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3 years ago

Nearlyof all serious occupational injuries and illnesses stem from overexertion of repetitive motion.

Engineering

1 answer:

Ostrovityanka [42]3 years ago

8 0

Answer:

a) 1/2

Explanation:

Overexertion accounted for more than half of all events that resulted in a disabling condition.

Furthermore, 30% of all overexertion cases were reported in the services industry, on the other hand, 25% of injuries resulting from contact with objects and equipment occurred in the manufacturing industry.

The above piece of information is taken from the bureau of labor statistics, Survey of Occupational Injuries and Illnesses

"LOST-WORKTIME INJURIES AND ILLNESSES: CHARACTERISTICS AND RESULTING DAYS AWAY FROM WORK, 2002"

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A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w

Mekhanik [1.2K]

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( $\frac{M1}{100}$ )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

5 0

2 years ago

A cylinder of aluminum-magnesium alloy 0.5 m long is subjected to an elastic tensile stress of 10.2 MPa. The measured elastic el

elena55 [62]

Answer:

E= 15 GPa.

Explanation:

Given that

Length ,L = 0.5 m

Tensile stress ,σ = 10.2 MPa

Elongation ,ΔL = 0.34 mm

lets take young modulus = E

We know that strain ε given as

$\varepsilon =\dfrac{\Delta L}{L}$

$\varepsilon =\dfrac{0.34}{0.5\times 1000}$

$\varepsilon =0.00068$

We know that

$\sigma = \varepsilon E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa$

Therefore the young's modulus will be 15 GPa.

8 0

2 years ago

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin

Leni [432]

Answer:

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^{-1}0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^{-1}0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

6 0

2 years ago

If a pilot-operated check valve (POC) does not check flow, you will see

Elan Coil [88]

if a pilot-operated check valve (POC) does not check flow, you will see The symptoms and symptoms of a failing swing test valve are regularly audible.

<h3>What happens when the pilot valve reaches set strain?</h3>

When the pilot valve reaches set strain it opens and releases the strain from the dome. The piston is then unfastened to open and the principle valve exhausts the machine fluid. The manage pilot opens both to the principle valve exhaust pipe or to the atmosphere.

Pilot operated test valves paintings through permitting unfastened go with the drift from the inlet port thru the opening port. Supplying a pilot strain to the pilot port permits go with the drift withinside the contrary direction. Air strain on the pinnacle of the poppet meeting opens the seal permitting air to go with the drift freely.