(1.24) Consumer Reports is doing an article comparing refrigerators in their next issue. Some of the characteristics to be included in the report are the brand name and model; whether it has a top, bottom, or side-by-side freezer; the estimated energy consumption per year (kilowatts); whether or not it is Energy Star compliant; the width, depth, and height in inches; and both the freezer and refrigerator net capacity in cubic feet. The "Height" is categorical variable, quantitative variable, or individuals

Answer: B. misinterpreted pressure tests to determine whether the well has been sealed.

Explanation:

Pressure test is a non-destructive test carried out to ensure there is integrity of the pressure shell on an equipment that is new or pressure and piping equipment installed earlier that has undergone some repairs.

After a crew carried out a variety of pressure tests to determine whether the well was sealed or not. The results of these tests were misinterpreted, so they thought the well was under control. This caused oil to spill.

Answer:

dimensions and materials information that are provided about composite wall (2.5mx6.5m 10 studs each 2.5m high)

to find:wall thermal resistance=??

properties:

Ka=0.094W/m.K

Kb=0.16W/m.K

Kc=0.17W/m.K

Kd=0.038W/m.K

using isothermal surface assumption:

(La/KaAa)=0.008/0.094(0.65x2.5)

=0.0524K/W

(LB/KbAb)=0.13/0.16(0.04x2.5)

=8.125K/W

(Ld/KdAd)=0.13/0.38(0.6x2.5)

=2.243K/W

core's equivalent resistance is:

Req=(1/Rb+1/Rd)⁻¹

=(1/8.125+1/2.243)⁻¹

=1.758K/W

For total resistance we will find sum of

Rtot=Ra+Req+Rc

=1.854K/W

Whereas 10 studs are used so:

Rtotal=(10x1/Rtot)⁻¹

=0.1854K/W

The correct verb forms to complete the given sentences are:

Members of the committee issued a mobile phone and one roll of stamps.

Neither the CEO nor the CFO has any recollection of the illegal funds transfer.

Mr. Adams and his assistant are planning to attend the annual technology showcase.

The finance committee is planning their vacations around the annual June meeting.

Everything in my home office seems to stop functioning when I need it the most.

In the first sentence, the verb "issued" is correct because it is in the past tense and agrees with the subject of the sentence, "Members of the committee."

In the second sentence, the verb "has" is correct because it is in the present tense and agrees with the subject of the sentence, "Neither the CEO nor the CFO." The verb "has" is also used when the subject is a singular noun that refers to two or more people or things joined by "neither" or "nor."

In the third sentence, the verb "are" is correct because it is in the present tense and agrees with the subject of the sentence, "Mr. Adams and his assistant." The verb "are" is also used when the subject is a plural noun.

In the fourth sentence, the verb "is" is correct because it is in the present tense and agrees with the subject of the sentence, "The finance committee." The verb "is" is also used when the subject is a singular noun that refers to a group of people or things.

In the fifth sentence, the verb "seems" is correct because it is in the present tense and agrees with the subject of the sentence, "Everything in my home office." The verb "seems" is also used to express an opinion or belief.

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Answer:

I = 5.0 A  V=0

Explanation:

Assuming that the ammeter is an ideal one (which means that the internal resistance is negligible compared with other resistors in the circuit), as the voltage through the capacitor can't change instantaneously, just after the switch is closed, will behave like a short, so applying Ohm's law, the current in the circuit will be as follows:

[tex]I =\frac{V}{R} = \frac{10 V}{2.0 \Omega} = 5.0 A[/tex]

As the voltage in the capacitor, can't change instantaneously, assuming an ideal voltmeter (infinite resistance) , the lecture on the voltmeter across the capacitor will be just 0.

As time goes by, the voltage measured will follow the following equation:

[tex]V = V0*( 1-e^{-\frac{t}{R*C}} )[/tex]

We see that when t=0, V=0.

The question involves analyzing an ideal Brayton cycle to find the exit temperature from a compressor, the back work ratio, and the thermal efficiency of a gas turbine engine with air as the working fluid, accounting for the variation in specific heats with temperature.

The student's question pertains to the ideal Brayton cycle, which is a thermodynamic cycle used to describe the workings of a gas turbine engine. Their question involves air with varying specific heats at different temperatures, typical in engineering thermodynamics analysis.

Although the appropriate formulas and calculations are not provided in the provided references, the typical approach would involve thermodynamic equations of state and specific heat relations to find the unknown temperatures, efficiency, and other parameters of interest. Analysis often includes using the Brayton cycle efficiency formula, temperature ratios, and the specific gas constant for air.

The back work ratio and thermal efficiency in particular can be calculated using the first and second laws of thermodynamics applied to each component of the cycle: the compressor and turbine. As specific heats vary with temperature, one typically uses mean values or functions that provide the specific heat values at the given temperatures.

Explanation of the Brayton cycle in gas turbines, calculation methods for compressor exit temperature, back work ratio, and thermal efficiency.

The Brayton cycle is a thermodynamic cycle that describes how gas turbines operate. It consists of four processes: two isentropic (reversible adiabatic) processes and two isobaric (constant-pressure) processes.

(a) The air temperature at the compressor exit can be calculated using the temperature ratios at the compressor and turbine, and the pressure ratio.

(b) The back work ratio is the ratio of the work required to drive the compressor to the work produced by the turbine.

(c) The thermal efficiency of the Brayton cycle can be determined by comparing the work output to the heat input.

To deliver 30 horsepower at the wheel for one hour with a 20% efficient internal combustion engine, approximately 2.88 gallons of gasoline are required.

The question at hand involves calculating the quantity of gasoline required to deliver 30 horsepower (HP) at the wheel for one hour, given that the energy conversion efficiency of an internal combustion engine is 20%. First, understand that 1 HP is equivalent to 746 Watts, and therefore, 30 HP equals 22,380 Watts or 22.38 Kilowatts. Since 1 HP for 1 hour is 0.746 kWh, 30 HP for 1 hour is 22.38 kWh. Knowing the energy content of gasoline and the engine's efficiency will allow us to calculate the required gallons of gasoline.

Given that a gallon of gasoline releases about 140 MJ (or 38.89 kWh) of energy when burned and considering the 20% efficiency, the useful energy per gallon of gasoline is approximately 7.778 kWh. To find out how many gallons of gasoline are needed to produce 22.38 kWh at 20% efficiency, we divide the total required energy output (22.38 kWh) by the useful energy per gallon (7.778 kWh), resulting in approximately 2.88 gallons of gasoline needed.

Answer:

0.267

Explanation:

The Mach number is given by the following formula:

[tex]Ma = \frac{v}{c}[/tex]

where v = speed of the object

           c= speed of air

Ma         = Mach number

   From the data:

Velocity of the object = 91.44 m/s

Velocity of air =  343 m/s        

 Therefore, the Mach number is given by the following formula:

Ma = [tex]\frac{91.44}{343}[/tex]

      = 0.267

Answer:

B.

Explanation:

For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.

The truth table  for a NAND gate with 2 inputs is as follows:

0 0    1

0 1     1

1  0    1

1  1    0

The  truth table for an OR gate, is as follows:

0 0    0

0 1     1

1  0    1

1  1     1

If we add two extra columns for inverted inputs, the truth table will be this one:

0 0    1  1       1

0 1     1  0      1

1  0    0  1      1

1  1     0  0     0

which is the same as for the NAND gate, not the opposite, so the statement is false.

This means that the right choice is B.

Answer:

[tex]BW=\Delta f= f_H-f_L=4300Hz-300Hz=4000Hz=4kHz[/tex]

Explanation:

Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. The bandwidth may be determined by use of the following formula:

[tex]BW=f_H-f_L\\\\Where:\\\\f_H=Upper\hspace{3}cutoff\hspace{3}frequency=4.3kHz=4300Hz\\f_L=Lower\hspace{3}cutoff\hspace{3}frequency=300Hz[/tex]

Hence the signal’s frequency bandwidth is:

[tex]BW=4300-300=4000Hz=4kHz[/tex]

Answer:

The coefficient of static friction, μₛ, between the trunk and turntable = 0.32

Explanation:

For this motion of the trunk B,

Initial velocity, v₀ = 0

Tangential Acceleration, a = 0.28 m/s²

Time taken, t = 10s

Using equations of motion,

v = v₀ + at

v = 0 + 0.28 × 10 = 2.8 m/s

Frictional force, Fᵣ = μₛN

μₛ = coefficient of static friction,

N = Normal reaction exerted on the trunk B as a result of its weight = mg

Doing a force balance on the trunk B,

Force keeping the trunk B in circular motion must balance the frictional forces.

Force keeping the trunk B in circular motion, F = mv²/r

Fᵣ = F

μₛN = mv²/r but N = mg

μₛmg = mv²/r

μₛg = v²/r

μₛ = v²/gr

μₛ = 2.8²/(9.8 × 2.5) = 0.32

Hope this helps!!!

The coefficient of static friction between the trunk and the turntable is 1.

To determine the coefficient of static friction between the trunk and the turntable, we can use the equation:

fs = m * at

where fs is the static friction, m is the mass of the trunk, and at is the tangential acceleration of the trunk.

Using the given values, we have:

0.28 = m * 0.28

Solving for m, we find:

m = 1 kg

Therefore, the coefficient of static friction between the trunk and the turntable is 1.

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Designing a DC power supply with a half-wave rectifier involves calculating the necessary transformer secondary RMS voltage, filter capacitor value, diode's peak inverse voltage, and average and peak diode currents considering the specific design requirements like output voltage and ripple.

The question relates to designing a DC power supply using a half-wave rectifier, including calculations for transformer secondary voltage, filter capacitor value, peak inverse voltage (PIV), and diode current parameters. Solving such a design problem involves understanding and applying principles of electrical engineering, particularly those regarding rectification, filtering, and electrical power supply designs.

The kitchen appliances, including a blender, phone charger, toaster oven, and lights, will pull a total current of 15 Amperes when operating on a 120-Volt household electrical system.

To calculate the total current pulled by various electrical devices in the kitchen, you add up their power ratings and divide the sum by the voltage of the electrical system. These devices, which include a 360 W blender, a 60 W phone charger, an 840 W toaster oven, and four 120 W lights, all operate on a standard household voltage of 120 Volts. The formula for current (I) when power (P) and voltage (V) are known is I = P/V.

First, calculate the total power consumed by all devices:
360 W (blender) + 60 W (phone charger) + 840 W (toaster oven) + (4 × 120 W) (lights) = 1800 W.

Then, to find the current, we use the formula with the total power and the household voltage: I = 1800 W / 120 V = 15 A. Therefore, these appliances will pull a total current of 15 Amperes.

Answer:

acceleration = 24.23 ms⁻¹

Explanation:

Let's gather the data:

The acceleration of the car is given by a = 0.5 [tex]e^{t}[/tex]

The acceleration is the change in the speed in relation to time. In other words:

[tex]\frac{dv}{dt}[/tex] = a = a = 0.5 [tex]e^{t}[/tex]            ...1

Solving the differential equation yields:

v = 0.5 [tex]e^{t}[/tex] + C₁

Initial conditions : 0 = 0.5 [tex]e^{0}[/tex] + C₁

                             C₁ = -5

at any time t, the velocity is: v= 0.5[tex]e^{t}[/tex]- 5

Solving for distance, s = 0. 5[tex]e^{t}[/tex] - 0.5 t - 0.5

                                    18 = 0.5 [tex]e^{t}[/tex] - 0.5 t - 0.5

                                       t = 3.71 s

Substitute t = 3.71 s

                   v= 0.5[tex]e^{t}[/tex]- 5

                     = 19.85 m/s

a = 0.5 [tex]e^{t}[/tex]             ...1

 = 20.3531

an = [tex]\frac{v^{2} }{p}[/tex]

     = [tex]\frac{(19.853)^{2} }{30}[/tex]

     = 13.1382

Magnitude of acceleration = [tex]\sqrt{ (a)^{2} + (an)^{2} }[/tex]

                                            = [tex]\sqrt{(20.3531)^{2} +(13.1382)^{2} }[/tex]

                                            = 24.23 ms⁻¹ Ans

Answer:

98.13kJ

Explanation:

Given that;

The rigid 10-L vessel initially contains a mixture of liquid water & vapor at

[tex]T_1 =100^0C\\T_2 = 180^0C[/tex]

We are to calculate the heat transfer required during the process by obtaining our data from the steam tables.

In order to do that, let start with our Energy Balance

So, Energy Balance for closed rigid tank system is given as:

[tex]\delta E_{system} = E_{in} - E_{out}[/tex]

Since the K.E and P.E are insignificant;

∴ K.E = P.E = 0

[tex]Q_{in}= \delta U + W\\Q_{in} = m(u_2-u_1)+ W[/tex]

Where;

m = mass flow rate of the mixture

[tex](u_2-u_1)[/tex] = corresponding change in the internal energy at state point 2 and 1

However, since we are informed that the vessel is rigid, then there is no work done in the system, then W turn out to be equal to zero .i,e

W = 0

we have our above equation re-written as:

[tex]Q_{in}= m (u_2-u_1)+0\\[/tex]

[tex]Q_{in}= m(u_2-u_1)[/tex]

We were told to obtain our data from the steam table, so were going to do just that

∴  At inlet temperature [tex]T_1 = 100^0C[/tex], the given quality of mixture of liquid water and vapor [tex](x_1)[/tex] = 123% = 0.123

Using the equation:

[tex]v_1 = v_f + x_1v_{fg}\\v_1 = v_f + x_1(v_g-v_f)[/tex]

where;

[tex]v_1[/tex] = specific volume at state 1

[tex]v_f[/tex] = specific volume of the liquid

[tex]v_g[/tex] = specific volume of the liquid vapor mixture

The above data from the steam table is given as;

[tex]v_f[/tex]  = 0.001043 m³/kg

[tex]v_g[/tex] = 1.6720 m³/kg

so; we have

[tex]v_1 = 0.001043 + 0.123(1.6720-0.001043)[/tex]

[tex]v_1 = 0.001043+0.123(1.670957)[/tex]

[tex]v_1 =0.001043+0.205527711[/tex]

[tex]v_1= 0.206570711m^3/kg[/tex]

[tex]v_1=0.2065m^3/kg[/tex]

At  [tex]T_1[/tex] = 100°C and [tex]x_1=0.123[/tex];

the following steam data from the tables were still obtained for the internal energy; which is given as:

Internal Energy [tex](u_1)[/tex] at the state 1

[tex]u_1= u_f + xu_{fg}[/tex]

where;

specific internal energy of the liquid  [tex](u_f)[/tex]  = 419.06 kJ/kg

The specific internal energy of the liquid vapor mixture [tex](u_{fg})[/tex] = 2087.0 kJ/kg

∴ since ; [tex]u_1= u_f + xu_{fg}[/tex]

[tex](u_1)[/tex]  = 419.06 + (0.123 × 2087.0)

[tex](u_1)[/tex]  = 675.761 kJ/kg

As the tank is rigid, so as the volume which is kept constant:

[tex]v_2=v_1\\=0.2065 m^3/kg[/tex]

Now, let take a look at when [tex]T_2 = 180^0C[/tex] from the data in the steam tables

Specific volume of the liquid [tex](v_f)[/tex] = 0.00113 m³/kg

specific volume of the liquid vapor mixture [tex](v_g)[/tex] = 0.19384 m³/kg

The quality of the mixture at the final state 2 can be determined  by using the  equation shown below:

[tex]v_2=v_f+x_2v_{fg}[/tex]

[tex]x_2=\frac{v_2-v_f}{v_{fg}}[/tex]

[tex]x_2=\frac{v_2-v_f}{v_g-v_f}[/tex]

[tex]x_2=\frac{0.2065-0.00113}{0.19384-0.00113}[/tex]

[tex]x_2=\frac{0.20537}{0.19271}[/tex]

    = 1.0657

From our usual steam table; we still obtained data for the Internal Energy when [tex]T_2=180^0C[/tex]

Specific internal energy of the liquid [tex](u_f)[/tex] = 761.92 kJ/kg

Specific internal energy of the liquid vapor mixture [tex]u_{fg}[/tex] = 1820.88 kJ/kg

Calculating the internal energy at finsl state point 2 ; we have:

[tex]u_2=u_f+u_{fg}[/tex]

= 761.92 + (1.0657 × 1820.88)

= 761.92 + 1940.511816

= 2702.431816

[tex]u_2[/tex] ≅ 2702.43 kJ/kg

Furthermore, let us calculate the mass in the system; we have:

[tex]m= \frac{V_1}{v_1}[/tex]

where V₁ = the volume 10 - L given by the system and v₁ = specific volume at state 1 as 0.2065

V₁ = the volume 10 - L = 10  × ( 0.001 m³/L)

v₁ = 0.2065

∴

mass (m) =  [tex]\frac{10(0.001m^3/L)}{0.2065}[/tex]

= 0.04842 kg

Now, we gotten all we nee do calculate for the heat transfer that is required during the process:

[tex]Q_{in}= m(u_2-u_1)[/tex]

[tex]Q_{in}= 0.04842(2702.43-675.761)[/tex]

[tex]Q_{in}= 0.04842(2026.669)[/tex]

[tex]Q_{in}= 98.13 kJ[/tex]

Therefore, the heat transfer that is required during the process = 98.13 kJ

There you have it!, I hope this really helps alot!

The heat transfer required for this process is 129.168 kilojoules.

According to steam tables, the initial state of the water inside the vessel is:

[tex]P = 101.42\,kPa[/tex], [tex]T = 100\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u = 675.761\,\frac{kJ}{kg}[/tex], [tex]x = 0.123[/tex] (Liquid-Vapor Mixture)

Given that process is isochoric, that is, at constant volume, we know the following information:

[tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], Superheated vapor

We need to know the pressure and the internal energy must be determined by using linear interpolation from steam tables twice. First, we use [tex]T = 180\,^{\circ}C[/tex] as our pivot, then we construct the following information by linear interpolation:

Lower bound

[tex]P = 800\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.24700\,\frac{m^{3}}{kg}[/tex], [tex]u = 2593.9\,\frac{kJ}{kg}[/tex]

Upper bound

[tex]P = 1000\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.19444\,\frac{m^{3}}{kg}[/tex], [tex]u = 2583.0\,\frac{kJ}{kg}[/tex]

Lastly, we find the expected pressure and internal energy by linear interpolation where specific volume is our pivot:

[tex]P = 952.207\,kPa[/tex], [tex]T = 180\,^{\circ}C[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u = 2585.605\,\frac{kJ}{kg}[/tex]

Once found all the required information, we can find the required heat transfer ([tex]Q[/tex]), in kilojoules, by means of this formula:

[tex]Q = \frac{V}{\nu}\cdot (u_{2}-u_{1})[/tex] (1)

Where:

If we know that [tex]V = 0.014\,m^{3}[/tex], [tex]\nu = 0.207\,\frac{m^{3}}{kg}[/tex], [tex]u_{1} = 675.761\,\frac{kJ}{kg}[/tex] and [tex]u_{2} = 2585.605\,\frac{kJ}{kg}[/tex], then the heat transfer required for this process is:

[tex]Q = \left(\frac{0.014\,m^{3}}{0.207\,\frac{m^{3}}{kg} } \right)\cdot \left(2585.605\,\frac{kJ}{kg}-675.761\,\frac{kJ}{kg} \right)[/tex]

[tex]Q = 129.168\,kJ[/tex]

The heat transfer required for this process is 129.168 kilojoules.

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