On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading be when you carry it up to the observation deck 500 ft above the base of the monument?

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:

a) 2,945 mC

b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

                           i (t) = 6*e^(-2*t)

                           v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

-  The amount of charge Q delivered can be determined by:                      

                                       dQ = i(t) . dt

                  [tex]Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt[/tex]

- Integrate and evaluate the on the interval:

                   [tex]= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C[/tex]

- The power can be calculated by using v(t) and i(t) as follows:

                 v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

                 v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

                 P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

                 P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

                 [tex]W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt[/tex]

- Integrate and evaluate the on the interval:

                  [tex]W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ[/tex]

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:

Explanation:

we would be analyzing this question with the important code given

#code :

from collections import namedtuple

#creating a named tuple named 'Player' with field names name, number, position and team

Player = namedtuple('Player',['name','number','position', 'team'])

cam = Player('Cam Newton','1','Quarterback','Carolina Panthers')

lebron = Player('Lebron James','23','Small forward','Los Angeles Lakers')

print(cam.name+'(#'+cam.number+')'+' is a '+cam.position+' for the '+cam.team+'.')

print(lebron.name+'(#'+lebron.number+')'+' is a '+lebron.position+' for the '+lebron.team+'.')

NB:

Lebron James (#23) rep. Small forward for the LA lakers

Cam Newton(#1) rep. a Quaterback for the Carolina Panthers

cheers i hope this helps

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

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:

1.57 KW

Explanation:

given data:

P= 1 atm

T= 12 min

power rating=??

solution:

latent heat of vaporization (L) of water at 1 atm = 2257.5 KJ/Kg

half of the water is evaporated in 12 min

so power rating is,

                                 =P×L/2.T

                                 =1×2257.5 /2×12×60

                                 =1.57 KW

The electrostatic energy in region A can be found by first calculating the electric field due to Plate 1, using the given charge density, and then determining the energy density. This energy density is then multiplied by the volume in region A, converted to SI units, to find the total electrostatic energy stored.

To find the electrostatic energy UE stored in a certain volume of region A, we must first determine the electric field in that region. For two infinite charged plates as described, we can use the principle of superposition. Each plate generates an electric field which is independent of the other. Plate 1, with a positive charge density, will generate an electric field directed away from itself, while Plate 2 will also generate an electric field directed away since it also has a positive charge density.

However, the student's scenario only involves calculating the energy in region A, where only the electric field due to Plate 1 exists since Plate 2's field does not extend into that region. The electric field E due to an infinite plate with charge density σ is E = σ / (2ε0) where ε0 is the permittivity of free space (ε0 ≈ 8.85 x 10-12 C2/N·m2). In this case, E for Plate 1 is calculated using σ = +86.8 μC/m2.

The electrostatic energy density u in a region with an electric field E is given by u = ε0E2/2. The energy density can then be multiplied by the volume to find the total energy UE. The volume in region A is 2.29 cm3. Hence, UE = u × Volume.

It's important to convert all measurements into SI units before performing the calculations. The area charge density has to be converted from μC/m2 to C/m2, and the volume from cm3 to m3.

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:

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:

Complex analysis = 682.2 V

Explanation:

1. Using a complex power approach:

Data:

Power drawn by the load, Q load = 120 kW

Power factor lagging = 0.85

The reactive power is solved as follows:

tan [arccos (power factor)] = [tex]\frac{Qload}{Pload}[/tex]

tan (cos⁻¹ [0.85])  = [tex]\frac{Qload}{120}[/tex]

solving the equation above gives Qload = 74. 4 kVar

The complex power drawn in by the load is given as:

S load = P load + jQload

           = 120 + j74.4 kVA

using the complex analysis above, we can solve for the current into the load like this: I = [tex]\frac{Sload}{Vload}[/tex]

                = [tex]\frac{120 + j74.4}{480/8}[/tex]

                = 294 (-31.8⁰) A

The power factor will be 682.2

2. Circuit power approach:

using the KVL:

Vsource = Zline + Vload

              = j 1(294 (-31.8⁰) + 480

              = 682.4

The power factor will be cos (21.5 - 31.79) = 0.598 lagging

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:

The gain of each antenna in decibels  is 353 .33 dB.

Explanation:

Given:

Frequency = 2 GHz

Diameter  =  1.2 m

To Find:

The gain of each antenna in decibels = ?

Solution:

Relationship between antenna gain and effective area

[tex]= \frac{4 \pi f^2 A_e}{c^2}[/tex]-----------------------------------(1)

Where

f is frequency

[tex]A_e[/tex]  is  effective area

c is the speed of light

[tex]A_e[/tex] = effective area of a parabolic antenna with a face area of A is 0.56A

[tex]A_e = 0.56(\pi r^2)[/tex]

r  is the radius

r = [tex]\frac{1.2}{2}[/tex]

r =  0.6

[tex]A_e = 0.56(\pi (0.6)^2)[/tex]

[tex]A_e = 0.56( 0.36 \pi)[/tex]

[tex]A_e = 0.2016 \pi[/tex]

Substituting the values in eq(1)

[tex]= \frac{4 \pi (2 \times 10^9)^2 \times 0.2016\pi}{(3 \times 10^8)^2}[/tex]

[tex]= \frac{4 \pi (4 \times 10^{18} ) \times 0.2016\pi}{(9 \times 10^{16})}[/tex]

[tex]= \frac{4 \pi (4 \times 10^{2} ) \times 0.2016\pi}{9}[/tex]

= [tex]\frac{31.80 \times 10^{2}}{9}[/tex]

= 353 .33 dB

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:

See attachment below

Explanation:

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.

Answer:

a) v = v₀(e^-kt)

b) x = -v₀(e^-kt)/k

c) v = -kx

Explanation:

a) a = dv/dt

a = -kv = dv/dt

dv/v = -kdt

Integrating the left hand side from v₀ to v and the right hand side from 0 to t

In (v/v₀) = -kt

v/v₀ = e^(-kt)

v = v₀(e^-kt)

b) v = dx/dt

dx/dt = v = v₀(e^-kt)

dx/dt = v₀(e^-kt)

dx = v₀(e^-kt)dt

Integrating the right hand side from 0 to t and the left hand side from 0 to x,

x = -v₀(e^-kt)/k

c) Since v = v₀(e^-kt)

x = -v₀(e^-kt)/k

x = -v/k

v = -kx

Hope this helps!

The velocity (v) expressed in terms of t is; v = v₀(e^(-kt))

The distance expressed in terms of t is; x = -v₀(e^-kt)/k

The velocity expressed in terms of x is; v = -kx

a) We know that acceleration is simply change in speed with respect to time. Thus;

a = dv/dt

We are told that a = -kv. Thus;

a = dv/dt = -kv

dv/v = -kdt

Integrating both sides with respective boundary conditions gives;

v₀ to v∫dv/v = 0 to t∫-kdt

⇒ In (v/v₀) = -kt

v/v₀ = e^(-kt)

v = v₀(e^(-kt))

b) We know that velocity is change in distance with respect to time. Thus;

v = dx/dt

From a above, we saw that v = v₀(e^(-kt))

Thus;

dx/dt = v₀(e^(-kt))

dx = v₀(e^(-kt))dt

Integrating both sides with respective boundary conditions gives;

0 to x∫dx = 0 to t∫v₀(e^(-kt))dt

x = -v₀(e^-kt)/k

c) Earlier we saw that v = v₀(e^-kt)

Since x = -v₀(e^-kt)/k, then;

x = -v/k

Thus, velocity is;

v = -kx

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Answer:

c. less than 60 mi/h

Explanation:

To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.

Total Distance Traveled = S = 100 mi + 100 mi

S = 200 mi

Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.

Total Time = t = Time from A to B + Time from B to C

t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)

t = 2 h + 1.43 h

t = 3.43 h

Now, the average speed of bus will be given as:

Average Speed = V = S/t

V = 200 mi/3.43 h

V = 58.33 mi/h

It is clear from this answer that the correct option is:

c. less than 60 mi/h

The average speed of the bus for the entire 200-mile trip is less than 60 mi/h (Option C)

Time = Distance / speed

Time = 100 / 50

Time = 2 h

Time = Distance / speed

Time = 100 / 70

Time = 1.43 h

Average speed = Total distance / total time

Average speed = 200 / 3.43

Average speed = 58.31 mi/h

Thus, we can conclude that the average speed is less than 60 mi/h

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Answer:

C = 292 Mbps

Explanation:

Given:

- Signal Transmitted Power P = 250mW

- The noise in channel N = 10 uW

- The signal bandwidth W = 20 MHz

Find:

what is the maximum capacity of the channel?

Solution:

-The capacity of the channel is given by Shannon's Formula:

                            C = W*log_2 ( 1 + P/N)

- Plug the values in:

                            C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)

                            C = (20*10^6)*log_2 (25001)

                            C = (20*10^6)*14.6096

                           C = 292 Mbps

Answer:

The correct code is given below:-

print("Predictions are hard.")

print("Especially about the future.")

user_num = 5

print("user_num is:", user_num)

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.

Answer: for the following process, I will explain each process and where the material is best to be used.

1. You need to connect a recently landed plane to the Earth in order to ground it and remove the built-up precipitation static. You would want to use this kind of material:

Answer: Conductor

Explanation: for you to ground the plane, you need a conductor that can be able to direct the current down to the earth. Because electron can only flow freely in a conductor.

2. You need to move a live power line out of a puddle of water. There is a lot of charge moving through this line, and if any makes it to your hands it's going to hurt. You would want to use this kind of material:

Answer: Inductor

Explanation: an Inductor resist the flow of electric current through it. You have to use an inductor, to avoid been electrocuted by the live wire. If a conductor is used current will flow through it, which may lead to electrocutions, as the water is also a conductor of electricity.

3. You need a smooth sphere to put a sensitive piece of equipment inside that will minimize any sparks between the sphere and pieces of equipment outside the sphere. You would want to use this kind of material:

Answer: Inductor

Explanation: to avoid spark, an inductor should be used, because when they is a friction between a conductors and an electric current, they will be a spark. So an inductor should be used to avoid spark. Inductors does not give a spark when in friction with an electric current

AN INDUCTOR IS ANY MATERIAL THAT RESIST THE FLOW OF ELECTRIC CURRENT THROUGH IT.

A CONDUCTOR IS ANY MATERIAL THAT ALLOWS THE FLOW ELECTRIC CURRENT THROUGH IT.

There are different kinds of material that conduct electricity. The answers are below;

A key difference  between a conductor and an inductor is that a conductor is known to be against an adjustment of voltage but an inductor only is against an adjustment of the current.

The inductor is known to stores energy because it has an attractive field. The conductor is said to stores energy because it serves as an electric field.

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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). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b).  the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

A = surface area

La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄

R-total = 0.0663 kw

From this we can calculate the rate of heat loss

using  q = Ti - To / R-total ..............(3)

given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C

from equation (3),

q = 27- (-10) / 0.0063 = 557.8 W

q = 557.8 W  

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

Answer:

Explanation:

Given

Charge [tex]q_1=13\ nC is placed at x=0[/tex]

Charge [tex]q_2=8\ nC is placed at x=5\ m[/tex]

Electric field because of both the charges will be away from them

Electric field because of charge [tex]q_1[/tex] at distance r from it

[tex]E_1=\frac{kq_1}{r^2}[/tex]

[tex]E_1=\frac{9\times 10^9\times 13\times 10^{-9}}{r^2}[/tex]

Electric Field due to charge [tex]q_2[/tex] at distance of 5-r from it

[tex]E_2=\frac{kq_2}{(5-r)^2}[/tex]

[tex]E_2=\frac{9\times 10^9\times 8\times 10^{-9}}{(5-r)^2}[/tex]

at this Point Net Electric field is zero i.e.

[tex]E_1=E_2[/tex]

[tex]\frac{9\times 10^9\times 13\times 10^{-9}}{r^2}=\frac{9\times 10^9\times 8\times 10^{-9}}{(5-r)^2}[/tex]

[tex]\frac{5-r}{r}=\sqrt{\frac{8}{13}}[/tex]

[tex]5-r=0.784 r[/tex]

[tex]5=1.784 r[/tex]

[tex]r=\frac{5}{1.784}[/tex]

[tex]r=2.80\ m[/tex]

Thus at [tex]x=2.8\ m[/tex] net electric field is zero

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.

To learn more on heat transfer, we kindly invite to check this verified question: https://brainly.com/question/12107378

Answer:

Source voltage (L-L) = 3479.50<2.13 V (in polar form)

Source voltage (L-L) = 3477.1 + j129.57 V (rectangular form)

Given Information:

A 3 phase source is feeding three loads connected in parallel.

Load voltage (L-N) = VLoad = 2000 V

Impedance of line = ZLine = 0.2 + j1.0 Ω

Load 1 = S1 = P + jQ = 150 + j120 kVA

Load 2 = S2 = Delta connected with Z2 = 150 - j48 Ω

Load 3 = S3 = 120 KVA at PF = 0.6 leading

Source voltage (L-L) = ?

Explanation:

The source voltage is = VLoad + total current*(ZLine)

Where total current is = I1 + I2 + I3

Lets first find current flowing in each of the loads

Load 1:

3 phase apparent power is given S1 = 150 + j120 kVA

Convert into per phase by diving by 3

S1 = (150 + j120)/3

S1 = 50 + j40 kVA

As we know, S = V1* (where * is the conjugate)

I1 = S1*/VLoad

I1 = (50,000 - j40,000)/2000  (notice minus sign due to conjugate)

I1 = 25 - j20 A

Load 2:

first convert delta impedance into wye by the relation

Zy = Zdelta/3

Zy = (150 - j48)/3

Zy = 50 - j16 Ω

I2 = V/Zy

I2 = 2000/(50 - j16)

I2 = 36.29 + j11.61 A

Load 3:

apparent power = 120 KVA

PF = cos(θ) = 0.6 leading

As we know, P = S*cos(θ) and Q = S*sin(θ)

θ = cos⁻¹(PF) = cos⁻¹(0.6) = 53.13°

P3 = 120*cos(53.13°) = 72 kW

Q3 = 120*sin(53.13°) = 96 kVAR

S3 = 72 + j96 kVA

Convert into per phase by diving by 3

S3 = (72 - 96)/3  (minus sign due to leading PF)

S3 = 24 - j32 kVA

I3 = S3*/VLoad

I3 = (24,000 + j32,000)/2000

I3 = 12 + j16 A

Total current = I1 + I2 + I3

Total current = (25 - j20) + (36.29) + j11.61) + (12 + j16)

Total current = 73.29 + j7.61

Source voltage (L-N) = VLoad + total current*(ZLine)

Source voltage (L-N) = 2000 + (73.29 + j7.61)*(0.2 + j1.0)

Source voltage (L-N) = 2007.05 + j74.81

Source voltage (L-L) = [tex]\sqrt{3}[/tex]*Source voltage (L-N)

Source voltage (L-L) =  [tex]\sqrt{3}[/tex] (2007.05 + j74.81)

Source voltage (L-L) = 3477.1 + j129.57 V

Source voltage (L-L) = 3479.50<2.13 V (in polar form)

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.