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3 years ago

When would working with machinery be a common type of caught-in and caught-between hazard?

Engineering

1 answer:

tigry1 [53]3 years ago

6 0

Answer:

A working with machinery be a common type of caught-in and caught-between hazard is described below in complete detail.

Explanation:

“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.

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A horizontal 2-m-diameter conduit is half filled with a liquid (SG=1.6 ) and is capped at both ends with plane vertical surfaces

luda_lava [24]

Answer:

Resultant force = 639 kN and it acts at 0.99m from the bottom of the conduit

Explanation:

The pressure is given as 200 KPa and the specific gravity of the liquid is 1.6.

The resultant force acting on the vertical plate, Ft, is equivalent to the sum of the resultant force as a result of pressurized air and resultant force due to oil, which will be taken as F1 and F2 respectively.

Therefore,

Ft = F1 + F2

According to Pascal's law which states that a change in pressure at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere, the air pressure will act on the whole cap surface.

To get F1,

F1 = p x A

= p x (πr²)

Substituting values,

F1 = 200 x π x 1²

F1 = 628.32 kN

This resultant force acts at the center of the plate.

To get F2,

F2 = Π x hc x A

F2 = Π x (4r/3π) x (πr²/2)

Π - weight density of oil,

A - area on which oil pressure is acting,

hc - the distance between the axis of the conduit and the centroid of the semicircular area

Π = Specific gravity x 9.81 x 1000

Therefore

F2 = 1.6 x 9.81 x 1000 x (4(1)/3π) x (π(1)²/2)

F2 = 10.464 kN

Ft = F1 + F2

Ft = 628.32 + 10.464

Ft = 638.784 kN

The resultant force on the surface is 639 kN

Taking moments of the forces F1 and F2 about the centre,

Mo = Ft x y

Ft x y = (F1 x r) + F2(1 - 4r/3π)

Making y the subject,

y = (628.32 + 10.464(1 - 4/3π)/ 638.784

y = 0.993m

7 0

4 years ago

How do you identify all sensors, functions, and where we can use them?

Alex17521 [72]

Sensor/Detectors/Transducers are electrical, opto-electrical, or electronic devices composed of specialty electronics or otherwise sensitive materials, for determining if there is a presence of a particular entity or function. Many vehicles including cars, trains, buses etc. all use sensors to monitor oil temperature and pressure, throttle and steering systems and so many more aspects.

7 0

3 years ago

on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain

Bess [88]

Answer:

The condition does not hold for a compression test

Explanation:

For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.

<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test

5 0

3 years ago

What type of companies would employ in mechanics engineering

Alex73 [517]

What do y’all do when ya girl go eat lunch and eat it and eat

3 0

3 years ago

A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k

goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = $\frac{ln\frac{r2}{r1}}{2 \pi *k * L}$ .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

heat loss is change in temperature divide thermal resistance

Q = $\frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}$

Q = $\frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }$

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0

4 years ago