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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.

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If forces acting on an object are unbalanced, which factor may result from an unbalanced force? A.The net force is negative. B.T

bezimeni [28]

Answer:

<h2>A. The net force is negative.</h2>

Explanation:

just took the test :)

4 0

3 years ago

Read 2 more answers

Define impulse and momentum.<br>No spam

Doss [256]

Impulse: a certain amount of force you apply for an amount of time.

Impulse: F*t where F= Force & t=time

Momentum: increasing forward motion.

A ball rolling down a slide gains momentum

p=mv where m=mass and v=velocity

Hope it helps!

~Just an emotional teen who listens to music

4 0

2 years ago

A 3 kg rock sits on a 0.8 meter ledge. If it is pushed off, how fast will it be going at the bottom?

andrey2020 [161]

As long as it sits on the shelf, its potential energy
relative to the floor is . . .

   Potential energy =    (mass) x (gravity) x (height) =

   (3 kg) x (9.8 m/s²) x (0.8m) = <u>23.52 joules</u> .

If it falls from the shelf and lands on the floor, then it has exactly that
same amount of energy when it hits the floor, only now the 23.52 joules
has changed to kinetic energy.

   Kinetic energy =    (1/2) x (mass) x (speed)²

   23.52 joules = (1/2) x (3 kg) x (speed)²

Divide each side by 1.5 kg :    23.52 m²/s² = speed²

Take the square root of each side: speed = √(23.52 m²/s²) = <em>4.85 m/s </em> (rounded)

6 0

3 years ago

a pool ball leaves a 0.60-meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. what is the

inysia [295]

Answer:

0.84 m

Explanation:

Given in the y direction:

Δy = 0.60 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

0.60 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.35 s

Given in the x direction:

v₀ = 2.4 m/s

a = 0 m/s²

t = 0.35 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.4 m/s) (0.35 s) + ½ (0 m/s²) (0.35 s)²

Δx = 0.84 m

5 0

3 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$