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kompoz [17]
2 years ago
13

A railroad runs form city A to city B, a distance of 800km, through mountainous terrain. The present one-way travel time (includ

ing time at intermediate yards) is 20 hours, and the rail freight rate is $20 per ton. There is a truck service that competes with the railroad, running over a roughly parallel road for approximately the same distance, at an average speed of 48km per hour and a rate of $30 per ton. A new highway is planned to replace the existing roads; it is expected that most of the traffic will be trucks (auto usage is expected to be negligible). The performance function of the new facility is t_T = t_0 + bV_T, where V_T is the flow in trucks per hour, t_0 = 10 hours, b = 0.08 hour per truck per hour. The railroad's estimate of the demand function is: V_T/V_R = a_0(t_r/t_R)^a_1 (c_T/c_R)^a_2 where t_T and t_R are the trip times (in hours) by truck and rail, respectively, c_T and c_R are the corresponding rates, V_T and V_R are the corresponding flows and a_0, a_1 and a_2 are parameters. The total demand is likely to remain constant at V_TOT = 200 tons per hour. The rail system is utilized at only a fraction of capacity, so its performance function is flat (travel time is constant, independent of volume). If a_0 = 1, a_1 = -1 and a_2 = -2, find the present flows of freight by truck and rail. Make an estimate of the equilibrium flows if the new highway were built. With the new highway built, what would the equilibrium flow be in each of the following two cases: if the railroad dropped its rate to $15 per ton? if truckers were taxed $5 per ton to help pay for the new highway?

Engineering
1 answer:
Verizon [17]2 years ago
4 0

Answer:

i) VT = 52.16

VR = 147.85

ii) VT = 61

VR = 138.99

Explanation:

The step by step solution is been done, please check the attached file below to see it

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Read 2 more answers
A 3-in-thick slab is 10 in wide and 15 ft long. The thickness of the slab is reduced by 20% and width increases by 3% in a hot-r
xxMikexx [17]

Answer: l = 2142.8575 ft

v = 193.99 ft/min.

Explanation:

Given data:

Thickness of the slab = 3in

Length of the slab = 15ft

Width of the slab = 10in

Speed of the slab = 40ft/min

Solution:

a. After three phase

three phase = (0.2)(0.2)(0.2)(3.0)

= 0.024in.

wf = (1.03)(1.03)(1.03)(10.0)

= 10.927 in.

Using constant volume formula

= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf

Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)

= 6750 /0.2625

= 25714.28in = 2142.8575 ft

b.

vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)

= (0.12)(424.36)/0.2625

= 50.9232/0.2625

= 193.99 ft/min.

4 0
3 years ago
I don’t understand this
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Answer:

Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of 160cm^{2} by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.

Have an amazing day and I hope this can somewhat help :)

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