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Soloha48 [4]
3 years ago
14

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 vacuum gage connected to a tank reads 30 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pr
navik [9.2K]

Answer:

Absolute pressure=70.72 KPa

Explanation:

Given that Vacuum gauge pressure= 30 KPa

Barometer reading =755 mm Hg

We know that barometer always reads atmospheric pressure at given situation.So  atmospheric pressure is equal to  755 mm Hg.

We know that P= ρ g h

Density of Hg=13600 \frac{kg}{m^3}

So P=13600 x 9.81 x 0.755

P=100.72 KPa

We know that

Absolute pressure=atmospheric pressure + gauge pressure

But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

Absolute pressure=atmospheric pressure + gauge pressure

Absolute pressure=100.72 - 30   KPa

So

Absolute pressure=70.72 KPa

8 0
3 years ago
No question but thx<br> reeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
jeka94

Answer:

why you doin this

Explanation:

is this so we get free points?

5 0
3 years ago
Read 2 more answers
The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in
amm1812

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be Q\frac{gallons}{minute}

Now replacing the gallon and the minute by the above values we get

Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}

Thus Q'=0.631Q\frac{liters}{second}

Now since we know that 1 liter = 0.0353ft^{3}

Using this in above relation we get

Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

3 0
3 years ago
Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch
Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 \frac{ft^{3}}{sec}

         D_{1}= 6 inch=0.5 ft

        D_{2}=2 inch=0.1667 ft

As we know that Q=AV

A_{1}\times V_{1}=A_{2}\times V_{2}

So V_{2}=\frac{Q}{A_2}

     V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}

     V_{2=0.687 ft/sec

We know that Head loss due to sudden contraction

           h_{l}=K\frac{V_{2}^2}{2g}

If nothing is given then take K=0.5

So head lossh_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}

                                    =0.00366 ft

So head loss=0.00366 ft

4 0
3 years ago
An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates wa
Phantasy [73]

Answer:

Yes

Explanation:

Given Data

Temprature of source=750°c=1023k

Temprature of sink =0°c=273k

Work produced=3.3KW

Heat Rejected=4.4KW

Efficiency of heat engine(η)=\frac{Work produced}{Heat supplied}

and

Heat Supplied {\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )

{Q_s}=3.3+4.4=7.7KW

η=\frac{3.3}{7.7}

η=42.85%

Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink

η=1-\frac{T_ {sink}}{T_ {source}}

η=1-\frac{273}{1023}

η=73.31%

Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.

5 0
3 years ago
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