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juin [17]
3 years ago
5

What is not an essential part of isolated soldier guidance

Physics
1 answer:
mel-nik [20]3 years ago
4 0

Answer:

Option B, visual sightings

Explanation:

Options for the question are

A) accountability mechanism

B) visual sightings

C)  intelligence

D) surveillance

E) reconnaissance  operations, or communications

Solutions

The Fundamentals of Army Personnel Recovery (PR) outlines certain circumstances under which a person has to undergo survival situation thereby taking the necessary steps to avoid capture and return safely to their respective unit.

An isolated soldier is expected to know where they are, upcoming route and rally points. They are supposed to know the near and far recognition signals, recovery site protocols, challenge and password etc. A proper preparation is to be done for this including planning, medicines, kits, etc.  

Visual sightings is not an essential part of isolated PR

Hence, option B is correct

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anyanavicka [17]
<h3><u>Answer and explanation;</u></h3>
  • <u><em>A neutral sodium atom contains 11 protons, 12 neutrons and 11 electrons. By removing an electron from this atom we get a positively charged Na+ ion that has a net charge of +1.</em></u> Thus; Sodium atom is neutral whereas sodium ion is a charged species with a charge of +1.
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3 0
3 years ago
PLEASE HELPPPPPPPPPPPPP
sertanlavr [38]

Answer: sliding

Explanation:

A snow skier slowing to a stop after skiing down a mountain is an example of sliding friction

8 0
2 years ago
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If the absolute pressure of a gas is 550.280 kPa, its gage pressure is
Gwar [14]
If the absolute pressure of a gas is 550.280 kPa, its gage pressure is <span> a. 101.325 kPa.
b. 448.955 kPa.
c. 651.605 kPa.
d. 277.280 kPa.</span>

The answer is B.
6 0
2 years ago
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Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of
schepotkina [342]
If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

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5 0
2 years ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}  = \frac{725}{\sqrt{725}}  = \sqrt{725} ft / sec

5 0
2 years ago
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