All categories

3 years ago

Explain why slopping roofs are used in areas with heavy rain or snow fall

Engineering

1 answer:

Savatey [412]3 years ago

8 0

Answer:

Slopping or slanting roofs are found in houses where there are snow or rain, because it helps the water or the snow to skid down easily. It minimizes the chances of waterlogging because water cannot rest on the roof surface for long.

Explanation:

Brainliest pls.

You might be interested in

Why is it important to review your plan when you write an algorithm?

user100 [1]

An algorithm is itself a general step-by-step solution of your problem. ... The most important point here is that you must use algorithms to solve problem, one way or the other. Most of the time it's better to think about your problem before you jump to coding - this phase is often called design.

7 0

3 years ago

The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

miss Akunina [59]

Answer:

a) Thermal conductivity of skin: $k_{skin}=1.5W/mK$

b) Temperature of interface: $T_{interface}=35.6\°C$

Heat flux through skin: $\frac{Q}{A}=2100W/m^2$

Explanation:

$k=\frac{QL}{A(T_{2}-T_{1})}$

Where: $k$ is thermal conductivity of a material, $\frac{Q}{A}$ is heat flux through a material, $L$ is the thickness of the material, $T_{1}$ is the temperature on the first side and $T_{2}$ is the temperature on the second side

$k_{skin}=\frac{QL}{A(T_{2}-T_{1})}$

$k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}$

$k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}$

$k_{skin}=1.5W/mK$

$k_{insulation}=\frac{k_{skin}}{2}$

$k_{insulation}=\frac{1.5}{2}$

$k_{insulation}=0.75W/mK$

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

$\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}$

$\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}$

$\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}$

$55500-1500T_{interface}=375T_{interface}-11250$

$1875T_{interface}=66750$

$T_{interface}=35.6\°C$

$\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}$

$\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}$

$\frac{Q}{A}=2100W/m^2$

3 0

3 years ago

One liter of oven-dry soil sampled from the Chorro Creek Ranch field requires 300 g of water to completely saturate it.

torisob [31]

Answer:

porosity=23.07%

x=5.974*10^9cm^3

Explanation:

One liter of oven-dry soil sampled from the Chorro Creek Ranch field requires 300 g of water to completely saturate it.

Calculate (a) its porosity, n, and (b) the volume of water required to saturate the top 20 cm of 1 hectare of the field.

-Porosity of a soil sample can be calculated in five ways but i shall use the formula below

porosity= $\frac{volume of pore}{volume of total} *100%$ %

volume pore volume of sample is equal to water volume used to saturate the soil

total volume=sample volume +pore volume

1L=1000cm^3

density of water=density=mass/volume

1g/cm^3=300g/volume

vol=300cm^3

total volume=300+1000

1300

porosity =(300/1300)*100%

porosity=23.07%

2. Recall that volume=area *height

1 hectare=1*10^9cm^2

the volume of soil sample=1*10^9cm^2*20cm

2*10^10cm^3

$porosity=\frac{x}{x+2*10^{10} } *100%$ %

23%=x/(x+2*10^10)*100%

x=5.974*10^9cm^3

8 0

3 years ago

A cylindrical tank is 50 inches long, has a diameter of 16 inches and contains 1.65 lbm of water. Calculate the density of water

CaHeK987 [17]

Answer:

0.285 lbm per cubic feet

Explanation:

length of tank = 50 inches = 50/12 feet = 4.17 feet

diameter of tank = 16 inches = 16/12 feet = 1.33 feet

weight of water = 1.65 lbm

density of water = ?

We know that the density of a substance is given as

ρ = w/v

where ρ is the density in Ibm per cubic feet

w is the weight in lbm

v is the volume in cubic feet

Volume of a cylinder = $\frac{\pi d^{2} l}{4}$

where d is the diameter

l is the length

volume = $\frac{3.142*1.33^{2}* 4.17}{4}$ = 5.79 cubic feet

Therefore, the density of water will be

ρ = w/v = 1.65/5.79 = 0.285 lbm per cubic feet

8 0

3 years ago

The Eads Bridge, which crosses the Mississippi River near St Louis, Missouri, was one of the first all steel bridges built in th

MrMuchimi

Answer: At 520 feet between the piers, the center arch of Eads Bridge was the longest rigid span ever built at the time of its construction (only a few suspension bridges had longer spans).

Explanation:

5 0

3 years ago