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3 years ago

When you are in charge of running a meeting, "winging it" with minimal planning is considered:

Engineering

1 answer:

amm18123 years ago

3 0

Answer:

The correct option is;

B) Counterproductive

Explanation:

An effective requires adequate preparations, with agenda of the meeting circulated among participants. The items to be discussed should be known before hand by the participants all of whom will have had adequate background information and relevant consultations with their constituencies so as to be able to add effectively to the meeting.

Other attributes of an effective meeting includes;

1. Ensure, notes of the meeting are taken down personally

2. Outside discussions should be made at the parking lot

3. Important decisions or points to be made should have been communicated to other members of the meeting

4. Members to attend the meeting should be well known and list of attendees reviewed

5. Meetings should be kept to schedule

6. Ensure meeting conclusions and decisions are followed up

7. Written agenda should be available

Therefore "winging it" with minimal planning is considered counterproductive.

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Answer: A maximum of 1 hour

Explanation:

Read your lesson buddy!!

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3 years ago

Read 2 more answers

About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa

Nana76 [90]

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

$K.E = \frac{2KZe^2}{r}$

where;

Z is the atomic number of aluminium = 13
e is charge
r is distance of closest approach = thickness of aluminium
k is Coulomb's constant = 9 x 10⁹ Nm²/C²

<h3>For 2.5 MeV electrons</h3>

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

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2 years ago

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3 years ago

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Nutka1998 [239]

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