Answer: 2.93 ft/sec
Explanation: Calculate the volume/sec entering from the two inlets (Pipes 1 and 2), add them, and then calculate the flow in Pipe 3.
The table illustrates the approach. I calculated the volume of each pipe for a 1 foot section with the indicated diameters, divided by 2 for the radius of each using V = πr²h. Units of V are in^3/foot length. Now we can multiply that volume by the flow rate, in ft/sec, to obtain the flow rate in in^3/sec.
Add the two rates from Pipes 1 and 2 (62.14 in^3/sec) to arrive at the flow rate for Pipe 3 necessary to keep the water level constant. Calculate the volume of 1 foot of Pipe 3 (21.21 in^3/foot) and then divide this into the inflow sum of 62.14 in^3/sec to find the flow rate of Pipe 3 (in feet/sec) necessary to keep the water level constant.
That is 2.93 ft/sec.
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.
Answer: Technician A
Explanation: A coolant recovery system Is one of the most important part of a vehicles cooling system. When a coolant gets too hot, it forces it way out of the spring loaded radiator cap so as to relieve pressure. Any coolant which has escaped is recovered back through the discharge tube located in the recovery tank. The system is usually air free
Answer:288 pm
Explanation:
Number of atoms(s) for face centered unit cell -
Lattice points: at corners and face centers of unit cell.
For face centered cubic (FCC), z=4.
- whereas
For an FCC lattices √2a =4r =2d
Therefore d = a/√2a = 408pm/√2a= 288pm
I think with this step by step procedure the, the answer was clearly stated.
Answer:
Explanation:
The rank of the magnitude of the diffusion coefficient from greatest to least is as follows:
C in Fe at 900°C > Cr in Fe at 900°C > Cr in Fe at 600°C
Reason
C in Fe is an interstitial impurity while Cr in Fe is a substutional impurity.Therefore interstitial impurity occurs in C in Fe systems,while substutitional diffusion occurs in Cr in Fe system.Interstitial is much faster than substitutional diffusion hence the order
Also with increasing temperature magnitude of diffusion coefficient increases,due to the relation.
D = D₀exp(-Qd/RT)
Where D₀=Temperature independent per exponential
Qd= The activation energy for diffusion
R= Universal gas constant
T=absolute temperature
