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3 years ago

Please someone help me lol

Physics

1 answer:

Makovka662 [10]3 years ago

7 0

From Largest to Smallest:

Earth > Biosphere > Biome > Ecosystem > Community > Population > Organism > Organelles

Hope this helps!

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Which of the following statements is TRUE about updating the exposure control plan?

iren2701 [21]

Statements that are true as regards exposure control plan and its updating are;

Updates must have the reflection of changes in tasks as well in procedures.

Updates must reflect changes in positions that affect occupational exposure.

Updates must have the cost of PPE that is needed and necessary to reduce exposure

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

3 0

2 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

3 years ago

3. When encountering low visibility from rain or fog, you

zimovet [89]

Answer:

c. low beams and fog lights

Explanation:

When encountering low visibility from rain or fog, use your low beams and fog lights. High beams will only increase the glare. If you can't see at least five seconds in front of you, don't drive. Pull over and put your hazards on until it clears up.

i just got the answer wrong and the drivers ed gave me this explanation !!

8 0

3 years ago

Read 2 more answers

50 POINTS

sweet-ann [11.9K]

Answer:

The acceleration is constant and positive

Explanation:

The straight line indicates that the acceleration is constant, while the positive slope indicates that the line is positive.

7 0

3 years ago

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A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the

sleet_krkn [62]

Answer:

1176.01 °C

Explanation:

Using Ohm's law,

V = IR................. Equation 1

Where V = Voltage, I = current, R = Resistance when the bulb is on

make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

Δθ = T₂-T₁

T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

5 0

3 years ago

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