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2 answers:
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Answer:
Answer with Explanation is in the following attachments.
Explanation:
Answer:
maximum speed for safe vehicle operation = 55mph
Explanation:
Given data :
radius ( R ) = 678 ft
old building located ( m )= 30 ft
super elevation = 0.06
<u>Determine the maximum speed for safe vehicle operation </u>
firstly calculate the stopping sight distance
m = R ( 1 - cos ) ---- ( 1 )
R = 678
m ( horizontal sightline ) = 30 ft
back to equation 1
30 = 678 ( 1 - cos (28.655 *s / 678 ) )
( 1 - cos (28.655 *s / 678 ) ) = 30 / 678 = 0.044
cos = 1.044
hence ; 28.65 * s = 678 * 0.2956
s = 6.99 ≈ 7 ft
next we will calculate the design speed ( u ) using the formula below
S = 1.47 ut + ---- ( 2 )
t = reaction time, a = vehicle acceleration, G1 = grade percentage
assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0
back to equation 2
6.99 = 1.47 * u * 2.5 +
3.675 u + 0.0958 u^2 - 6.99 = 0
u ( 3.675 + 0.0958 u ) = 6.99
Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S =
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M