All categories

3 years ago

Which EOC organizational structure is familiar and aligns with the on-scene incident organization?

Physics

1 answer:

dimulka [17.4K]3 years ago

3 0

Answer:

Incident Command Structure, ICS or ICS-like EOC Structure is familiar and aligns with the on-scene incident organization. ICS or ICS-like EOC Structure is familiar and aligns with the on-scene incident organization.

You might be interested in

A boy with a mouse is 80 kg stands on a scale in an ascending elevator with an acceleration of 3 m/s What is the resultant upwar

Rus_ich [418]

Answer:

Upward force on the boy will be 1024 N

Explanation:

We have given mass of the boy m = 80 kg

Acceleration due to gravity $g=9.8m/sec^2$

It is given that elevator is ascending with acceleration of $3m/sec^2$

So net acceleration on the boy = 9.8+3 = 12.8 $m/sec^2$ ( As the car elevator is moving ascending )

So upward force on the boy will be equal to F = m(g+a)

So $F=80\times 12.8=1024N$

So upward force on the boy will be 1024 N

3 0

3 years ago

Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. If the en

satela [25.4K]

Answer:

The fence is 5feet less.

Explanation:

We need to determine

The less amount of fence required, if the enclosure has full width and reduced length, compared to full length and reduced width.

Approach & WorkingArea of lawn = 30 × 403/4th of the area of lawn = ¾(30 × 40) = 30 * 30

When full width will be fenced, and reduced length will be fenced.

Width = 30 feet30 * L = 30 * 30Hence, length = 30 feetLength of fence needed = 2(30 + 30) = 120 feet

When full length will be fenced, and reduced width will be fenced

Length = 40 feet40 * W = 30 * 30W = 22.5 feetLength of fence needed = 2(40 + 22.5) = 125 feet

Difference in length of fence needed = 125 – 120 = 5 feet.

4 0

4 years ago

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep

Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

m1 = 0.0225 Kg

V1 = (0.869, -0.283) m / s

Px = m Vx

Px1 = 0.0225 0.869

Px1 = 0.01955 Kg m

Py = m Vy

Py1 = 0.0225 (-0.283)

Py1 = -0.006368 Kg m

P1 = (0.0196, -0.00637) Kg m

Mouse 2

m2 = 0.0223 Kg

Px2 = 0.0223 (-0.883) = -0.0196 Kg m

Py2 = 0.0223 (-0.253) = -0.00564 Kg m

P2 = (-0.0196, -0.00564) Kg m

Mouse 3

m3 = 0.0197

Px3 = 0.0197 0.345 = 0.00680 Kg m

Py3 = 0.0197 0.803 = 0.0158 Kg m

P3 = (0.00680, 0.0158) Kg m

Mouse4

m4 = 0.0127 Kg

Px4 = 0.0127 (-0.555) = -0.00705 Kg m

Py4 = 0.0127 0.205 = 0.00260 Kg m

P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

Px = Px1 + Px2 + Px3 + Px4

Py = py1 + Py2 + Py3 + Py4

Px = 0.0196 -0.0196 +0.00680 -0.00705

Px = -0,000250 Kg m

Py = -0.00637 -0.00564 +0.0158 +0.00260

Py = 0.00639 Kg m

P = (-0.000250, 0.00639) Kg m

7 0

3 years ago

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

$K_t=\dfrac{1}{2}mv^2$

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?

arlik [135]

Answer: 0.25 m/s

Explanation: Speed = wavelengt · frequency

v = λf and frequency is 1/period f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

6 0

3 years ago