Consider two houses that are identical, except that the walls are built using bricks in one house, and wood in the other. The walls of the brick house are twice as thick. Which house do you think will be more energy efficient?

Answers

Answer 1

Answer:

Walls Built Using Bricks and Wood

The brick house is more energy-efficient than the one built with wood.

Explanation:

Because of their high thermal mass, which gives bricks the ability to absorb heat and release it over time, bricks remain more energy-efficient than other building materials, including wood.  In summer, bricks leave your home cool.  In winter, they make it warm.  With these two advantages provided by bricks over other building materials, bricks are the most energy-efficient building material.


Related Questions

ow Pass Filter Design 0.0/5.0 points (graded) Determine the transfer function H(s) for a low pass filter with the following characteristics: a cutoff frequency of 100 kHz a stopband attenuation rate of 40 dB/decade. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency Write the formula for H(s) that satisfies these requirements:

Answers

Answer:

H(s) = 20 / [ 1 + s / 10^5 ]^2

Explanation:

Given data:

cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20 dB

new nominal passband gain at cutoff = 14 dB

Represent the transfer function H(s)

The attenuation rate show that there are two(2) poles

H(s) = k / [ 1 + s/Wc ]^2  ----- ( 1 )

where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20

Input values into equation 1

H(s) = 20 / [ 1 + s / 10^5 ]^2

A 0.82-in-diameter aluminum rod is 5.5 ft long and carries a load of 3000 lbf. Find the tensile stress, the total deformation, the unit strains, and the change in the rod diameter.

Answers

Answer:

Tensile stress = 0.1855Kpsi

Total deformation = 0.0012243 in

Unit strain =  1.855 *10^-5   or  18.55μ

Change in the rod diameter = 5.02 * 10^ -6 in

Explanation:

Data given: D= 0.82 in

                   L = 5.5 ft * 12 = 66 in

load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm

Area = (π/4) D² = (π/4) 0.82²  = 0.502655 in²

∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²

                           Rt = 0.1855 Kpsi

∴ Total deformation = PL / AE = Rt * L/ Eal

                                 = 0.1855 * 10³  * 66 / 10000 * 10³

                                 = 0.0012243 in

∴the unit strains = total deformation / L = 0.0012243/ 66

                          =0.00001855 = 1.855 *10^-5

                         = 18.55μ

∴ Change in rod   Δd/ d = μ ΔL/L

                           = (0.33) 1.855 *10^-5 * 0.82

                           = 5.02 * 10^ -6 in

Draw a sinusoidal signal and illustrate how quantization and sampling is handled by
using relevant grids.

Answers

Answer: Your mums gay

Explanation: Your mums gay

The temperature gradient in a spherical (or cylindrical) wall at steady state will always decrease (in magnitude) with increasing distance from the center (line), i.e. radial distance.
A. True
B. False

Answers

Answer:

True

Explanation:

Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.

We have this cylinder equation as

[T1-T2 / ln(r1-r2)]2πKL

The radial distance is r2-r1

The gradient of temperature is T1-T2

From the equation,

The temperature gradient has a direct and proportional relationship to radial distance

T1-T2 ∝ ln(r2-r1)

1/T1-T2 = k(r2-r1)

This inverse relationship above confirms that the statement is true

Activity 1. Fill the blank with the correct answer. Write your answer on the blank. 1. ___________________ is a regular pattern of dots displayed on the screen which acts as a visual aid and also used to define the extent of your drawing. 2. Ortho is short for ___________________, which means either vertical or horizontal. 3. Tangent is a point where two _______________________ meet at just a single point. 4. If you want to create a new drawing, simply press ___________________ for the short cut key. 5. There are _______________Osnap that can help you performs your task easier.

Answers

Answer:

1. Drawing grid.

2. Orthogonal.

3. Geometries.

4. CTRL+N.

5. Thirteen (13).

Explanation:

CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.

Some of the features of an AutoCAD software are;

1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.

2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.

3. Tangent is a point where two geometries meet at just a single point.

4. If you want to create a new drawing, simply press CTRL+N for the short cut key.

5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.

Imagine a cantilever beam fixed at one end with a mass = m and a length = L. If this beam is subject to an inertial force and a uniformly distributed load = w, what is the moment present at a length of L/4?

Answers

Answer:

jsow

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Explanation:

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The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the outside surface of the solid shaft. Determine the principal stresses, the maximum in-plane shear stress and average normal stress using Mohr's Circle.

Answers

Answer: Attached below is the missing detail and Mohr's circle.

i) б1 =  9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii)  -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

      = -20 / 0.785 * 5^2  = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

  = (16 * 250 ) / π * 5^3  = 010.19 Ksi

Calculate the Principal stress, maximum in-plane shear stress and average normal stress

Using Mohr's circle ( attached below )

i) principal stresses:

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

ii) maximum in-plane shear stress

ζ  = radius of Mohr's circle

   = 5.1 cm = 10.2 Ksi   ( Given that ; 1 cm = 2Ksi )

iii) average normal stress

 = 9.6 + ( - 10.7 ) / 2

  = -0.51Ksi

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution.
x dx/dy−y=x^2sinx

Answers

Answer:

Interval:  x∈ ( 0, ∞ )

There are no transient terms

Explanation:

x (dy/dx) – y= x^2sinx

Attached below is the detailed solution of the Given problem

There are no transient terms found in the general solution

Interval:  x∈ ( 0, ∞ )

Do you know who Candice is

Answers

Answer: Can these nuts fit in your mouth?

Explanation:

im just here for the points >:)

Given resistance 30ohms Inductance 200mH is connected to a 230v,50hZ supply. Impedance 69.6ohms Calculate current consumed?

Answers

Answer:

the current consumed is 3.3 A

Explanation:

Given;

resistance, R = 30 ohms

inductance, L = 200 mH

Voltage supply, V = 230 V

frequency of the coil, f = 50 Hz

impedance, Z = 69.6 Ohms

The current consumed is calculated as;

[tex]I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A[/tex]

Therefore, the current consumed is 3.3 A

how does load transfer of space needle​

Answers

Answer:

The Space Needle is a cut away with minimal residual deflection due to load transfer.

what type of slab and beam used in construction of space neddle​

Answers

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Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.

Answers

Answer:

(a) the reversible power output of turbine is 5810 kw

(b) The second-law efficiency of he turbine = 86.05%

Explanation:

In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.

h1 = 3658 kJ/kg s1=7.167 kJ/kgK

In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state

h2 = 2682kl/kg S2= 7.694 kJ/kg

Assuming that the energy balance equation given  

Wout=m [h1-h2+(v1²-v2²) /2]

Let

W =5 MW

V1= 80 m/s  V2= 140 m/s

h1 = 3658kJ/kg  h2 = 2682 kJ/kg

∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)

m = 5.158kg/s

Consider the energy balance equation given  

Wrev,out =Wout-mT0(s1-s2)

Substitute Wout =5 MW m = 5.158kg/s 7

s1=  7.167 kJ/kg-K            s2= 7.694kJ/kg-K and 25°C .

Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)

= 5810 kW

(a) Therefore, the reversible power output of turbine is 5810 kw.

The given values of quantities were substituted and the reversible power output are calculated.

(b) Calculating the second law efficiency of the turbine:  

η=Wout/W rev,out

Let Wout =  5 MW and Wrev,out = 5810 kW  

η=(5 MW x 1000 kW)/(1 MW *5810)  

η= 86.05%

bending stress distribution is a.rectangle b.parabolic c.curve d.i section​

Answers

B parabolic

Hope this helps :))))))))))

A 1m3 tank containing air at 25℃ and 500kPa is connected through a valve to
another tank containing 5kg of air at 35℃ and 200kPa. Now the valve is opened,
and the entire system is allowed to reach thermal equilibrium, which is at 20℃
(Take: Ru = 8.314 kJ / kg.K).

Answers

Answer:

The right answer is "2.2099 m³".

Explanation:

Given:

Mass,

m = 5 kg

Temperature,

T = 35℃

or,

  = 35 + 273

Pressure,

P = 200 kPa

Gas constant,

R = 0.2870 kj/kgK

By using the ideal gas equation,

The volume will be:

⇒ [tex]PV=mRT[/tex]

or,

⇒    [tex]V=\frac{mRT}{P}[/tex]

By substituting the values, we get

          [tex]=\frac{5(0.2870)(35+273)}{200}[/tex]

          [tex]=\frac{441.98}{200}[/tex]

          [tex]=2.2099 \ m^3[/tex]

In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference between the inlet and test-section for a fixed ratio of inlet-to-test section cross-sectional area.
a. True
b. false

Answers

Answer:

Hence the given statement is false.

Explanation:

For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.

The speed within the test section is decided by the planning of the tunnel.  

Thus by adjusting the pressure difference won't change the worth of test section velocity.

Answer:

The given statement is false .

A designer needs to select the material for a plate under tensile stress. Assuming that the applied tensile force is 13,000 lb and the area under the stress is 4 square inches, determine which material should be selected to assure safety. Assume safety factor is 2. Material A: Ultimate Tensile stress is 8000 lb/in2Material B: Ultimate Tensile stress is 5500 lb/in2

Answers

Wow Muy bien no se que dice pero bien.

A flow inside a centrifuge can be approximated by a combination of a central cylinder and a radial line source flow, giving the following potential function:
Ø= a2/r -cosØ + aßlnr = r
Where a is the radius of the central base of the centrifuge and ß is a constant.
a) Provide expressions for the velocities Vr and vo .
b) Find the expression for the stream function.

Answers

Answer:

a)  Vr = - a^2/r cosθ  + aß / r

    Vθ = 1/r [ -a^2/r * sinθ ]

b) attached below

Explanation:

potential function

Ø= a^2 /r  cosØ + aßlnr ----- ( 1 )

a = radius ,  ß = constant

a) Expressions for Vr and Vθ

Vr =  dØ / dr  ----- ( 2 )

hence expression : Vr = - a^2/r cosθ  + aß / r

Vθ = 1/r dØ / dθ ------ ( 3 )

back to equation 1

dØ / dr = - a^2/r sinθ + 0  --- ( 4 )

Resolving equations 3 and 4

Vθ = 1/r [ -a^2/r * sinθ ]

b) expression for stream function

attached below

A signalized intersection has a sum of critical flow ratios of 0.72 and a total cycle lost time of 12 seconds. Assuming a critical intersection v/c ration of 0.9, calculate the minimum necessary cycle length.

Answers

Answer:

[tex]T_o=82.1sec[/tex]

Explanation:

From the question we are told that:

Lost Time [tex]t=12secs[/tex]

Sum of critical flow ratios [tex]X=0.72[/tex]

Generally the Webster Method's equation for Optimum cycle time is is mathematically given by

 [tex]T_o=\frac{1.5t+5}{1-x}[/tex]

 [tex]T_o=\frac{1.5*12+5}{1-0.72}[/tex]

 [tex]T_o=82.1sec[/tex]

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