Answer:
distance = 22.57 ft
superelevation rate = 2%
Explanation:
given data
radius = 2,300-ft
lanes width = 12-ft
no of lane = 2
design speed = 65-mph
solution
we get here sufficient sight distance SSD that is express as
SSD = 1.47 ut +
..............1
here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here
so put here value and we get
SSD = 1.47 × 65 ×2.5 +
solve it we get
SSD = 644 ft
so here minimum distance clear from the inside edge of the inside lane is
Ms = Rv ( 1 -
) .....................2
here Rv is = R - one lane width
Rv = 2300 - 6 = 2294 ft
put value in equation 2 we get
Ms = 2294 ( 1 -
)
solve it we get
Ms = 22.57 ft
and
superelevation rate for the curve will be here as
R =
..................3
here f is coefficient of friction that is 0.10
put here value and we get e
2300 = 
solve it we get
e = 2%
Answer: True.
Explanation: Coolant is flammable and toxic.
Answer:
Ponding will occur in 40mins
Explanation:
We say that the infiltration rate is the velocity or speed at which water enters into the soil. This often times is measured by the depth (in mm) of the water layer that can enter the soil in one hour. An infiltration rate of 15 mm/hour means that a water layer of 15 mm on the soil surface, will take one hour to infiltrate.
Consider checking attachment for the step by step solution.
Answer:
www stands for world wide web
Explanation:
It will really help you thank you.
Answer:
14.506°C
Explanation:
Given data :
flow rate of water been cooled = 0.011 m^3/s
inlet temp = 30°C + 273 = 303 k
cooling medium temperature = 6°C + 273 = 279 k
flow rate of cooling medium = 0.02 m^3/s
Determine the outlet temperature
we can determine the outlet temperature by applying the relation below
Heat gained by cooling medium = Heat lost by water
= ( Mcp ( To - 6 ) = Mcp ( 30 - To )
since the properties of water and the cooling medium ( water ) is the same
= 0.02 ( To - 6 ) = 0.011 ( 30 - To )
= 1.82 ( To - 6 ) = 30 - To
hence To ( outlet temperature ) = 14.506°C