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liq [111]
3 years ago
9

In the contemporary approach to control systems, benefits of continuous monitoring include which one of the following? Multiple

Choice dramatically increased time lags
early detection of changes in the competitive environment
reduced organizational flexibility
increased organization response time
Engineering
1 answer:
tangare [24]3 years ago
3 0

Answer:

increased organization response time

Explanation:

The main objective of continuous monitoring is to ensure that customers get access to all the necessary information they need. This also includes the speed and flexibility of the customers' access to vital information. The customers will be able to make valuable decision and the company will benefit immensely.

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Answer:

The level of service of of compound grade freeway is LOSB.

Explanation:

Find the provided attachments for explanation

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Q10: Technician A says that nearly all brands of scan tools will pull DTCs from the ABS
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3 years ago
is released from under a vertical gate into a 2-mwide lined rectangular channel. The gate opening is 50 cm, and the flow rate in
SVEN [57.7K]

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 = C_{c} y_{g} ----------- ( 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

4 0
3 years ago
A 10MHz clock frequency is applied to a cascaded counter consisting of a modulus-5 counter, a modulus-8 counter and two modulus-
rewona [7]

If a clock frequency is applied to a cascaded counter, The lowest output frequency available will be

  • The lowest output frequency will be = 2.5kHz

<h3>Cascade Counter</h3>

For a cascade counter,

Overall frequency = 5*8*10*10

Overall frequency = 4000

<h3>Lowest Frequency</h3>

Therefore,

the lowest frequency

F = \frac{10*10^6}{4*10^3}\\\\F = 2500Hz\\\\F = 2.5kHz

For more information on frequency, visit

brainly.com/question/17029587?referrer=searchResults

4 0
2 years ago
A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
Marrrta [24]

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

  diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

3 0
2 years ago
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