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hichkok12 [17]
3 years ago
10

Reposition the hazardous items to reduce the impact of disaster.​

Geography
2 answers:
Semenov [28]3 years ago
7 0

Answer:

Image result for Reposition the hazardous items to reduce the impact of disaster.​

Awareness, education, preparedness, and prediction and warning systems can reduce the disruptive impacts of a natural disaster on communities. Mitigation measures such as adoption of zoning, land-use practices, and building codes are needed, however, to prevent or reduce actual damage from hazards.

Triss [41]3 years ago
7 0

Answer:

<em>Hope</em><em> </em><em>it</em><em> </em><em>help</em><em> </em><em>for</em><em> </em><em>you</em><em>.</em><em>Stay</em><em> </em><em>safe</em><em>,</em><em>Stay</em><em> </em><em>healthy</em><em>.</em><em>Have</em><em> </em><em>a</em><em> </em><em>nice</em><em> </em><em>day</em><em>:</em><em>)</em>

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The correct answer is C.
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A force of 15N acts on angle of 60degree to the horizontal of a body of mass 3kg initially at rest on a smooth horizontal plain.
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Explanation:

KEY POINTS

Understanding work is quintessential to understanding systems in terms of their energy, which is necessary for higher level physics.

  • Work is equivalent to the change in kinetic energy of a system.
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Here are a few example problems:

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(1.b) Since the box is displaced 5 meters and the force is 2 N, we multiply the two quantities together. The object’s mass will dictate how fast it is accelerating under the force, and thus the time it takes to move the object from point a to point b. Regardless of how long it takes, the object will have the same displacement and thus the same work done on it.

(2.a) Consider the same box (M = 3 kg) being pushed by a constant force of four newtons (F = 4 N). It begins at rest and is pushed for five meters (d = 5m). Assuming a frictionless surface, calculate the velocity of the box at 5 meters.

(2.b) We now understand that the work is proportional to the change in kinetic energy, from this we can calculate the final velocity. What do we know so far? We know that the block begins at rest, so the initial kinetic energy must be zero. From this we algebraically isolate and solve for the final velocity

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