerator at the same time and see if the vehicle moves

7. For an average driver, driving 55 MPH on dry pavement, it will take about _____ to bring The vehicle to a stop:

a. Twice the length of the vehicle

b. Half the length of a football field

c. The length of a football field

8. You are driving a vehicle with a light load, traffic is moving at 35 MPH in a 55 MPH zone. The safest speed for your vehicle in this situation is most likely:

a. 30 MPH

b. 35 MPH

c. 40 MPH

9. Which of these is a good rule to follow when driving at night:

a. Keep your speed slow enough to stop within the range of your headlights

b. Look directly at oncoming headlights

c. Keep your instrument lights bright

10. A moving vehicle ahead of you has a red triangle with an orange center on the rear. What does this mean?

a. The vehicle is hauling hazardous materials

b. It may be a slow-moving vehicle

c. It may be oversized

11. You wish to turn right form a two lane two way street to make the turn. Which of these Drawings show how the turn should be made.

12. You are driving a heavy vehicle and must exit a highway using an offramp that curves downhill:

a. Use the posted speed limit for the offramp

b. Slow down to a safe speed before the turn

c. Wait until you are in the turn before downshifting

13. Which of these statements about using mirrors is true:

a. You should look at a mirror for several seconds at a time

b. There are “blind spots” that your mirror cannot show you

c. A lane change requires you to look at the mirrors twice

14. You must park on the side of a level, straight, two-lane road. Where should you place the three reflective triangles?

a. one within 10 feet of the rear of the vehicle, one about 100 feet to the rear and one about 200 feet to the rear.

b. One with 10 feet of the rear of the vehicle, one about 100 feet to the rear and one about 100 feet from the front of the vehicle

c. One about 50 feet from the rear of the vehicle, one about 100 feet to the rear and one about 100 feet from the front of the vehicle

15. Your vehicle is in a traffic emergency and may collide with another vehicle if you do not take action. Which of these is a good rule to remember at such a time?

a. Stopping is always the safest action in a traffic emergency

b. Heavy vehicles can almost always turn more quickly than they can stop

c. Leaving the road is always more risky than hitting another vehicle

16. The most important reason for being alert to hazards is:

a. Law enforcement personnel can be called

b. You will have time to plan your escape if the hazard becomes an emergency

c. You can help impaired drivers

17. You are traveling down a long, steep hill. Your brakes begin to fade and then fail. What should you do?

a. Downshift

b. Pump the brake pedal

c. Look for an escape ramp or escape route

18. The most common cause of serious vehicle skids is:

a. Driving too fast for road conditions

b. Poorly adjusted brakes

c. Bad tires

19. To avoid a crash, you had to drive onto the right shoulder. You are now driving at 40 MPH on the shoulder. How should you move back onto the pavement?

a. If clear, come to a complete stop before steering back onto the pavement

b. Brake had to slow the vehicle, then steer sharply onto the pavement

c. Keep moving at the present speed and steer very gently back onto the pavement

20. If

a. Slide sideways and spin out

b. Go straight ahead but will turn if you turn the steering wheel

c. Go straight ahead even if the steering wheel is turned

Answer: Because of the role the base region play in the transistor.

Explanation:

The base region of BJT transistor - an opposite polarity charge carrier from emitter region to collector region, plays a vital role in triggering for a sufficient emiter - to - collector current.

The current received by the base region of BJT determines the effect of the continue flow of current into the collector region which will eventually determine the output current.

Answer:

The minimum mass flow rate will be "330 kg/s".

Explanation:

Given:

For steam,

[tex]m_{s}=5.55 \ kg/s[/tex]

[tex]\Delta h=2491 \ kg/kj[/tex]

For water,

[tex]\Delta T=10^{\circ}C[/tex]

[tex](Cp)_{w}=4.184 \ kJ/kg^{\circ}C[/tex]

They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,

⇒  [tex]m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T[/tex]

On putting the estimated values, we get

⇒  [tex]5.55\times 2491=M_{w}\times 4.184\times 10\\[/tex]

⇒  [tex]13825.05=M_{w}\times 41.84[/tex]

⇒  [tex]M_{w}=330 \ kg/s[/tex]

Every one deals with stress. Stress or anxiety is usually when your scared or anxious because you know you could have done better then what you did. For example on a test, you could be scared about what grade you get. You can also have stress when you feel something bad is going to happen.

Answer:

it is the component of stress coplanar with a material cross section. It arises from the shear force, the component of force vector parallel to the material cross section

Explanation:

hope this helps and have a good day :-)

Answer:

If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

Explanation:

Reynolds number is an important dimensionless parameter in fluid mechanics.

It is calculated as;

[tex]R_e__N} = \frac{\rho vd}{\mu}[/tex]

where;

ρ is density

v is velocity

d is diameter

μ is viscosity

All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.

In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

Answer:

4.2ms

Explanation:

Calibrated time= 0.3+0.2+0.5+1.4= 2.4

Measured time= 6.6ms

Accuracy is closeness of measurement to an observed or true value

Accuracy= 6.6-2.4= 4.2ms

Answer:

0.09

Explanation:

Packet switching involves breaking a message into packets and sending them independently. Since the user only needs to transmit 10 percent of the time, the probability that a given (specific) user is transmitting = 10% = 0.1

The  probability that a user is not transmitting = 100% - 10% = 90% = 0.9

Therefore, the probability that a given (specific) user is transmitting, and the remaining users are not transmitting = 0.1 * 0.9 = 0.09