The structure of a house is such that it loses heat at a rate of 5400 kJ/h per degree Cdifference between the indoors and outdoors. A heat pump that requires a power input of 6 kW isused to maintain this house at 21 C. Determine the lowest outdoor temperature for which the heatpump can meet the heating requirements of this house

Answer:

pay off the parking tickets

Explanation:

In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.

Answer:

C) Decreases with an increase in temperature

Explanation:

As the temperature of a transistor increases, the thermal runaway property of the transistor becomes more significant and the transistors, conducting more freely as a result of the rise in temperature, causes an increase in the collector current or leakage current. The transistor base-emitter voltage decreases as a result.

With increased heating due to heavy current flow, the transistor is damaged.

Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

Cp = 1007 j/kg⁰c

k = 0.02735 W/m⁰c

Pr = 0.7228

v  = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 *  7 = 0.24 kg/s

Answer:

  resistance

Explanation:

A strain gauge changes resistance with applied strain.

Answer:

B) gate-source junction is reverse-biased

Explanation:

FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow"   in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".

In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".

Answer:

Correct Answer:

A. water pump

Explanation:

Timing belt in a vehicle helps to ensure that crankshaft, pistons and valves operate together in proper sequence. Timing belts are lighter, quieter and more efficient than chains that was previously used in vehicles.

Most car manufacturers recommended that, when replacing timing belt, tension assembly, water pump, camshaft oil seal should also be replaced with it at same time.

Answer:

Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.

The uncertainty has always spawned a certain fear inside creators. The fear of creating something no one will enjoy. Spending hundreds of dollars and hours building something which might not bring back any real tangible results. The fear of losing our investment to a poor concept is daunting but not random. But simple app ideas are actually pretty easy to come by.

Great app idea generation is not a gift given to a selected few, instead, it is a process by which any of us are able to carefully explore step by step methods to find our own solution to any problem. Whether you are a seasoned creator or a novice, we have provided a few recommendations to challenge and aid you as you create your next masterpiece.

if I am right then make me brainliest

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

Answer:

As the asteroid falls closer to the Earth's surface its Gravitational Potential energy decreases and its Kinetic energy increases.

Answer: answer provided in the explanation section.

Explanation:

Weather phenomenons that would impart Aviation Operations in Santa Barbara -

1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.

2.  The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.  

3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.

4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.

This and more are some factors to look into when considering wheather conditions that would affect aviation operations.

I hope this was a bit helpful. cheers

Answer:

thread depth = 2.5 mm

thread width = 2.5 mm

mean diameter = 27.5 mm

root diameter = 25 mm

lead of screw = 5 mm

Explanation:

given data

power screw diameter D = 30 mm

thread pitch  P = 5 mm

solution

First, we get here thread depth fr square thread

thread depth = [tex]\frac{P}{2}[/tex]   ......................1

thread depth = [tex]\frac{5}{2}[/tex]

thread depth = 2.5 mm

and

thread width for square thread

thread width = [tex]\frac{P}{2}[/tex]   ......................2

thread width = [tex]\frac{5}{2}[/tex]

thread width = 2.5 mm

and

mean diameter is

mean diameter = D - [tex]\frac{P}{2}[/tex]    ................3

mean diameter = 30 - [tex]\frac{5}{2}[/tex]

mean diameter = 27.5 mm

and

root diameter is

root diameter = D - P   ....................4

root diameter = 30 - 5

root diameter = 25 mm

and

lead of screw for single thread so n = 1

so lead of screw = 1 × 5

lead of screw = 5 mm

Answer:

A) Flow depth = 2.46 m, Froude number after jump = 0.464

B) head loss = 0.572 m

C) dissipation ratio = 0.173

Explanation:

Given data :

Velocity before jump ( v1 ) = 7 m/s

flow depth before jump ( y1 ) = 0.8 m

g = 9.81 m/s

Esi = 3.3 m ( calculated )

attached below is a detailed solution of the problem

Answer:

IV > [tex]I^{2} R[/tex]

Explanation:

The current in the power line = I

The voltage in the power line = V

The resistance of the power line = R

Power supplied from the power house = P

power delivered to the households = [tex]p[/tex]

We know that the power supplied to a power line system is proportional to

P = IV    ....1

we also know that according to Ohm's law, the relationship between the voltage, resistance, and current through an electrical system is given as

V = IR    ....2

substituting equation 2 into equation 1, the power delivered to the households is proportional to the square of the current.

[tex]p[/tex] = [tex]I^{2} R[/tex]    ....3

The problem is that when power is delivered across a transmission line, some of the power is loss due to Joules heating effect of the power lines. This energy and power loss is proportional to [tex]I^{2}[/tex] therefore, the electrical power delivered to the households will be less than the electrical power supplied from the power station. This means that

P > [tex]p[/tex]

equating these two powers from equations 1 and equation 3, we have

IV > [tex]I^{2} R[/tex]

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = [tex]\frac{\pi }{4} * D^2 * l[/tex] = [tex]\frac{\pi }{4} * 5.5^2 * 5[/tex] = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

Answer:

30 feet

Explanation:

Given data :

design speed = 30 miles/h

super elevation = 0.08

determine the width of the turning roadway

calculate the value of R = V^2 / 15( e + p)

e = 0.08 , p = 0.2 , v = 30

R = (30)^2 / 15 ( 0.08 + 0.2 )

  = 900 / 15 ( 0.28 )

  ≈ 215 ft

pavement width from the calculation above = 28 ft

width of the turning roadway = pavement width + 2 = 30 feet ( because there are two vertical widths joining up the main road at the T junction )

Answer:

a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min

b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²

first we find the volume and Area;

Volume V = πD²t / 4

Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³

Area A = 2πD²/ 4 + πDt

Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}

Area A = 392,699.08 + 31,415.93

Area A = 424,115 mm²

a)

Chvorinov’s rule

T(aluminium) = Cm (V/A) ²

T(aluminium) =  2.0 × (3,926,991 / 424,115) ²

T(aluminium) = 171.5 s = 2.86 min

∴ the minimum time (minutes) for the casting to solidify is 2.86 min

b)

For cast iron

Cm (mold constant = 1.488 sec/mm²)

Chvorinov’s rule

T(iron) = Cm (V/A) ²

T(iron) = 1.488 × (3,926,991 / 424,115) ²

T(iron) = 127.5720s = 2.13 min

Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Answer:

[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy at constant pressure:

[tex]ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}[/tex]

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

[tex]\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,[/tex]

We obtain the expression to compute the specific entropy change:

[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]

Best regards.

Answer:

The rate of heat transfer has increased.

Explanation:

Heat transfer rate is the rate at which heat energy is dissipated to the ambient from a hot body. The rate of heat transfer is proportional to the available surface area for heat exchange. This means that the greater the exposed surface area for heat exchange, the greater the rate at which heat is lost to the ambient. In introducing the fins to the heat exchange system (fins have a large surface area to volume ratio for maximum exposure to the ambient), one maximizes the available surface area for heat exchange between the material and the ambient, increasing the rate of heat transfer.

Answer:

Explanation:

MOSFET: The acronym for  "metal oxide semiconductor field-effect transistor"  are mostly used in switching circuits.

There are two classes of MOSFET

1. Depletion mode

2. Enhancement mode

          Generally a  MOSFET is a kind of transistor, it is actually a field effect transistor with tree terminals gate, source and drain terminals, also the MOSFET can be used as an amplifier for the amplification of electronic signals in the electronic circuit/devices

Answer:

Explanation:

The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.

At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.

Then link CE has 15 N tension.

The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.

In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.

At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.

At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.

The horizontal forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.

In summary, the link forces are ...

_____

Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.

Answer:

a) Mt = 0.0023229

b) = U1 = 214.07

c) = V₁  = 0.861 m³/kg

d) = Vr1 = 621.2

Explanation:

Given that

R = 0.287 KJ/kg.K, T1 = 300 K , P1 = 100 kPa , V1 = 500 cm³, r = 10 , Q = 60 kW , Speed N = 5600 RPM, Number of cylinders K = 4

specific heat at constant volume Cv = 0.7174 kJ/kg.K

Specific heat at constant pressure is 1.0045 Kj/kg.K

a)  To determine the total mass (kg) of air in the engine.

we say

P1V1 = mRT1

we the figures substitute

(100 x 10³) ( 500 x  10⁻⁶) = m ( 0.287 x  10³) ( 300 )

50 = m x 86100

m = 0.00005 / 86100 = 0.0005807 ( mass of one cylinder)

Total mass of 4 cylinder

Mt = m x k

Mt = 0.0005807 x 4

Mt = 0.0023229

b) To determine the specific internal energy (kJ/kg) at state 1

i.e at T1 = 300

we obtain the value of specific internal energy U1 at 300 K ( state 1) from the table ideal gas properties of air.

U1 = 214.07

c) To determine the specific volume (m³/kg) at state 1.

we say

V₁ = V1/m

V₁ = (500 x  10⁻⁶) / 0.0005807

V₁  = 0.861 m³/kg

d) To determine the relative specific volume at state 1.

To obtain the value of relative specific volume at 300 K ( i.e state 1) from the table ideal gas properties of air.

At T1 = 300 k

Vr1 = 621.2

Answer: A prototype

Explanation:

The scale model that proves the initial concept is called a domain model.

A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.

A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.

A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.

Therefore, a domain model is the scale model that proves the initial concept.

To learn more about the scale model, refer to the below link:

https://brainly.com/question/14341149

#SPJ2

Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)

Answer:

Average heat transfer coefficient =  31 kw/m^2 k

Heat transfer rate per meter length of pipe =  116.808 KW

Explanation:

water temperature = 20⁰c,  

free-stream velocity = 1.5 m/s

circular pipe diameter = 2.0 cm = 0.02 m

surface temperature = 80⁰c

A) calculate average heat transfer coefficient

we apply the formula below :

m = αAv

A (area) = [tex]\pi /4 (d)^2[/tex]

m = 10^3 * [tex]\pi / 4 ( 0.02)^2[/tex] * 1.5

   = 10^3 * 0.7857( 0.0004) * 1.5

   = 0.4714 kg/s

Average heat transfer coefficient  

h = [tex]\frac{m(cp)}{A}[/tex]  ,  A = [tex]\pi DL[/tex]

L = 1 m , m = 0.4714 kgs , cp = 4.18

back to equation

h = [tex]\frac{0.4714*4.18}{\pi * 0.02 }[/tex]   = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k

B) Heat transfer rate per meter length of pipe

Q = ha( ΔT ),  a = [tex]\pi DL[/tex]

   = 31 * 0.0628 * ( 80 - 20 )

  = 31 * 0.0628 * 60 = 116.808 KW

Answer:

A) 251.8 kj/kg

B) 0.9150 kj/kg-k

C) 155.4 kj/kg

F) 1.50

G) 3.95 kw

H) 2.6 kw

Explanation:

Given conditions :

air conditioner : R -134a

compressor efficiency (nc) = 90%.

T1 = 9⁰c,  T3 = 39⁰c, mass flow rate = 0.027 kg/s

A) Specific enthalpy at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

from the R-134a property table

h1 = 251.8 kj/kg

B ) specific entropy ( kj/kg-k) at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

s = 0.9150 kj/kg-k ( from the R-134a property table )

C) specific enthalpy at the compressor exit

at T3 = 39⁰c , s2 = s1

has = 165.12 kj/kg

h2 = 155.4 kj/kg

attached below is the remaining solution to some of the problems

Answer:

Move the buret clamp to a ring stand with a larger base.

Explanation:

A right stand is used for titration experiments in the laboratory. It holds the burette firmly during experiments so that accurate readings can be taken.

The right stand is made up of support base, vertical stainless steel, clamp with adjustable screw that holds on to the vertical rod.

The clamp is used to hold the burette in place.

If after clamping a buret to a ring stand, you notice that the set-up is tippy and unstable, the best action will be to move the buret clamp to a ring stand with a larger base.

The larger base provides a better center of gravity and stabilises the setup

Answer:

Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.

Explanation:

In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about  1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.

There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.

Answer: hello attached below is the diagram which is part of your question

Total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k  it violates Clausius increase of entropy which is Sgen > 0

Explanation:

Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .  

applying the increase in entropy principle to prove this

temp of cold reservoir (t hot)= 600 k

temp of hot reservoir(t cold) = 1220 k

energy (q) = 100 kj

total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k

entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k

entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k

hence it violates  Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0

Answer:

New electric field = 18 N/C

Explanation:

Given:

Length (E1) = 2 cm

New length (E2) = 4 cm

Electric field =  36 N/C

Find:

New electric field

Computation:

New electric field = 36 [2 / 4]

New electric field = 36 [1/2]

New electric field = 18 N/C