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3 years ago

What is the outcome of the Training and Exercise Planning Workshop (TEPW)?

Physics

1 answer:

balandron [24]3 years ago

8 0

Answer / Explanation:

The result of the Training and Exercise Planning Workshop (TEPW) is to set the foundation for the strategy and pattern for a proposed exercise program. The TEPW purpose is to engage elected and selected officials in identifying exercise program priorities and planning a schedule of training and exercise events to meet those priorities.

An essential factor for the exercise management process is to create a collaborative environment where a whole community stakeholders can engage in a forum to discuss and coordinate training and exercise activities across local organizations to maximize the use of available resources and prevent duplication of effort.

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A body weighs 10newton in air and 9.5 in water. How will it weigh in alchohol if density 0.8gcm-³

Katen [24]

Answer:

The weight of object in alcohol is, W₀ = 9.6 N

Explanation:

Given,

The weight of body in air, Wₐ = 10 N

The weight of the body in water, Wₓ = 9.5 N

The density of the alcohol, ρ = 0.8 g/cm³

= 800 kg/m³

The weight of body in water = weight of object in air - weight of water displaced

9.5 N = 10 N - weight of water displaced

Weight of water displaced = 0.5 N

The mass of the displaced water, m = 0.5 / 9.8

= 0.051 kg

Therefore the volume of the water displaced,

V = m / ρ

= 0.051 / 1000

= 5.1 x 10⁻⁵ m³

The volume of water displaced is equal to the volume of alcohol displaced

Therefore, the mass of the alcohol displaced, = V x ρ

= 5.1 x 10⁻⁵ x 800

= 0.0405 kg

The weight of the alcohol displaced, w = 0.0405 x 9.8

= 0.4 N

Therefore,

The weight of body in alcohol = weight of object in air - weight of alcohol displaced

W₀ = W - w

= 10 N - 0.4 N

= 9.6 N

Hence, the weight of object in alcohol is, W₀ = 9.6 N

4 0

2 years ago

A football kick returner catches the ball just as a player from the opposing team dives to tackle him. At the time of impact, th

pochemuha

The total momentum of the players after collision is 130 kgm/s.

The given parameters:

Initial momentum of the returner, $P_i_1$ = 0 kgm/s
The initial momentum of the diving player, $P_i_2$ = 130 kgm/s

The total momentum of the players after collision is determined by applying the principle of conservation of linear momentum as follows;

$P_f = P_i_1 + P_i_2\\\\P_f = 0 + 130\\\\P_f = 130 \ kgm/s$

Thus, the total momentum of the players after collision is 130 kgm/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

6 0

2 years ago

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

konstantin123 [22]

Answer:

$I=2.71\times 10^{-5}\ A$

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

$C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}$ , r is radius

Let I is the displacement current. It is given by :

$I=C\dfrac{dV}{dt}$

Here, $\dfrac{dV}{dt}$ is rate of increasing potential difference

$I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A$