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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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In a river reach, the rate of inflow at any time is 350 cfs and the rate of outflow is 285 cfs. After 90 min, the inflow and out
Semenov [28]

Answer:

change in storage =  -310,500 ft^3

intital storage= 3.67 acre ft

Explanation:

Given data:

Rate of inflow = 350 cfs

Rate of outflow = 285 cfs

After 90 min,  rate of inflow = 250 cfs

Rate of outflow = 200 cfs

final storage = 10.8 acre-ft

calculating the average inflow and outflow

average inflow  = \frac{(350+250)}{2} = 300 cfs

average outlow  = \frac{(285+200)}{2} = 242.5 cfs

total amount of water drain during the period of one hour

= (average outflow - average inflow) *60*90

= (242.5 - 300)*60*90 = -310,500

change in storage is calculate as

= -310,500 ft^3

in cubic meter

= -310500/35.315 = 8792.30 cm^3

in acre-ft

= -310,500/43560 = 7.13 acre ft

initial storage = 10.8 - 7.13 = 3.67 acre ft

3 0
3 years ago
What is the primary responsibility of ABET?
nalin [4]

Answer:

C

Explanation:

The ABET (Accreditation Board for Engineering and Technology) is a non-governmental organization that accredits programs in applied science, computing, engineering, and engineering technology, both in the United States and elsewhere.

Give Brainliest pls

5 0
3 years ago
1. A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with S = 1.5. The manometer readi
julia-pushkina [17]

Answer:

1) The pressure difference is 4.207 kilopascals.

2) 2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

Explanation:

1) We can calculate the gas pressure difference from the U-tube manometer by using the following hydrostatic formula:

\Delta P = \frac{S\cdot \rho_{w}\cdot g \cdot \Delta h}{1000} (Eq. 1)

Where:

S - Relative density, dimensionless.

\rho_{w} - Density of water, measured in kilograms per cubic meter.

g - Gravitational acceleration, measured in meters per square second.

\Delta h - Height difference in the U-tube manometer, measured in meters.

\Delta P - Gas pressure difference, measured in kilopascals.

If we know that S = 1.5, \rho_{w} = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and \Delta h = 0.286\,m, then the pressure difference is:

\Delta P = \frac{1.5\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.286\,m)}{1000}

\Delta P = 4.207\,kPa

The pressure difference is 4.207 kilopascals.

2) From Physics we remember that a pound per square unit equals 2.036 inches of mercury and 2.307 feet of water and we must multiply the given pressure by corresponding conversion unit: (p = 2.5\,psi)

p = 2.5\,psi\times 2.037\,\frac{in\,Hg}{psi}

p = 5.093\,in\,Hg

p = 2.5\,psi\times 2.307\,\frac{ft\,H_{2}O}{psi}

p = 5.768\,ft\,H_{2}O

2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

4 0
4 years ago
What is not required for current to flow through a conductor
musickatia [10]

Answer: Something that has air molecules.

Explanation:

5 0
3 years ago
What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-
Hitman42 [59]

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}

Where \sigma_{m} is the magnitude of he maximum stress at the tip of the crack, \sigma_{p} is the magnitude of the tensile stress, l_{c} is 1/2 the length of the internal crack, and r_{c} is the radius of curvature of the crack.

We have:

r_{c}=1.9*10^{-4} [mm]

l_{c}=3.8*10^{-2} [mm]

\sigma_{c}=140 [MPa]

We replace:

\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}

We get:

\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]

5 0
3 years ago
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