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2 years ago

1. The reference OSHA standard requires several important characteristics of a PFAS:

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The reference OSHA standard requires several important characteristics of a PFAS. The employer must ensure that personal fall arrest systems (PFAs) must:

a. Limit the maximum arresting force on the employee to 1,800 pounds (8 kN). The figure used above must be based on the use of a full body harness.

Maximum arresting force is the highest amount of force that the fall protection system and the individual attached to the system will face as gotten by the deceleration device.

OSHA regulations gives the system performance criteria for a personal fall arrest system.

From the above, we can therefore say that The reference OSHA standard requires several important characteristics of personal fall arrest systems (PFAs) as employers must reduce the maximum arresting force on the employee to 1,800 pounds (8 kN) using full body harness.

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3. Which of the following is an example of qualitative data?

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What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

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Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin

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Answer:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

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4 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603 225

.203 368

.162 442

answer : 191 mPa

Explanation:

Determine the stress present at the time of fracture for the original plate

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603 bending stress = 225

at 0.203 bending stress = 368

at 0.162 bending stress = 442

applying equation 1 determine the value of n for several combinations

( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) = ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x = 0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

= 225 * 0.84747

= 191 mPa 

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