Answer:
the elevation at point X is 2152.72 ft
Explanation:
given data
elev = 2156.77 ft
BS = 2.67 ft
FS = 6.72 ft
solution
first we get here height of instrument that is
H.I = elev + BS ..............1
put here value
H.I = 2156.77 ft + 2.67 ft
H.I = 2159.44 ft
and
Elevation at point (x) will be
point (x) = H.I - FS .............2
point (x) = 2159.44 ft - 6.72 ft
point (x) = 2152.72 ft
Answer:
hello your question is incomplete attached below is the complete question
answer: There is a hydraulic jump
Explanation:
First we have to calculate the depth of flow downstream of the gate
y1 =
----------- ( 1 )
Cc ( concentration coefficient ) = 0.61 ( assumed )
Yg ( depth of gate opening ) = 0.5
hence equation 1 becomes
y1 = 0.61 * 0.5 = 0.305 m
calculate the flow per unit width q
q = Q / b ----------- ( 2 )
Q = 10 m^3 /s
b = 2 m
hence equation 2 becomes
q = 10 / 2 = 5 m^2/s
next calculate the depth before hydraulic jump y2 by using the hydraulic equation
answer : where y1 < y2 hence a hydraulic jump occurs in the lined channel
attached below is the remaining part of the solution
Answer:
1) For preparation of plan :
It is to gather the team and creating the timeline. Gathering the inputs for the plan and confirming the mission and vision then launching the strategy.
2) A mission would describe about the company and the specification of the company
A vision is the long term or future state of the organization to achieve in a time frame.
Both would include the OAS statements and strategic shifts.
3) The OAS and strategic shift would help to create a balanced score card that would help for data gathering and analysis
This would include the financial, customers goal, Process goals, people's goals etc.
4) To create and develop the plan's framework:
To set the objective for high level organizational goals
To measure the and accomplish the objectives
To take initiatives to achieve the objectives
Explanation:
Answer:
transmission bandwidth required is very large.
Explanation:
Answer:
A) After 2.5 milliseconds
Explanation:
Given that :
In a CAN bus, there are three computers: Computer A, Computer B, and Computer C.
Length in time of the message sent are 2 milliseconds long
As the computer start sending the message; the message are being sent at a message duration rate of 2 milliseconds
250 microseconds later, Computer B starts to send a message; There is a delay of 250 microseconds = 0.25 milliseconds here at Computer B
and 250 microseconds after that, Computer C starts to send a message
Similarly; delay at Computer C = 0.25 milliseconds
Assuming
is the retransmit time for Computer A to retransmit its message, Then :
= 
= 2 milliseconds + 0.25 milliseconds + 0.25 milliseconds
= 2.5 milliseconds
Thus; the correct option is A) After 2.5 milliseconds