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prohojiy [21]
4 years ago
6

When a project is completed on time, the project team must accomplish all of the following activities EXCEPT: ​

Physics
1 answer:
gregori [183]4 years ago
5 0

Answer:

The answer is C

Explanation:

When the project finishes under a certain agreement. the project team must capture and share lessons that learn to not repeat same mistakes that made while project ongoing. Therefore, when the similar work need to be redone new project team saves time and labor force with those feedback.

The project team must secure customer feedback and approval to get their claims and progress payments.

The project team must plan a smooth transition of deliverable into ongoing operations in order to get feedback or approval from them in a certain of time. With those feedback of approvals the project team can plan their rework or payment progress.

Alternative dispute resolution is not a must work. It depends on the agreement.

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A 2.0 kg particle moving along the z-axis experiences the
DochEvi [55]

At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Given that a 2.0 kg particle moving along the z-axis experiences the  force shown in a given figure.

Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.

If the particle's velocity is  3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Learn more here: brainly.com/question/20366032

6 0
3 years ago
A proton moves with a speed of 1.17 105 m/s through Earth's magnetic field, which has a value of 50.0 µT at a particular locatio
vlabodo [156]

a) Southward you need to apply right hand rule. If you close your hand to the east, your thumb will indicate south.

b) Given the equation for Magnetic Force

F= qVB

Replacing

F= (1.16*10^{-19})(1.17*10^5)(50*10^{-6})

F=9.36*10^{-19}

c) Given the second Newton's Law by

F_g = 1.67*10^{-27}*9.81

F_g = 1.64*10^{-26}

Given the electric force by,

F_e = 1.6*10^{-19}*1.5*10^2

F_e = 2.4*10^{-17}N

F=9.36*10^{-19}N

7 0
3 years ago
If a substance has a pH of 10 -11, it is considered a ...
Mrac [35]

Answer:

Base

Explanation:

anything with a pH between 0-6 is an acid.  pH of 7 means neutral.  8-14 means it is a base

hope I helped :)

8 0
4 years ago
Solve -7= sqrt 2x-9​
Sauron [17]

Answer:

x = 2

Explanation:

if it was -7 = the square root of both 2x-9 together, it would be false.

if it was square root of just 2x in the equation, the answer is:

x = 2

°°°°°°°°°

-7 = √2x - 9

-√2x = -9 + 7

√-2x = -2

√2x = 2

2x = 4

x = 2

4 0
3 years ago
Two points charge of 4\mu C and 2\mu C are placed at theopposite corners of a rectangle. What is the potential difference Va- Vb
bulgar [2K]

Answer:

Va-Vb=168KV

Explanation:

From the question we are told that

Two points charge of 4\mu C and 2\mu C

Generally we find the  Va and Vb individually to find there difference

Given a rectangle with two equal sides each,Assume lengths for bot sides

Length L=0.3

Breath B=0.4

Diagonal D=\sqrt{0.3^2+0.4^2} =0.5

at  opposite sides

Mathematically Va can represented as

Va =k(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )

Va =9*10^9(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )

Va =9*10^9(0.00001333333-0.000004} )

Va =84000V

Va =84KV

Mathematically Vb is  represented as

Va =k(\frac{-4*10^_-_6}{0.3} +\frac{2*10^_-_6}{0.5} )

Va =9*10^9(\frac{-4*10^_-_6}{0.3} +\frac{+2*10^_-_6}{0.5} )

Va =9*10^9(-0.00001333333+0.000004} )

Va =-84000V

Va =-84KV

Therefore

Va-Vb=84-(-84)\\Va-Vb=84+84\\Va-Vb=168KV

7 0
3 years ago
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