What type of plans have to do with earth, soil, excavation, and location<br> of a house on a lot?

Help please if you don't know don't give wrong answer please

mrs_skeptik [129]

4.75cm * 5.22cm is 24.795cm²

but let's do this with m² instead:

0.0475m * 0.0522m = 0.0024795

now we can compare it with the 49780 much easier.

devide. that's the <u>roughly</u> 20,000,000fold of the plot, but it's still squared, so let's take the root

so the representative fraction is 1:4,480.

the exact value is in the screens

have a nice day:)

6 0

3 years ago

Name the seven physical qualities for which standards have been developed.

SIZIF [17.4K]

Answer:

HUMAN DEVELOPMENT

MOTOR BEHAVIOR

EXERCISE SCIENCE

MEASUREMENT AND EVALUATION

HISTORY AND PHILOSOPHY

UNIQUE ATTRIBUTES OF LEARNERS

CURRICULUM THEORY AND DEVELOPMENT

Explanation:

6 0

3 years ago

A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a te

Artyom0805 [142]

Explanation:

a converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures

100% (3 ratings)

A_2 = 0.001 m^2 P_1 = 1 MPa, T_1 = 360 k P_2 = 500 kpa p^gamma - 1/gamma proportional T (1000/500)^1.4 - 1/1.4 = (360/T_2) 2^4/14 = 360/T_2 T_2

3 0

2 years ago

Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th

Sergeeva-Olga [200]

Answer:

HEAT LOST

polycarbonate = 252 W

soda lime glass = 1680 W

aerogel = 16.8 W

COST associated with heat loss

polycarbonate = $ 262.08

soda lime glass = $ 1,747.2

aerogel = $ 17.472

The cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

Explanation:

Given that;

surface area for each window = 0.4m * 0.4m = 0.16m^2

DeltaT = 90°C, L = 12mm = 0.012m

thermal conductivity of soda line can be gotten from tables in FUNDAMENTALS OF HEAT AND MASS TRANSFER

so at 300K

KsL = 1.4 W/mK

Kag = 0.014 W/mK

Kpc = 0.21 W/mK

Now HEAT LOSS

for polycarbonate;

Qpc = -KA dt/dx

NOTE ( heat flows from high temperature region to low temperature regions. so the second temperature would be smaller compared to the initial causing a negative in the change in temperature)

so Qag = (0.21 * 0.16 * 90) / 0.012

= 252 W

for soda lime glass;

Qsl = (1.4 * 0.16 * 90) / 0.012

= 1680 W

for aerogel

Qaq = (0.014 * 0.16 * 90) / 0.012

= 16.8 W

Now for COST associated with heat lost

for polycarbonate;

cost = Qpc * 130 * 8 * 1/1000

= 252 * 130 * 8 * 1/1000

= $ 262.08

for soda lime glass;

cost = 1680 * 130 * 8 * 1/1000

= $ 1,747.2

for aerogel

cost = 16.8 * 130 * 8 * 1/1000

= $ 17.472

Therefore the cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

6 0

3 years ago

The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air

Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

4 0

3 years ago

What type of plans have to do with earth, soil, excavation, and location of a house on a lot?