Match the scenario to the related government program.

Using Java..

uysha [10]

Answer:

The source code files for this question have been attached to this response.

Please download it and go through each of the class files.

The codes contain explanatory comments explaining important segments of the codes, kindly go through these comments.

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7 0

2 years ago

As the asteroid falls closer to the Earth's surface its _______ energy decreases and its _______ energy increases.

Levart [38]

Answer:

As the asteroid falls closer to the Earth's surface its Gravitational Potential energy decreases and its Kinetic energy increases.

4 0

3 years ago

6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right

dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0

3 years ago

The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i

koban [17]

Answer: (a) 9.00 Mega Newtons or 9.00 * 10^6 N

(b) 17.1 m

Explanation: The length of wall under the surface can be given by

$b=25m/sin(60)\\=28.867$

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

$F(resultant) = Pavg ( A) = (Patm + \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N$

Noting from the Bernoulli equation that

$Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\$

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

$Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m$

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0

2 years ago

In the engineering design process, testing is the least important step. True False

Wittaler [7]

Answer:

false

Explanation:

7 0

2 years ago

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