Halite or sulfur or gold or silver
Answer:

Explanation:
Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.
.
Now, let's use the ideal gas equation to the initial and the final state:

Let's recall that the term nR is a constant. That is why we can match these equations.
We can find a relation between the volumes of the initial and the final state.

Combining this equation with the first equation we have:


Now, we just need to solve this equation for T₂.

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.
Here,
Finally, T2 will be:

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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It’s around the g force so it’s gonna be around 54 km/h
Answer:
D. 12.4 m
Explanation:
Given that,
The initial velocity of the ball, u = 18 m/s
The angle at which the ball is projected, θ = 60°
The maximum height of the ball is given by the formula
h = u² sin²θ/2g m
Where,
g - acceleration due to gravity. (9.8 m/s)
Substituting the values in the above equation
h = 18² · sin²60 / 2 x 9.8
= 18² x 0.75 / 2 x 9.8
= 12.4 m
Hence, the maximum height of the ball attained, h = 12.4 m