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3 years ago

______________ help protect the lower legs and feet from heat hazards like molten metal and welding sparks.

Engineering

1 answer:

astraxan [27]3 years ago

8 0

Answer:

leggings

Explanation:

they allow the metal or sparks to not hurt you can the leggings can be easily and quickly removed.

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If you owned a business, what are some of the ways you could follow green computing recommendations with regards to recycling an

krek1111 [17]

Answer: E-cycle used computer equipment. Recycle Printer Cartridges and Drum Units. Properly Dispose of Old Technology.

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8 0

3 years ago

How to plot 0.45 gradation chart for sieve analysis ?

LekaFEV [45]

Answer:

532235w3r35w3r

Explanation:

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2 years ago

A room is cooled by circulating chilled water through a heat exchanger located in the room. The air is circulated through the he

Zanzabum

Answer:

input power of the geothermal motor will be 0.4166 hP

Explanation:

We have given shaft output = 0.25 hP

hP is known as horse power which is unit of measuring the power of motor

Efficiency of motor is given as efficiency = 60 % = 0.6

We have to fond the heat supply by the fan motor Q

We know that efficiency of any motor is the ratio of shaft power and input power

So $0.6=\frac{0.25}{P_0}$

So $P_0=\frac{0.25}{0.6}=0.4166hP$

So input power of the geothermal motor will be 0.4166 hP

6 0

3 years ago

8. What are used by the project architect to depict different building systems and to show how they correlate to one anothe

grigory [225]

Explanation:

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6 0

2 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago