She is

kilometers away from her starting point
When one body(sun) exerts a force on a second body(planet), the second body simultaneously exerts a force equal in magnitude and opposite in direction of the first body. Which makes the planet orbit in path C.
Hope this helps!!
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.
brainly.com/question/1203927?referrer=searchResults
A process known as fixation<span>. the majority of nitrogen is fixed by </span>bacteria<span>, most of which are </span>symbiotic<span> with plants</span>
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct