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Brilliant_brown [7]
2 years ago
5

Which of the following describes one of an employee's responsibilities under OSHA's rules?

Engineering
2 answers:
Ira Lisetskai [31]2 years ago
6 0

Answer:

First off, employees must follow all OSHA safety and health standards AND all rules issued by their employers that are intended to comply with OSHA's health and safety standards and the OSH Act. Report all occupational injuries and illnesses. Report hazardous conditions. Cooperate during an OSHA inspection.

kow [346]2 years ago
5 0

Explanation:

what are the options for this question?

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grandymaker [24]

Answer:

Explanation:

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2 years ago
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Consider laminar, fully developed flow in a channel of constant surface temperature Ts. For a given mass flow rate and channel l
Lisa [10]

Answer:

Please see attachment.

Explanation:

6 0
3 years ago
I want to solve the question
TEA [102]
Hope it is clear :)

8 0
3 years ago
At the instant shown car A is travelling with a velocity of 24 m/s and which is decreasing at 4 m/s2 along the highway. At the s
SVEN [57.7K]

(a) V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

<u>Explanation:</u>

We will solve with respect to Cartesian vector form.

So,

V(A)= (24i) m/s

acc(A) = (4i) m/s²

There are two components of Car B, cos 60⁰ and sin 60⁰

V(B) = 20 cos 60° i + 20 sin 60° j

V(B) = (10 i + 17.32 j ) m/s

The car B moves along a curve, so it will have a tangential acceleration and a normal acceleration.

The tangential acceleration, a(t) = 5 m/s²

Normal acceleration, a(n) = \frac{v^2}{p} \\\\

So,

a(n) = \frac{(20)^2}{250}\\ \\a(n) = 1.6 m/s^2

For the tangential acceleration, the acceleration is slowing down. So,

a(t) = (-5 cos 60° i - 5 sin 60° j ) m/s²

a(t) = ( -2.5 i - 4.33 j) m/s²

For normal acceleration, it towards center. So,

a(n) = (1.6 sin 60° i - 1.6 cos 60° j) m/s²

a(n) = (1.39 i - 0.8 j ) m/s²

Total acceleration of Car B:

acc(B) = a(t) + a(n)

acc(B) = ( -2.5 i - 4.33 j) m/s² + (1.39 i - 0.8 j ) m/s²

acc(B) = (-1.11i - 5.13 j ) m/s²

(a) V(A/B) = ?

V(A) = V(B) + V(A/B)

(24i) m/s = (10 i + 17.32 j ) m/s + V(A/B)

V(A/B) = (14 i - 17.32 J) m/s

(b) acc(A/B) = ?

acc(A) = acc(B) + acc(A/B)

(4i) m/s² = (-1.11i - 5.13 j ) m/s² + acc(A/B)

acc(A/B) = ( 5.11 i + 5.13 j ) m/s²

3 0
2 years ago
Electrical circuits must be locked-out/tagged-out before electricians work on any equipment. Is this true or false?
dybincka [34]

Answer:

true

Explanation:

Equipment that are "locked-out/tagged-out" <em>prevent the electrician from being electrocuted</em> or attaining a serious injury in relation to it. Locking out an equipment prevents it from releasing its energy because such energy can be <em>hazardous</em> to the electrician. There are instances when the equipment accidentally starts up, thus, it is essential that the equipment's source of energy is<em> isolated.</em>

8 0
3 years ago
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