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3 years ago

Who is responsible for conducting a hazard assessment?

Engineering

1 answer:

slava [35]3 years ago

5 0

Answer:

The Employee

Explanation:

Because it is there responsibility

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Kipish [7]

Answer:

tHE answer is b

Explanation:

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3 years ago

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The denominator of a fraction is 4 more than the numenator. If 4 is added to the numenator and 7 is added to the denominator, th

kati45 [8]

Answer:

$\frac{3}{7}$

Explanation:

Lets take the numerator of the fraction to be = x

So the denominator of the fraction is 4 more than the numerator = x+4

The fraction is ; $\frac{x}{4+x}$

Now add 4 to the numerator and add 7 to the denominator as;

$\frac{x+4}{4+x+7} =\frac{x+4}{x+11}$

This new fraction is equal to 1 half =1/2

write the equation as;

$\frac{x+4}{x+11} =\frac{1}{2}$

perform cross-product

2(x+4 )=1( x+11 )

2x+8 = x + 11

2x-x = 11-8

x=3

The original fraction is;

$\frac{x}{4+x} =\frac{3}{3+4} =\frac{3}{7}$

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3 years ago

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Business plan, a good strategy to help pin point the house best fetchers

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What is acid mine drainage

avanturin [10]

Acid mine drainage is the formation and movement of highly acidic water rich in heavy metals. This acidic water forms through the chemical reaction of surface water (rainwater, snowmelt, pond water) and shallow subsurface water with rocks that contain sulfur-bearing minerals, resulting in sulfuric acid.

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3 years ago

About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa

Nana76 [90]

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

$K.E = \frac{2KZe^2}{r}$

where;

Z is the atomic number of aluminium = 13
e is charge
r is distance of closest approach = thickness of aluminium
k is Coulomb's constant = 9 x 10⁹ Nm²/C²

<h3>For 2.5 MeV electrons</h3>

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

$r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m$

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

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2 years ago