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UkoKoshka [18]
3 years ago
11

When do you need to apply for program completion and review?

Engineering
1 answer:
Leya [2.2K]3 years ago
8 0

Answer:

1/2 semesters before completion

Explanation:

Program completion and review are both a necessary part and parcel of a successful completion of any program. The recommended period for applying for program completion and review is one to two semesters before the program is to be completed. This is mandatory, so that there is different time span for review and any other issues that are needed to be carried out. And then afterwards, finally the certificates are then issued to the qualified person.

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Air at atmospheric pressure and at 300K flows with a velocity of 1.5m/s over a flat plate. The transition from laminar to turbul
Savatey [412]

Answer:3.47 m

Explanation:

Given

Temperature(T)=300 K

velocity(v)=1.5 m/s

At 300 K

\mu =1.846 \times 10^{-5} Pa-s

\rho =1.77 kg/m^3

And reynold's number is given by

Re.=\frac{\rho v\time x}{\mu }

5\times 10^5=\frac{1.77\times 1.5\times x}{1.846\times 10^{-5}}

x=\frac{5\times 10^5\times 1.846\times 10^{-5}}{1.77\times 1.5}

x=3.47 m

5 0
3 years ago
What would happen if an exposed film was accidentally placed in the fixer before being placed in the developer
Eduardwww [97]
Do you still want this answered
5 0
3 years ago
An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8
marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

a=18.66\ m/s^2

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0
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
Which of the following describes how the author introduces dust storms in the text??​
Oduvanchick [21]

Answer:

B-as deadly storms that claim lives in th great pines

Explanation:

I hope it helped if it did pls make me brainliest

8 0
3 years ago
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