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iragen [17]
3 years ago
13

A minor road intersects a major 4-lane divided road with a design speed of 50 mph and a median width of 12 ft. The intersection

is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 5%.
Engineering
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

minimum sight distance = 699 ft

Explanation:

given data

road lane = 4 divided road

median width = 12 ft

grade road = 5%

solution

we take here time gap factor for minor road vehicle when enter to major road  from table

time gap = 8.1  sec

and for median width of 12 ft

time gap = 8.2 + 0.7 ( 1 + \frac{12}{12}  )  

time gap = 9.5 second

so minimum sight distance will be

minimum sight distance = 1.47 × design speed × time gap  

minimum sight distance = 1.47 × 50  × 9.5

minimum sight distance = 699 ft

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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:

a)

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

b)

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

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3 years ago
# 17
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Answer:

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Answer and Explanation:

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theta(n) =-(360/(2*pi))*atan(b(n)./a(n));  

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Answer:

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How can you do this 5.2.4: Rating?
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Answer:

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Explanation:

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