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ValentinkaMS [17]
3 years ago
8

What type of foundation do engineers use for a small and light building and when the load of the building is borne by columns? A

. individual footings B. strip footings C. mat foundations D. end-bearing piles E. friction piles
Engineering
1 answer:
ikadub [295]3 years ago
8 0

Answer:

A.

Explanation:

Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected

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To make 1000 containers of ice cream you need: 600 gallons of milk, 275 gallons of cream, and 120 gallons of flavor. Each ingred
Kamila [148]

Answer:

For the cream, 32 gallons should be reduced and 12 gallons should be decreased for flavor.

Explanation:

To prepare a total of 1000 gallons of ice cream you need 600 gallons of milk, 275 gallons of cream and 120 gallons of flavor, therefore we must calculate the percentages of each ingredient, as follows:

%milk=(600/1000)x100=60%

%cream=(275/1000)x100=27.5%

%flavor=(120/1000)x100=12%

If you reduce the amount of milk by 10% you have:

Milk quantity=600 gallons-(600x0.1)=540 gallons

To maintain the same percentages of each ingredient, you must make a rule of three to know the amount of cream and flavor that would need to be used with the 540 gallons of milk. The rule of three is as follows:

540 gallons of milk------------------60%

x gallons of cream--------------------27.5%

Clearing the x:

x gallons of cream=(540x27.5)/60=243 gallons

In the same way for flavor:

540 gallons of milk------------------60%

x gallons of flavor--------------------12%

Clearing the x:

x gallons of flavor=(540x12)/60=108 gallons

Verifying that they meet the percentages that were calculated before:

Total amount of ice cream=540+243+108=891 gallons

Calculate the percentages of each ingredient:

%milk=(540/891)x100=60.6%

%cream=(243/891)x100=27.3%

%flavor=(108/891)x100=12.1%

As can be seen, it is found that approximately the same percentages calculated above are met. Therefore, we can already calculate the amount by which the cream should be reduced and the flavor.

For the cream:

Gallons of cream=275-243=32 gallons

For the flavor:

Gallons of flavor=120-108=12 gallons

3 0
3 years ago
The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt
Step2247 [10]

Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

Gs = 759.99W/m² ------- Approximated

3 0
3 years ago
1. If a Gear with a 3 inch Diameter is being turned by a Gear with a 6 inch Diameter, which Gear will rotate at a higher Rate?
pshichka [43]

Answer:

The smaller gear will rotate faster.

Explanation:

If a larger gear is driven by a smaller gear, the large gear will rotate slower than the smaller gear but will have a greater moment. For example, a low gear on a bike or car. If a smaller gear is driven by a larger gear, the smaller gear will rotate quicker than the larger gear but will have a smaller moment.

I hope this helps! :)

5 0
2 years ago
A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a
nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, g_{c} = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

c_{p}= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence v_{a} ^{2} = 0

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} +  \frac{v_{1} ^{2} }{2c_{p} }

T_{1} = 20+460 = 480°R

T_{a}  =480+  \frac{(800)(800}{2(0.240)(25037}= 533.25°R

Pressure at the inlet of compressor at isentropic condition

P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}

P_{a} = (10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}= 14.45 psia

P_{2}= 8P_{a} = 8(14.45) = 115.6 psia

4 0
3 years ago
Read 2 more answers
Which of the following is NOT a line used on blueprints?
jonny [76]

Answer: Photo lines

Explanation: made more sense

4 0
3 years ago
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