Consider the attic of a home located in a hot climate. The floor of the attic is characterized by a width of L1= 10 m while the roof makes an angle of theta=30 degrees from the horizontal direction. The home owner wishes to reduce the heat load tot he home by adhering bright aluminum foil (of emissivity 0.07) onto the surfaces of the attic space. Prior to installation of the foil the surfaces are of emissivity 0.85
a) consider installation on the bottom of the attic roof only. Determine the ratio of the radiation heat transfer after to before the installation of the foil.
b) installation on the top of the attic floor
c) installation on both the roof bottom and floor top

Answer:

\(W_s =\) 283.181 hp

Explanation:

Given that:

Air enters a compressor operating at steady state at a pressure \(P_1\) =  176.4 lbf/in.^2  and Temperature \(T_1\) at 260°F

Volumetric flow rate V = 424 ft^3/min

Air exits at a pressure \(P_2\)  = 15.4 lbf/in.^2 and Temperature \(T_2\) at 80°F.

Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:

\(Q_{cv}\) = -6800 Btu/h  = - 1.9924 kW

Using the steady  state  energy in the process;

\(h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}\)

where;

\(g(z_2-z_1) =0\)  and  \(\dfrac{1}{2}(v^2_2-v_1^2) = 0\)

Then; we have :

\(h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}\)

\(h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}\)

\({m}(h_2 - h_1) ={Q_{cv} - W_s}\)

\(W_s ={Q_{cv} + {m}(h_2 - h_1)\) ----- (1)

Using the relation of Ideal gas equation;

P₁V₁ = mRT₁

Pressure \(P_1\) =  176.4 lbf/in.^2   = ( 176.4 ×  6894.76 ) N/m² = 1216235.664 N/m²

Volumetric flow rate V = 424 ft^3/min = (424 ×  0.0004719) m³  /sec

= 0.2000856 m³  /sec

Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K

Gas constant R=287 J/kg K

Then;

1216235.664 N/m² × 0.2000856 m³  /sec = m × 287 J/kg K × 399.817 K

\(m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }\)

m = 2.121 kg/sec

The change in enthalpy:

\(m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)\)

\(= 2.121* 1.005* ( 399.817 -299.817)\)

= 213.1605 kW

From (1)

\(W_s ={Q_{cv} + {m}(h_2 - h_1)\)

\(W_s =\)  - 1.9924 kW + 213.1605 kW

\(W_s =\) 211.1681  kW

\(W_s =\) 283.181 hp

The power input is \(W_s =\) 283.181 hp

Answer:

YES

Explanation:

NO

;-;

Answer: a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.

Explanation:

Giving a way to the law enforcement vehicle and a medical emergency vehicle is necessary. If one approaches an emergency vehicle parked along the roadway one should change the lane as the vehicle may not move and the driver may also waste his or her time also one should also slow down his or her speed while approaching the vehicle as most of the emergency vehicle are in rush to reach the hospital so the driver should maintain some distance with the medical emergency vehicle.

Answer:

there's no photo? but I'm willing to help

Answer:

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

Explanation:

We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:

\(\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}\)

Where \(\Delta V_{storage}\) is the monthly storage change of the lake, measured in cubic feet.

Monthly inflow

\(V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{inflow} = 77.76\times 10^{6}\,ft^{3}\)

Monthly outflow

\(V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{outflow} = 66.98\times 10^{6}\,ft^{3}\)

Seepage losses

\(V_{seepage} = s_{seepage}\cdot A_{lake}\)

Where:

\(s_{seepage}\) - Seepage length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{seepage} = 1.5\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{seepage} = 2.86\times 10^{6}\,ft^{3}\)

Evaporation losses

\(V_{evaporation} = s_{evaporation}\cdot A_{lake}\)

Where:

\(s_{evaporation}\) - Evaporation length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{evaporation} = 6\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{evaporation} = 11.44\times 10^{6}\,ft^{3}\)

Precipitation

\(V_{precipitation} = s_{precipitation}\cdot A_{lake}\)

Where:

\(s_{precipitation}\) - Precipitation length gain, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{precipitation} = 4.25\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{precipitation} = 8.10\times 10^{6}\,ft^{3}\)

Finally, we estimate the storage change of the lake during the month:

\(\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}\)

\(\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}\)

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

The volume of water gained and the loss of water through flow,

seepage, precipitation and evaporation gives the storage change.

Response:

Given parameters:

Area of the lake = 525 acres

Inflow = 30 ft.³/s

Outflow = 27 ft.³/s

Seepage loss = 1.5 in. = 0.125 ft.

Total precipitation = 4.25 inches

Evaporator loss = 6 inches

Number of seconds in a month is found as follows;

\(30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds\)

Number of seconds in a month = 2592000 s.

Volume change due to flow, \(V_{fl}\) = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³

1 acre = 43560 ft.²

Therefore;

525 acres = 525 × 43560 ft.² =  2.2869 × 10⁷ ft.²

Volume of water in seepage loss, \(V_s\) = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³

Volume gained due to precipitation, \(V_p\) = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³

Volume evaporation loss, \(V_e\) = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³

Which gives;

ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123

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Ford requested inclusion so that the firm wouldn't suffer, but the Treasury Department lent money and purchased GM and Chrysler stock instead.

Companies that relied on credit suffered greatly during the 2008 financial crisis, commonly known as the "credit crisis," and as a result, the automotive industry experienced a crisis from 2008 to 2010. Prior to the crisis, the automotive industry had been suffering from the rise in car fuel prices and the energy crisis, which had deterred consumers from purchasing fuel-hungry vehicles. Although the 2008 financial crisis had an impact on the European and Asian automobile industries, it had the greatest impact on the American industry because it was the source of the financial catastrophe. Therefore, the "Big Three" automakers—GM, Chrysler, and Ford—asked Congress for a financial bailout and issued a bankruptcy warning for GM and Chrysler.

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Answer:

Except what? I'm confused

All of these are true about using adhesive except Bilateral. A bilateral contract is defined as an agreements between two parties in which each side agrees to fulfill his or her side of the bargain.

A bilateral contract is defined as an agreements between two parties in which each side agrees to fulfill his or her side of the bargain. According to my research on the different terms used when referencing an insurance contract, I can say that all of the answers provided except for Bilateral are considered typical characteristics describing the nature of an insurance contract.

Since an insurance contract is a fund that the insurance company pays in the case of an accident in which the person is injured, there is only one party that agrees to fulfill their side of the bargain and that is the insurance company.

Therefore, All of these are true about using adhesive except Bilateral. A bilateral contract is defined as an agreements between two parties in which each side agrees to fulfill his or her side of the bargain.

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The Last Address In A SRAM Chip With Pins AO - A15 And DO - D7 (15 Points) E.G. 2CD4H: Ao-A15 represent the address lines and Do-D7 represent the Data lines.

Ao-A15 represent the address lines and Do-D7 represent the Data lines.

Number of address lines=16 bits.

For last address in SRAM all 16 bits are 1's.

Last address=1111111111111111.

In hexa-decimal it is FFFF.

Thus, the Last Address In A SRAM Chip With Pins AO - A15 And DO - D7 (15 Points) E.G. 2CD4H: Ao-A15 represent the address lines and Do-D7 represent the Data lines.

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When it comes to identifying transistor materials, it is important to understand which elements can be used as a dopant in a p-type semiconductor. Of these options, gallium and antimony are typically used as dopants in p-type semiconductors.

In this case, the options are gallium, arsenic, antimony, and phosphorus.This is because these elements have fewer valence electrons than the base semiconductor material, which allows for the creation of electron holes in the p-type material. By adding these dopants, it becomes easier to control the conductivity of the semiconductor, making it possible to create p-type transistors. Overall, understanding the properties of different semiconductor materials and dopants is essential for developing advanced electronic devices.

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ᴄᴏᴍᴘʟᴇᴛᴇɴᴇꜱꜱ ᴍᴇᴀɴꜱ ᴛʜᴇ ꜱᴛᴀᴛᴇ ᴏꜰ ʙᴇɪɴɢ ᴄᴏᴍᴘʟᴇᴛᴇ ᴀɴᴅ ᴇɴᴛɪʀᴇ; ʜᴀᴠɪɴɢ ᴇᴠᴇʀʏᴛʜɪɴɢ ᴛʜᴀᴛ ɪꜱ ɴᴇᴇᴅᴇᴅ.

ʙᴇᴀᴜᴛʏ ᴍᴇᴀɴꜱ combination of qualities, such as shape, colour, or form, that pleases the aesthetic senses, especially the sight.

Answer:

or is the strongest evenger she hulk

Explanation:

?????????

Answer:

Thor!

Explanation:

In Thor: Ragnarok he beat the Hulk in order for Hulk to win thor had to be electrocuted and in Avengers: Endgame Thor is seen holding open the  "Floodgates" and withstanding the radiation from a dying star, also the fact that Thor is a god means that he is all powerful and the rightful heir to the throne to Asgard, plus the fact that he has defeated Loki multiple times a feat that not even the Hulk has done.

Until emergency assistance arrives, portable fire extinguishers can put out or manage a fire.

The use of portable fire extinguishers can put out or contain a fire until emergency assistance is received.

Utilize portable units exclusively for small flames or fires that are still in the early stages because their discharge times are often only a few seconds. Make a plan for a getaway. Avoid breathing in the extinguishing agent and smoke by keeping low to the ground.

Only use a portable fire extinguisher when a fire is small, restricted to a single area, and not made up of extremely flammable materials. If the fire becomes out of control, leave the premises right away and get assistance.

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Solution :

The dimensionless conduction heat rate, \($q_{ss}^*$\)

\($q_{ss}^*=\frac{q\times L_c}{K A_s(T_1-T_2)}$\)   ...........(1)

where \($L_c$\)  = characteristic length

                \($=\left(\frac{A}{4\pi} \right)^{1/2}. \sqrt{\frac{D^2}{4}}$\)

              A is surface area

          q = heat transfer

\($q=Sk(T_1-T_2)$\)   ..................(2)

where, S = conductor shape factor

Now substituting (2) in (1),

\($q_{ss}^* = \frac{Sk(T_1-T_2)L_c}{kA(T_1-T_2)}$\)

\($q_{ss}^* = \frac{S \times L_c}{A}$\)

\($q_{ss}^* = \frac{S \times D/2}{\pi D^2}$\)

\($q_{ss}^* = \frac{S \times D}{2\pi D^2}$\)   ...................(3)

For a sphere, we know S = 2πD

Putting this in (3),

\($q_{ss}^* = \frac{2 \pi D \times D}{2\pi D^2}$\)

\($q_{ss}^* = \frac{2 \pi D^2}{2\pi D^2}$\)

\($q_{ss}^* = 1$\)

Therefore, the dimensionless heat conduction rate for a sphere is 1.

The correct statement of the domain of (f+g)(x) is that it is restricted to all non-negative real numbers, or [0,∞).

The error in stating the domain of (f + g)(x) as all real numbers is that the domain of the function (f+g)(x) is determined by the intersection of the domains of the functions f(x) and g(x).

In the given equations, the domain of f(x) is restricted to non-negative real numbers as the square root of a negative number is undefined in the real number system. However, the domain of g(x) is all non-negative real numbers.

To find the domain of (f+g)(x), we need to find the intersection of the domains of f(x) and g(x). Since the domain of g(x) is already included in the domain of f(x), the domain of (f+g)(x) is also restricted to all non-negative real numbers.

The correct statement of the domain of (f+g)(x) is that it is restricted to all non-negative real numbers, or [0,∞).

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A) The horizontal displacement of point A is 0.144 mm. B) The shear strain Yx'y between the lines in the x' direction and the y' direction is 0.00096 m.

Shear strain is a type of deformation caused by a force applied to an object. It occurs when two opposite parallel forces are applied in opposite directions, causing the object to elongate in one direction and compress in the other. Shear strain is different from tensile strain, which happens when a force is applied in one direction only.

(a) The horizontal displacement of point A due to the shear stress τ is given by the equation da = τYxy. Substituting the given values, we obtain da = τ(0.0012 m) = (120 N/mm2)(0.0012 m) = 0.144 mm.

(b) The shear strain Yx'y between the lines in the x' direction and the y' direction is given by the equation Yx'y = da/δ', where δ' is the distance between the lines in the x' direction. Substituting the given values, we obtain Yx'y = (0.144 mm)/(150 mm) = 0.00096 m.

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Answer:

Explanation:

A spectrogram is a Visual representation of the spectrum of frequencies in a sound or other signal as the vary with time or some other variable, hope this helps :)

The term for the rotational movement of the nose of a plane is known as pitch. This term refers to the motion of the airplane's nose up and down, relative to the horizontal line. It is one of the three primary rotational movements of an aircraft, along with roll and yaw.

The pitch of an airplane is controlled by the elevators, which are located on the trailing edge of the airplane's horizontal stabilizer. These elevators move in opposite directions, with one elevating and the other depressing, to control the pitch angle of the airplane. When the elevators move upward, the nose of the plane pitches up, while when the elevators move downward, the nose pitches down.The pitch angle of an airplane is an essential parameter for flight control, as it affects the airplane's speed, altitude, and angle of attack. A high pitch angle can cause the airplane to lose altitude rapidly and decrease its airspeed, while a low pitch angle can cause the airplane to climb and increase its speed.

Therefore, pilots must maintain a proper pitch angle to ensure safe and efficient flight operations.

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Shanika needs to move the orange coaster car to point B. This movement will result in the greatest potential energy increase.

Potential energy is stored energy that is influenced by the relative position of various system components. The potential energy of a spring increases when it is compressed or stretched. When a steel ball is raised above ground, it has more potential energy than when it falls to Earth.

In the elevated position, it can do more work. Potential energy is a property of a system, not an individual body or particle; for example, as the Earth and raised ball are separated, the system has more potential energy.

Potential energy exists in systems with parts that exert varying magnitudes of forces on each other based on their configuration, or relative position.

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Answer:

a) the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b) the flow rate is 5.8 ml/min

Explanation:

Given the data in the question;

Length of blood vessel L = 1 m

radius r = 1 mm = 0.001 m

blood viscosity μ = 2.7 × 10⁻³ pa.s = 2.7 × 10⁻³ × 10⁻⁹ Gpa.s = 2.7 × 10⁻¹² Gpa.s

Now, we know that Resistance = 8μL / πr⁴

so we substitute

Resistance = [8 × (2.7 × 10⁻¹²) × 1] / [π(0.001)⁴]

Resistance = [2.16 × 10⁻¹¹] / [3.14159 × 10⁻¹²]

Resistance = 6.8755 ≈ 6.88 Gpa.s/m³

Therefore, the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b)

blood pressure at the inlet of the vessel = 43 mm Hg

blood pressure at the outlet of the vessel = 38 mm Hg

flow rate = ?

we know that;

flow rate Q = ΔP / R

where ΔP is change in pressure and R is resistance.

ΔP = Inlet pressure - Outlet pressure = 43 - 38 = 5 mm Hg =  665 pa

R = 6.8755 Gpa.s/m³ = { 6.8755 × 10⁹ / 60 × 10⁶ } = 114.5916 pa.min.ml⁻¹

so we substitute

Q =  665 pa / 114.5916 pa.m.ml⁻¹

Q = 5.8 ml/min

Therefore, the flow rate is 5.8 ml/min

Answer:

safety glasses, hearing protection, respirator, work gloves, and work boots.

Explanation:

please give brainliest!! <3

Answer: true

Explanation:

Under the standard, ethernet switches can provide enough power over utp for wireless access points.

An access point is a device that forms a wireless local area network, or WLAN, in a building or workplace. An access point uses an Ethernet connection to connect to a wired router, switch, or hub and broadcasts a WiFi signal to a specific region.

Routers can provide wired or wireless connectivity to a variety of end-user devices, whereas APs primarily service wireless devices such as phones, laptops, and tablets. An AP, in essence, adds wireless capacity to a wired network.

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Full Question:

Under the standard, Ethernet switches can provide enough power over UTP for ________.

A) wireless access points

B) voice over IP telephones

C) both A and B

D) neither A nor B

It depends on the age and condition of the car's AC unit, as well as the cost of the compressor replacement. If the car is relatively new and the AC unit is in good condition, replacing the compressor may be a worthwhile investment to extend the life of the unit.

However, if the car is older or the AC unit has other issues, it may be more cost-effective to replace the entire unit instead of just the compressor.

Ultimately, the decision to replace the compressor on an AC unit in a car depends on a variety of factors and should be based on a careful assessment of the costs and benefits. It is recommended to consult with a qualified mechanic or automotive technician to determine the best course of action for your specific situation.

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To determine the total quantity of nitrogen (N2) needed to fill a vehicle's four tires to a gage pressure 4 x 0.6 m3 x (180 kPa + 1 atm) x 0.028 kg/L x 0.028 kg/L = 48.48 kg is the needed mass of N2.

We must first know the capacity of each tire, which is 0.6 m3, in order to calculate the total amount of nitrogen (N2) needed to inflate all four tires of a car to a gage pressure of 180 kPa at a temperature of 25oC. The total pressure in each tire is 181 kPa when the ambient pressure of 1 atm is taken into account. The number of moles of nitrogen in each tire may then be determined using the ideal gas equation, PV = nRT, where n denotes the number of moles, R denotes the ideal gas constant, T is the temperature in Kelvin, and P and V denote the pressure and volume of the tire, respectively. We can determine the mass of nitrogen using its molar mass (28 g/mol).

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Answer:

Explanation:

To determine the 3Ø rating of the transformer bank required to supply a 3Ø, 45 KVA load, we can use the following formula:

KVA = √3 x VL x IL

Where:

√3 = 1.732 (the square root of 3)

VL = line voltage

IL = line current

For the load:

KVA = 45 KVA (given)

For the transformer bank:

VL1 = 480 V (primary line voltage)

VL2 = 120/208 V (secondary line voltage)

IL2 = KVA / (√3 x VL2) = 45 KVA / (1.732 x 208 V) = 122.5 A (secondary line current)

Since the transformer bank is a 3Ø transformer, the KVA rating of the transformer bank can be calculated using the following formula:

KVA = √3 x VL1 x IL1

Where:

VL1 = primary line voltage

IL1 = primary line current

To solve for IL1:

IL2 = IL1 / √3 (assuming balanced loads and ideal transformer)

IL1 = IL2 x √3 = 122.5 A x √3 = 212.1 A (primary line current)

Therefore, the required 3Ø rating of the transformer bank is:

KVA = √3 x VL1 x IL1 = 1.732 x 480 V x 212.1 A ≈ 174 KVA

So, the transformer bank must have a 3Ø rating of at least 174 KVA to supply a 3Ø, 45 KVA load with a 3Ø, 480 Δ to 120/208-volt "Y" transformer bank.

The scale that is based on the tenths and hundredths of an inch is c. Decimal.

The decimal scale is a measurement system that is based on tenths and hundredths of an inch. It is a precise measurement system commonly used in various fields such as engineering, manufacturing, and construction.

In the decimal scale, one inch is divided into 10 equal parts, with each part being referred to as a "tenth". Each tenth is further divided into 10 equal parts, resulting in 100 equal parts, with each part being referred to as a "hundredth". Therefore, one inch is equal to 10 tenths or 100 hundredths.

For example, a measurement of 2.45 inches would mean 2 whole inches and 45 hundredths of an inch. This can be represented as 2.45 on the decimal scale. The decimal scale is used for precision measurements where accuracy is important, and it is commonly used in applications such as machining, metrology, and inspection.

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Answer:

True

Explanation:

Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true