Statements that are true as regards exposure control plan and its updating are;
<em>Updates must have the reflection of changes in tasks as well in procedures.</em>
<em>Updates must reflect changes in positions that affect occupational exposure.</em>
<em>Updates must have the cost of PPE that is needed and necessary to reduce exposure</em>
An exposure control plan can be regarded as the framework for compliance between the employer and the workers.
- This framework give room for the employer to creates a written plan that will help in protecting their workers from bloodborne pathogens.
- This plan gives hope to workers in term of protection when working with their Employer.
- There are some elements that is associated with Exposure Control Plan, and theses are;
- Health hazards as well as risk that is attributed to each product in the worksite.
- Statement of purpose.
- procedures and practices in a written form
- Responsibilities from the Manager, CEO, designated resources and employer.
Therefore, exposure control plan is avenue to protect workers from bloodborne pathogens.
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C. The higher the altitude the less gravity affects you
When the surface of the comb rubs on your hair, the comb is electrically charged. When the comb comes close to the paper, the charge on the comb causes charge separation on the paper bits. Since paper is neutral, positive and negative charges are equivalent. The charge on the comb charges the area of the bit of paper nearest the comb to the opposite. Thus, the bits of paper become attracted to the comb.
Answer:
304.89m
Explanation:
Given
acceleration a = 2.52m/s²
final speed v = 39.2m/s
initial speed = 0m/s (car accelerates from rest)
Using the equation of motion below to get the distance of Doc brown from Marty;
v² = u²+2as
substitute the given parameters
39.2² = 0²+2(2.52)s
1536.64 = 0+5.04s
divide both sides by 5.04
1536.64/5.04 = 5.04s/5.04
rearrange the equation
5.04s/5.04 = 1536.64/5.04
s = 304.89m
Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?
With acceleration
and initial velocity
the velocity at time <em>t</em> (b) is given by
We can get the position at time <em>t</em> (a) by integrating the velocity:
The particle starts at the origin, so 
.
Get the coordinates at <em>t</em> = 8.00 s by evaluating 
at this time:
so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).
Get the speed at <em>t</em> = 8.00 s by evaluating 
at the same time:
This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:
