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zzz [600]
2 years ago
10

What is the difference between engineering controls and administrative controls?

Physics
1 answer:
nadezda [96]2 years ago
5 0

Engineering controls could involve altering the weight of the products, adjusting the heights of the work surfaces, or getting lifting equipment. Administrative controls are workplace policies, practices, and processes that reduce the likelihood of risky situations for employees.

<h3>What are Engineering controls?</h3>
  • Engineering controls are methods for preventing dangerous situations for workers by erecting a barrier between them and the danger or by ventilating the area to remove any hazardous materials.
  • Instead of depending on employee behavior or mandating that employees wear safety gear, engineering controls entail making physical changes to the workplace itself.
<h3>What is Administrative controls?</h3>
  • Administrative controls are policies, procedures, shift designs, or training that reduce the threat of a hazard to a person.
  • Typically, administrative controls aim to alter people's behavior (such as that of manufacturing workers) rather than eliminating the real risk or provide personal protective equipment (PPE).

Learn more about Administrative controls here:

brainly.com/question/17628591

#SPJ4

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nydimaria [60]

Answer:

A, D, E

Explanation:

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3 0
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If a rooster lays and egg on the middle of the roof, which way does the egg role?
Mekhanik [1.2K]

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8 0
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Novosadov [1.4K]

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8 0
3 years ago
A football is kicked into the air from an initial height of 4 feet. The height, in feet, of the football above the ground is giv
kakasveta [241]

Answer: 0.5 seconds or 2.625 seconds

Explanation:

At t = 0, The ball is 4 ft above the ground.

The height of the football varies with time in the following way:

s(t) = -16 t² + 50 t + 4

we need to find the time in which the height would of the football would be 25 ft:

⇒25 = -16 t² + 50 t + 4

we need to solve the quadratic equation:

⇒ 16 t² - 50 t + 21 = 0

t = \frac{50 \pm \sqrt{50^2-4\times 16\times 21}}{2\times16}

⇒ t = 0.5 s or 2.625 s

Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.

3 0
3 years ago
Read 2 more answers
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ozzi

Answer:

The position of my house is a little uphill as compared to the position of my school. The distance I have to travel from my house to school is nearly 2 kilometers. The displacement is in the 2000 m towards the left from my house. The speed of the bus which I usually take is 40 km/ hour.

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