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kolezko [41]
3 years ago
8

Since the beginning of the 21st century, Bogota has successfully developed what two alternatives to the automobile as methods of

moving people around the city? a. Ride Sharing Apps and Subway Lines b. Subway Lines and Bus Rapid Transit c. Bicycle Lanes and Ride Sharing Apps d. Bus Rapid Transit and Bicycle Lanes
Geography
2 answers:
malfutka [58]3 years ago
6 0

Answer:

D. Bus fast lanes and bicycle lanes

Explanation:

Bogota's modern transportation system is good planning and highly organized, making travel to and in the capital city of Colombia easy and efficient. The main and most productive form of mass transportation in the city is to have two bus systems namely the traditional system and TransMilenio. The traditional system uses a combination of large city buses, mini buses and mini vans to transport passengers through city streets. TransMilenio is a fast transit system, articulated by modern buses that operate on the bus road only with smaller feeder buses carrying passengers from residential areas to the main network. TransMilenio, which is still being expanded, is expected to cover the entire city by 2030, and by that time it will completely replace the traditional bus system.

70 percent of residents in Bogota use public transport. Since the Trans Millenium, in 2005 the use of private vehicles has dropped from 14% to 8%. So the correct answer is D. Bus Rapid Transit and Bicycle Lanes

Learn More

How to calculate length on a map brainly.com/question/8072603

Details

- Grade : Senior High

- Subject : Geography

Keywords

- Bogota

- Bus Rapid Transit

Anika [276]3 years ago
5 0

Hi !!

CORRECT ANSWER = D

D- Bus rapid Transit and Bicycle lanes is the correct answer

D- BRT y cyclovia es la respuesta correcta

☺☺☺

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a rural kansas watershed that is ungauged has an area of 475 acres and a main channel length of 6870 feet with an average slope
dlinn [17]

Answer:

\mathbf{Q_p  =682 \ \  ft^3/s}

Explanation:

Given that:

Area = 475 acres

The length of the channel (L) = 6870 feet

The average water shield slope (S) = 100 feet/mile

Since; 1 mile = 5280 feet

Burst duration D = 15 min

∴

100 feet/mile = 100/5280

The average water shield slope (S) = 5/264

Using hydrograph method:

The time of concentration t_c = 0.0078L^{0.77} S^{-0.385}

where;

L = 6870

S = 5/264

t_c = 0.0078(6870)^{0.77} (\dfrac{5}{264})^{-0.385}

t_c =32.34 min

Since 60 min = 1 hour

32.34 min will be (32.34*1)/60

= 0.539 hour

Lag time T_l = 0.67\times t_c

T_l = 0.67\times 32.34

T_l = 21.6678\ min

The time to peak i.e

T_p = \dfrac{D}{2}+ T_L \\ \\  T_p = \dfrac{15}{2}+ 21.6678 \\ \\  T_p = 29.168 \ min

T_r = \dfrac{T_p}{5.5} \\ \\  T_r = \dfrac{29.1678}{5.5} \\ \\ T_r = 5.30 \ min

Since D = 15 min is not equal to T_r, then we hydrograph apart from T_r duration lag time.

Then;

T_p \ ' = T_p + \dfrac{D-t_r}{4} \\ \\ T_p \ ' = 29.168 + \dfrac{15-5.30}{4} \\ \\ T_p \ ' = 31.593

Now, we need to determine the peak discharge Q_p by using the formula:

Q_p  = \dfrac{484 \times A}{T_p \ '}

where

484 = peak factor

Recall that A = 475 acres, to miles, we have:

A = 0.7422 mile²

T_p \ ' = 31.593/60

∴

Q_p  = \dfrac{484 \times 0.7422}{\dfrac{31.593}{60}}

\mathbf{Q_p  =682 \ \  ft^3/s}

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Answer:

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