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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

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The answer would be B..

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

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Note that the scale bar under the photo is labeled 1 μm (micrometer). The scale bar works the same way as a scale on a map, wher
AnnZ [28]

Answer:

Hello your question has some missing parts attached below is the complete question

answer : 4 μm

Explanation:

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3 years ago
What type of device forms images through refraction?
Tcecarenko [31]

Answer:

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3 0
2 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

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3 years ago
The word focus means _______.
rusak2 [61]
D.to bring light together
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3 years ago
Read 2 more answers
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