The heat loss from a boiler is to be held at a maximum of 900Btu/h ft2 of wall area. What thickness of asbestos (k= 0.10 Btu/h ft ℉) is required if the inner and outer surfaces of the insulation are to be 1600 and 500℉, respectively? Now if a 3-in.-thick layer of kaolin brick (k= 0.07 Btu/h ft ℉) is added to the outside of the asbestos, what heat flux will be result if the outside surface of the kaolin is 250℉? What will be the temperature at the interface between the asbestos and kaolin for this condition?

Answers

Answer 1

Answer:

a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F

Explanation:

a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.

Now [dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft

b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where  [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.

Substituting these values into the equation, we have

-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft

-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

collecting like terms, we have

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

Answer 2

The thickness of asbestos required is 0.122 ft.

The heat flux will be -70 Btu/h ft²

And the temperature of the interface is 633.33 °F.  

(i) the rate of heat loss :

dQ/dt = kAΔT/d

where k = thermal conductivity, A = area, ΔT = temperature gradient, and

d = thickness of insulation.

[dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft²,

k = 0.10 Btu/h ft ℉,

ΔT = 500 °F - 1600 °F = -1100 °F

We have to find the thickness of asbestos that is d.

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft is the thickness required.

(ii) a 3-in thick Kaolin is added to the outside of the asbestos

outside temperature of the asbestos is 250℉,

the heat loss due to the Kaolin is:

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

(iii) temperature at the interface

the total heat flux :

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d'

where  [dQ/dt]/A = -900 Btu/h ft²,

k = 0.10 Btu/h ft ℉  (for asbestos),

k' = 0.07 Btu/h ft ℉  (for Kaolin),

T₁ = 1600 °F and T₃ = 250℉.

-900 = 0.10(T₂ - 1600 °F)/0.122 + 0.07(250℉ - T₂)/0.25

-900 = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

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(-ℏ^2/2m) * [(Ae^(-αx))(α^2x - 2α + 2α^2x^2 - 2αx)] + (-xq^2)(Axe^(-αx)) = E(Axe^(-αx))

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-ℏ^2α^2 = Eq^2   ... (1)

By substituting α = ℏ^2/(2mq^2) into equation (1), we can verify that the relationship holds. Therefore, the wavefunction satisfies the one-dimensional time independent Schrödinger equation with the given potential term.

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Bumpers, collapsible frames, airbags and seat belts all come into play when an an accident occurs. True or false

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Bumpers, collapsible frames, airbags, and seat belts are safety features in vehicles designed to protect occupants in the event of an accident.

Bumpers help absorb impact and minimize damage to the vehicle, collapsible frames absorb and distribute energy, airbags rapidly inflate to provide a cushioning effect, and seat belts restrain occupants and prevent them from being thrown forward. All of these safety features work together to enhance occupant safety during accidents.

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I just got asked a question about quantum entanglement. Can someone please give me a basic understandable of what it is​

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If you want the definition I will tell you the definition of what it means.

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Explanation:

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Question 2 10 pts A company has designed and built a new air compressor section for our advanced Gas turbine engine used in electrical power generation. They state that their compressor operates adiabatically, and has a pressure ratio of 30. The inlet temperature is 35 deg C and the intet pressure is 100 kPa. The mass flow rate is steady and is 50 kg/s The stated power to run the compressor is 24713 kW Co = 1.005 kJ/kg K k-1.4 What is the isentropic temperature at the compressor exit? 815.2 K 752K 800 K 93 deg C

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Given the pressure ratio of 30, the inlet temperature of 35 °C (308 K), and the inlet pressure of 100 kPa, we can calculate the isentropic temperature using the equation. Considering the specific heat ratio for air as 1.4, we find that the isentropic temperature at the compressor exit is approximately 815.2 K.

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The probability distribution for a
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Answer:

Explanation:

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Answer:

Option 3. The tennis ball began from rest and rolls at a rate of 14.7 m/s safer 1.5 seconds.

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To know the the correct answer to the question, it is important that we know the definition of acceleration.

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With the above information in mind, let us consider the options given in the question above to know which conform to the difinition of acceleration.

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We were told that the tennis ball has the following:

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Answers

Responder:

490 julios

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d = 0 + 1/2 (2) (7) ²

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Explanation:

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Answer:

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Explanation:

Help pls ;) and thank you

Help pls ;) and thank you

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Answer:

1a) Transverse wave.

1b) Ripples on the surface of water.

1c) See first attachment.

1d)  See below for explanation.

2a) Longitudinal wave.

2b) Sound waves.

2c) See second attachment.

2d)  See below for explanation.

Part 2) 54 m/s

Explanation:

Part 1

Question 1

a)  The wave in the picture is a transverse wave.

In transverse waves, the oscillations are at right angles to the direction of wave travel.

b)  A real world example of a transverse wave is ripples on the surface of water.

c)  See first attachment.

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d)  The amplitude is the maximum displacement of a point of a wave from its resting position.  It is measured by calculating the difference in height between a crest and the resting position (mid-line).

Question 2

a)  The wave in the picture is a longitudinal wave.

In longitudinal waves, the oscillations are parallel to the direction of wave travel.

b)  A real world example of a longitudinal wave is sound waves.

c)  See second attachment.

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d)  The amplitude is the maximum displacement of a point of a wave from its resting position.  It is measured by calculating the distance between the particles in the areas where it is compressed.

Part 2

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Answers

Answer:

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Explanation:

You want to know your weight if the Earth were 3 times as massive and had 5 times the present radius. Your weight is 710 N.

Weight

Your weight is proportional to the mass of the Earth and the square of the radius between your mass and the center of the Earth. The revised dimensions of the earth would multiply your weight by ...

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Answer:

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We can determine whether a change is physical or chemical by analyzing the what type of change is occurred. The change in color or formation of new product is a chemical change.

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A chemical change is a type of change which include the breaking or making of chemical bonds and the formation of new products which has properties different from the initial substances.

Phase changes, change in size, shape, dissolving, conductivity etc. are  physical changes. They do not involve any change in chemical bonds or the formation of new product.

The change in color indicates the formation of a new product. All the chemical reactions like the corrosion of metals reaction of metals with acids or water etc . are all chemical changes.

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One molecule of calcium oxide, Cao, and one molecule of carbon dioxide, CO2, combine in a chemical reaction to form one
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CaCO3 is the product of this reaction

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The chemical formula of the product of this reaction is ; CaCO3.

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The law states that matter( which is the building block of mass) can neither be created nor destroyed.

Therefore, in a reaction where one molecule of calcium oxide, CaO, and one molecule of carbon dioxide, CO2, combine in a chemical reaction to form one substance;

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Answer:

All of these

Explanation:

a space vehicle is traveling at 2600 km/h relative to earth when the exhausted rocket motor is disengaged and sent backward. the relative speed between the motor and the command module is then 73 km/h. the mass of the motor is four times the mass of the module. what is the speed of the command module relative to earth just after the separation?

Answers

The speed of the command module relative to earth just after the separation is  2585.4 km/h

Since we don't know the masses of the motor and the module, So considering the mass of the module M, and the mass of the motor  to be  4M and  v be the velocity of the motor

So, The total mass of the vehicle is 5M( in kg), before separation,the momentum of the vehicle is:

2600 * 5M = 13000 M kg-km/h.

since the velocity of the module will be v + 73. After the separation, the momentum of the motor is 4Mv, and the momentum of the module is

M(v + 73) = Mv + 73M.

therefore, The total momentum is 4MV + Mv + 73M = 5Mv + 73M.

which will be the same as the initial momentum.

So, it  gives us the equation :

13000 M = 5Mv + 73M

Dividing  both sides by M:

=>13000 = 5v + 73

=>12927 = 5v

=>2585.4= v

But v is the velocity of the motor, so we have to add 93 to get the velocity of the module, which is  2658.4

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The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential difference between the two plates.

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Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases

The relation between potential difference and the electric field is given by ΔV = E.d

Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.

The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.

The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.

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Calculate the KE in joules of a 1500 kg car moving at 29 m/s?

Answers

Answer:

x J = (1500 kg)(29 m/s)(y m/s)

Explanation:

x J = (1500 kg)(29 m/s)(y m/s)

To know that a 1500 kg car is moving at 29 m/s is not enough.

The value of x depends on y. You’re missing a number of meters and a quantity of ‘per seconds’ somewhere in your problem statement and you need to find them in order to solve the problem

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Answer:

They would be pointing in the same direction

Explanation:

If they were facing each other then it may seem like they are pointing in different directions they would still point the same way.

A radio station's channel, such as 100.7 FM or 92.3 FM, is actually its frequency in megahertz (MHz),where 1 MHz =106 Hz and 1 Hz = 1 s−1. Calculate the broadcast wavelength of the radio station 95.90 FM

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The radio station's broadcast wavelength. 299792458/95700000=3.133m . λ=\(\frac{3.10^{8}m/s }{95.90.10^{6}Hz }\) =3.38m

In the actual world, what is frequency?

Frequency of a wave is defined as the total amount of waves generated in a second. Frequency is the measurement of the quantity of vibrations per second. As an example, consider the following: If five full waves are created in a second, their frequency is 5 hertz (Hz), or 5 cycles every second.

Which two frequencies are there?

The two primary frequency distribution types utilized in data analysis are cumulative and relative frequency distributions. Both depend on incidence, which in descriptive and inferential statistics refers to how frequently a particular event happens within a particular data set.

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Which planetary body has the fastest orbit, and which has the slowest orbit? Do you notice a general pattern here? Briefly explain a relationship between orbital velocity and orbital radius.

Answers

The planetary body with the fastest orbit is Mercury, and the one with the slowest orbit is Neptune.

There is a general pattern between orbital velocity and orbital radius known as Kepler's second law of planetary motion. According to this law, a planet sweeps out equal areas in equal times as it orbits the Sun. This implies that planets closer to the Sun have smaller orbital radii and must travel faster to cover the same area in the same amount of time.

The relationship between orbital velocity and orbital radius can be expressed as v ∝ 1/r, where v represents the orbital velocity and r denotes the orbital radius. This relationship shows that as the orbital radius increases, the orbital velocity decreases. In other words, planets farther from the Sun have slower orbital velocities compared to those closer to the Sun.

This pattern is consistent with observations in our solar system. The inner planets, such as Mercury, have smaller orbital radii and faster orbital velocities, while the outer planets, like Neptune, have larger orbital radii and slower orbital velocities.

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a soccer ball is kicked at an angle of 64° to the horizontal with an initial speed of 15 m/s. assume for the moment that we can neglect air resistance. (a) for how much time is the ball in the air?

Answers

Time taken by the soccer ball in the air when we neglect the air resistance is 2.75s

projectile motion - It is the motion of an object which projected into the air  and moving under the gravity.

a soccer ball is kicked at an angle of 64° to the horizontal

initial speed = 15 m/s

Initial velocity in vertical direction is given as

\(V_{v} = V_{0}sin \theta\)

\(V_{v} = 15sin(64)\)

\(V_{v}\) = 15× 0.8987

    = 13.48m/s

time taken to reach the maximum height is given by

T =\(V_{v}\) /g

where

\(V_{v}\)  is the velocity

g is the gravity

Ball comes downward with the same speed

then that time period is given by.

2\(V_{v}\) /g

= 2× 13.48 /9.8

T = 2.75s

Time taken by the soccer ball in the air when we neglect the air resistance is 2.75s

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a reaction that happens when one substance breaks down into two or more substances

Answers

Answer:

decomposition reaction

Explanation:

decomposition reactions occurs when one substance breaks down, or decomposes, into two or more substances

You walk 100m due north. You then turn and walk 55m due east. You then make another turn and walk 12m due south. What is the resultant vector for your walk?

Answers

Answer:

Explanation:

Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:

\(A_x=100cos90\\A_x=0\\B_x=55cos0\\B_x=55\\C_x=12cos270\\C_x=55\)

Add them all together to get the x component of the resultant vector, V:

\(V_x=55\)

Do the same to find the y components of the part of this journey:

\(A_y=100sin90\\A_y=100\\B_y=55sin0\\B_y=0\\C_y=12sin270\\C_y=-12\)

Add them together to get the y component of the resultant vector, V:

\(V_y=88\)

One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.

We find the final magnitude:

\(V_{mag}=\sqrt{55^2+88^2}\) and, rounding to 2 sig dig's as needed:

\(V_{mag}=\) 1.0 × 10² m; now for the direction:

\(\theta=tan^{-1}(\frac{88}{55})=\) 58°

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