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3 years ago

15

A zone reconnaissance involves a directed effort to obtain detailed information on all routes, obstacles, terrain, enemy forces,

or specific civil considerations within a zone defined by boundaries to include obstacles (existing and reinforcing) as well as areas of CBRN contamination. True or false?

1 answer:

son4ous [18]3 years ago

7 0

Answer:

It is True

Explanation:.

A commander assigns a zone reconnaissance mission when he seeks additional information on a zone before committing other forces in the zone. It is appropriate when the enemy situation is vague, existing knowledge of the terrain is limited, or combat operations have altered the terrain. A zone reconnaissance could include several route or area reconnaissance missions assigned to subordinate units.

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A student is measuring the circumferences of pine trees for an experiment.

Artemon [7]

Answer:

Her measurements will be more accurate if she uses a cloth tape

measure instead of a ruler.

Explanation: (apex)

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3 years ago

What is all matter made up of?

aliya0001 [1]

All matter is made up of atoms

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3 years ago

Read 2 more answers

This is due by 11:59 PM tonight.

svetoff [14.1K]

Answer:

1. increases

2. increases

3. increases

Explanation:

Part 1:

First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:

F1 - fs = 0.

And this friction force fs is:

fs = Nμs,

where μs is the static coefficient of friction, and N is the normal force.

Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:

N = mg + F2.

So, F2 is increasing, that means fs is increasing too.

Part 2:

As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.

Part 3:

In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.

6 0

4 years ago

Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

3 years ago

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in

bezimeni [28]

1) Potential difference: 1 V

2) $V_b-V_a = -1 V$

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

$\Delta U=q\Delta V$

where

q is the charge's magnitude

$\Delta V$ is the potential difference between the initial and final position

In this problem, we have:

$q=4.80\cdot 10^{-19}C$ is the magnitude of the charge

$\Delta U = 4.80\cdot 10^{-19}J$ is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

$\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V$

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

$V_b-V_a = - 1V$

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3 years ago