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vodomira [7]
2 years ago
13

Describe the risks associated with their working environment (such as the tools, materials and equipment that they use, spillage

s of oil, chemicals and other substances, not reporting accidental breakages of tools or equipment and not following laid-Down working practices and procedures)
Engineering
1 answer:
Llana [10]2 years ago
3 0

The risks which are associated with such working environment and not reporting accidental breakages of tools is pollution and damage to the body system of individuals.

<h3>What is Risk?</h3>

This is defined as the possibility of something causing a bad event to occur and is common in workplace when tools are left carelessly which can cause injuries etc.

In the case of a working environment with oil spills, it can lead to slipping of the individuals present in the area and not reporting accidental breakages of tools or equipment could also lead to pollution and different forms of accident which thereby making it the most appropriate choice in this context.

Read more about Risk here brainly.com/question/1224221

#SPJ1

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Sun of first 1 nayural numbers​
marin [14]

Answer:

the answer is

n(n + 1)  \div 2

n(n+1)/2

4 0
3 years ago
Read 2 more answers
John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa
IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= 35000\times (\frac{F}{P}, 4 \ percent,5 )

= 35000\times 1.216 7

= 42584.50

Now,

time (t) will be:

= 5\times 12

= 60 \ monthly \ payments

So, monthly payment will be:

= 42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )

= 42584.50\times 0.0193

= 821.88

6 0
3 years ago
Find the time-domain sinusoid for the following phasors:_________
sattari [20]

<u>Answer</u>:

a.  r(t) = 6.40 cos (ωt + 38.66°) units

b.  r(t) = 6.40 cos (ωt - 38.66°) units

c.  r(t) = 6.40 cos (ωt - 38.66°) units

d.  r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = \sqrt{a^2 + b^2}

Ф = direction = tan⁻¹ (\frac{b}{a})

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{-5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{-5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

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3 years ago
The period of an 800 hertz sine wave is
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Explanation:

White Board Activity: Practice: A sound has a frequency of 800 Hz. What is the period of the wave? The wave repeats 800 times in 1 second and the period of the function is 1/800 or 0.00125.

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3 years ago
Infinitivo de vivia kkk xd
blagie [28]

Answer:

pls put a question not random letters

Explanation:

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