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4 years ago

Whose responsibility is it to identify confined spaces and reduce hazards? A The employee B The rescuer C The employer D OSHA

Engineering

1 answer:

natali 33 [55]4 years ago

7 0

Answer:

C. The Employer

Explanation:

Under OSHA's Confined Space Standard, it is the responsibility of the "employer" to identify or evaluate the spaces whether they are "permit spaces" or "confined spaces." These spaces have serious safety hazards or health hazards. Examples: tanks, vessels, pits, underground vaults, and the like.

The employers need to disclose the information of the permit spaces and their risks to the employees. Thus, he should have a written permit of the permit space if he will be allowing the employees to enter. It is also the employer's responsibility to reduce the hazards imposed to the employees such as verifying the entry conditions that are acceptable, making sure the permit space is isolated, providing barriers, etc.

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Burka [1]

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3 years ago

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Tasya [4]

Answer:

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Explanation:

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2 years ago

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- What will happen if high voltage from the HV battery or motor-generator is shorting to frame ground?

Novosadov [1.4K]

Answer:

Unlike a low voltage battery such as 12V, high voltage from a High Voltage battery should not be grounded to the chassis for several numbers of reason which are;

- HV up to 350V have a corresponding high current which generate unwanted magnetic field and causes magnetic interference. This can be reduced by using a twisted conductor so that the interference can be cancelled.

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3 years ago

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

natima [27]

Answer:

1) A=282.6 mm

2) $a_{max}=60.35\ m/s^2$

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm ------------1

V=2.5 m/s at x=225 mm ------------2

Where x measured from mid position.

We know that velocity in simple harmonic given as

$V=\omega \sqrt{A^2-x^2}$

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

$3.5=\omega \sqrt{A^2-0.15^2}$ ------3

$2.5=\omega \sqrt{A^2-0.225^2}$ --------4

Now by dividing equation 3 by 4

$\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}$

$1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}$

So A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

$3.5=\omega \sqrt{A^2-0.15^2}$

$3.5=\omega \sqrt{0.2826^2-0.15^2}$

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

$a_{max}=\omega ^2A$

$a_{max}=14.61 ^2\times 0.2826\ m/s^2$

$a_{max}=60.35\ m/s^2$

Time period T

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{14.609}$

T=0.42 sec

8 0

4 years ago

Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has

trasher [3.6K]

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

8 0

3 years ago