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nirvana33 [79]
3 years ago
14

Stream Piracy – Kaaterskill, NY. Check and double-click the Problem 15 folder. The dark blue and orange streams highlight the pr

ocess of stream piracy/stream capture (i.e., they are pirating/capturing the headwaters of the cyan and magenta streams, respectively). Which of the following helps promote this?
a. The pirating streams are larger and can hold more water than the other streams.
b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.
c. The land to the west is being uplifted.
d. Landslides have diverted the water into the stream pirates.
Engineering
1 answer:
baherus [9]3 years ago
5 0

Answer:

b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.

Explanation:

From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.

The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.

However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.

The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.

It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.

As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.

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Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons
mixas84 [53]

Answer:

work done = 48.88 × 10^{9} J

Explanation:

given data

mass = 100 kN

velocity =  310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time  - force ×velocity ×time  - mass ×descends )  10³ J

work done = ( 100 × 310 × 1800  - 12×310 ×1800  - 100 × 2200 )  10³ J

work done = 48.88 × 10^{9} J

6 0
3 years ago
A flow rate sensing device used on a liquid transport pipeline functions as follows. The device provides a 5-bit output where al
marysya [2.9K]

Answer:

Explanation:

The step by step analysis is as shown in the attached files.

8 0
3 years ago
An airline ticket counter forecasts that 220 people per hour will need to check in. It takes an average of 2 minutes to service
Nadusha1986 [10]

A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents

B) The number of agents required on duty to reduce cost = 9 agents

<u>Given data : </u>

Arrival rate of customers ( β ) = 220 per hour

Service rate ( mu ) = 60 minutes / 2 minutes = 30 customer per hour

utilization ( rho ) = 220 / 30 ≈ 7

at least 8 server personnel are required for stability of the queue

A<u>) Determine the number of agents required to achieve a wait time of 10 minutes or less per customer</u>

waiting time = 10 - 2 = 8 minutes

number of customers waiting ( ∝ ) = 7 and required server = 8

assuming   Lq = 5.2266

Hence the waiting time in line = Lq / arrival rate

                                                  = 5.2266 / 220 = 0.0238 hour

                                                  = 0.0238 * 60 = 1.428 minutes

Since the waiting time ( 1.428 minutes ) is less than the original waiting time ( 2 minutes ) the number of agents that will achieve a wait time of 10 minutes or less is = 8 agents

<u>B) Determine the number of</u><u> ticket agents</u><u> that should be on duty to minimize cost </u>

salary of ticket agent = £12 per hour

cost of customer waiting in queue = £5 per hour per customer

<em> </em><u>i) When 8 agents are used </u>

waiting time of customers = 0.0238 * 220 = 5.236

waiting cost for customers = 5.236 * 5 = £26.18

employee cost = 8 * 12 = £96

∴ Total cost = 96 + 26.18

                    = £ 122.18

<u>ii) When 9 agents are used </u>

waiting time for customers = 0.0074 * 220 = 1.628

Wq = 1.6367 / 220 = 0.0074

waiting cost for customers = 1.6367 * 5 = £ 8.1835

assuming Lq = 1.6367

employee cost = 9 * 12 = £ 108

∴ Total cost = 108 + 8.1835 = £ 116.18

From the calculations in ( i ) and ( ii ) the Ideal number of ticket agents that should be on duty to minimize cost should be 9 agents.

Hence we can conclude that A) Number of agents required to achieve a wait time of 10 minutes or less = 8 agents and The number of agents required on duty to reduce cost = 9 agents.

Learn more about cost reduction : brainly.com/question/14115944

6 0
3 years ago
Engineers will redesign their products when they find flaws. TRUE O False​
nataly862011 [7]

Answer:

true

Explanation:

6 0
3 years ago
A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a
OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS        A                B                    C

Benzene 1             4.72583     1660.652        -1.461

Toluene  2            4.07827     1343.943         -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5%    = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis     for all components

So for Benzene ; p × y1=x1 × p1s   ------let this be equation 1

for Toluene ; p × y2=x2 × p2s   ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0
2 years ago
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