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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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in a car moving at constant acceleration, you travel 230 mm between the instants at which the speedometer reads 40 km/hkm/h and
Ronch [10]

The acceleration of the car is 0.8049m/s^{2}.It takes 13.802s to travel the 230 m.

<h3>What is acceleration?</h3>

In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it. Newton's Second Law states that the combined effect of two factors determines how much an item accelerates: 

(i) It follows that  the magnitude of the net balance of all external forces acting on the object is directly proportional to the magnitude of this net resulting force, and

(ii) the mass of the thing, depending on the materials out of which it is constructed, is inversely proportional to the mass of the thing.

Calculations:

40 km/hr ----- 11.11m/s

80 km/hr ----- 22.22m/s

v^{2} -u^{2} =2as\\22.22^{2} - 11.11^{2} = 2 x a x 230\\ 370.296=460a\\ a= 0.8049m/s^{2} \\

Time taken

v-u=at

22.22-11.11= 0.8049 x t

t=13.802s

To learn more about acceleration ,visit:

brainly.com/question/2303856

#SPJ4

4 0
2 years ago
What is the acceleration of a Ford Mustang GT that can go from 0.00 to 27.8<br> m/s in 5.15 seconds?
bixtya [17]
Finding acceleration= final speed-initial speed/time taken (or A=V-U\T)

Finial speed= 27.8s
Initial speed= 0s
Time taken= 5.15

So..

27.8-0/5.15= 5.40m/s (rounded to two decimal places)
4 0
3 years ago
How to find the magnitude and direction of a resultant velocity?
mixas84 [53]
Find the horizontal components vcos30 ...one goes right and one goes left so they cancel each other.
Find vertical components vsin30.....there are two of them.... so 2vcos30....hey presto... resultant velocity = 2vCos30
5 0
3 years ago
An object is hanging by a string from the ceiling of an elevator. The elevator is moving upward with a constant speed. What is t
xeze [42]

Answer:

The magnitude of the tension in he string is equal to the magnitude of the weight of the object.

Explanation:

According to the Newton's 1st law, An object will remain at rest or in uniform motion in a straight line unless acted upon by an unbalanced force.

In here, the elevator is moving with a constant speed. So the object must have the equal constant speed. Which means, it has a uniform motion. According to Newton's 1st law, the total unbalanced force on the object must be zero . As we know, there are only two forces are on the object and they are,

The tension in string(T)   ,   The weight of the object(W) .

                                  ∴ F    =   0

                                 T  -  W  =   0

So to balanced those forces, the magnitude of the tension in the string must be equal to the magnitude of the weight of  the object.  

       

8 0
3 years ago
A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the
aleksandrvk [35]

Answer:

θ= 5 radian

Explanation:

Given data:

Radius r = 0.70 m

Initial angular speed ω_i = 2rev/s

Time t = 5 s

Final angular speed ω_f =0

so we have angular displacement

\theta= \frac{\omega_f-\omega_i}{2}\times t

putting values

\theta= \frac{0-2}{2}\times5 = 5 rad

8 0
4 years ago
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