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3 years ago

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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If 0.5 A is flowing through a household light

stiks02 [169]

Answer:

60W

Explanation:

P=IV=0.5x120

P =60W

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3 years ago

It requires 2,500 joules to raise a certain amount of water (c = 4.186 J/g C) from 20 C to 60 C

Ksivusya [100]

The mass of the water is 14.9 g

Explanation:

When a certain amount of a susbtance is heated, the temperature of the substance increases according to the equation

$Q=mC_s \Delta T$

where

Q is the amount of energy supplied to the substance

m is the mass of the substance

$C_s$ is its specific heat capacity

$\Delta T$ is the change in temperature

In this problem, we have:

Q = 2500 J of energy supplied to the water

$C_s = 4.186 J/gC$ is the specific heat capacity of water

$\Delta T=60 C - 20 C = 40^{\circ}C$ is the change in temperature of the water

Therefore we can solve for m to find the mass of the water:

$m=\frac{Q}{C_s \Delta T}=\frac{2500}{(4.186)(40)}=14.9 g$

Learn more about specific heat capacity:

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7 0

3 years ago

a graduated cylinder contains 60 ml of water. A rock is added and the water level rises to 90 ml. if the rock has a mass of 0.6k

Margarita [4]

Answer:

20g/ml

Explanation:

the mass must be in gram and 90-60 =30

so 600g/30ml =20g/ml

3 0

2 years ago

The body weighing 2 kg moves through the horizontal surface and crosses the path x = 75 cm The coefficient of friction of the bo

Taya2010 [7]

The kinetic energy of the body in definitive position is 4.24 J.

Explanation:

As per the work energy theorem, the work done on any system or object to move it from one position to another is equal to the change in kinetic energy of the object. In this case, the body weighing 2 kg is moved over an horizontal surface for a distance of 75 cm. As there will be frictional force acting on the body while moving over the surface. This frictional force multiplied by the distance the object is moved will give the work done on the body.

Frictional force = Coeffficent of friction × Normal force.

As the weight of the body is 2 kg, the normal force acting on it will be mass multiplied with acceleration due to gravity.

Frictional force = - 0.8×9.8 × 2 =-15.68 N

So the work done will be the product of frictional force with the displacement of 75 cm or 0.75 m.

Work done = Frictional force × Displacement

Work done = -15.68×0.75 = -11.76 J.

So the work is done by the object.

If the kinetic energy of the body at starting is 16 J, then the kinetic energy of the body at definitive position will be obtained as below.

Work done = change in kinetic energy

-11.76 J = Final kinetic energy-16 J

Final Kinetic energy = - 11.76+16

Final kinetic energy = 4.24 J

Thus, the kinetic energy of the body in definitive position is 4.24 J.

3 0

3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

7 0

3 years ago