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NeTakaya
3 years ago
12

True/False An anemometer displays wind direction, wind speed, altitude and type of precipitation

Engineering
1 answer:
inessss [21]3 years ago
3 0

Answer:FALSE

Explanation:

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A 0.5 m^3 container is filled with a mixture of 10% by volume ethanol and 90% by volume water at 25 °C. Find the weight of the l
svp [43]

Answer:

total weight of liquid = 4788.25 N or 488.09 kg

Explanation:

given data

total volume = 0.5 m³

volume of ethanol = 10 %  of volume = 0.10 × 0.5 = 0.05 m³

volume of water = 90 % at 25 °C of volume = 0.90 × 0.5 = 0.45 m³

to find out

weight of the liquid

solution

we know that density of water at 25  is 997 kg/m³

and density of ethanol is 789 kg/m³

so weight of water is = density × volume × g

put here value and we take g = 9.81

weight of water is = 997 × 0.45 × 9.81

weight of water = 4401.25 N     ......................1

weight of ethanol is = density × volume × g

put here value and we take g = 9.81

weight of ethanol is = 789 × 0.05 × 9.81

weight of ethanol = 387.00 N       ...............2

so total weight of liquid = sum of equation 1 add 2

total weight of liquid =  4401.25 + 387

total weight of liquid = 4788.25 N or 488.09 kg

7 0
3 years ago
(a) For BCC iron, calculate the diameter of the minimum space available in an octahedral site at the center of the (010) plane,
Romashka-Z-Leto [24]

Answer:

a) diameter available = 0.0384 nm

b)The space  is smaller than the carbon atom which has a  radius of  0.077 nm and this simply means that the carbon atom will not conquer these sites

Explanation:

For BCC iron

From Appendix B given,select the lattice parameter ( a ) as = 0.2866 nm

The BCC iron has 4 atomic radii and therefore the body diagonal length = a(3)^\frac{1}{2}

expressing the atomic radius of the BCC iron

4r = a(3)^\frac{1}{2}

insert the value of (a) from appendix B which is = 0.2866 nm

4r = 0.2866 nm (3)^\frac{1}{2}

therefore  r =  0.4964 nm / 4 = 0.1241 nm

Refer again to appendix C given select the atomic radius of the BCC iron as = 0.1241 nm   assuming the atomic radius of the iron are the same

then the radius ratio = 0.62

Refer to the Figure 3.2 given, the amount of space required for an interstitial at the BCC position is between the atoms at the FCC position and also in this space there are two atoms that are equal to a radius of 0.2482 nm

The diameter of the minimum space available

d_{a} = a - r_{a}

r_{a}  = atomic radii = 0.2482 nm

a = 0.2666 nm

therefore

d_{a} = 0.2866 nm - 0.2482 nm = 0.0384 nm

comparing this to the diameter of a carbon atom

The space  is smaller than the carbon atom which has a  radius of  0.077 nm and this simply means that the carbon atom will not conquer these sites

7 0
3 years ago
What resources did Margaret Hutchinson Rousseau use ?
blagie [28]

Answer:

Explanation:

During the Second World War, she oversaw the design of production plants for the strategically important materials of penicillin and synthetic rubber.Her development of deep-tank fermentation of penicillium mold enabled large-scale production of penicillin.She worked on the development of high-octane gasoline for aviation fuel.Her later work included improved distillation column design and plants for the production of ethylene glycol and glacial acetic acid.

3 0
3 years ago
Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state
lord [1]

Answer:

W=-940.36 KJ

Explanation:

Given that

P_1=1\ bar,V_1=4.25 {m^3}

P_2=6.8\ bar

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

W=P_1V_1\ln \dfrac{P_1}{P_2}

Now by putting the values                (1.4 bar =140 KPa)

W=P_1V_1\ln \dfrac{P_1}{P_2}

W=140\times 4.25 \ln \dfrac{1.4}{6.8}

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

7 0
4 years ago
Rod of steel, 200 mm length reduces its diameter (50 mm) by turning by 2 mm with feed speed 25 mm/min. You are required to calcu
diamong [38]

Answer:

125 cm³/min

Explanation:

The material rate of removal is usually given by the formula

Material Rate of Removal = Radial Depth of Cut * Axial Depth of Cut * Feed Rate, where

Radial Depth of Cut = 25 mm

Axial depth of cut = 200 mm

Feed rate = 25 mm/min

On multiplying all together, we will then have

MRR = 25 mm * 200 mm * 25 mm/min

MRR = 125000 mm³/min

Or we convert it to cm³/min and have

MRR = 125000 mm³/min ÷ 1000

MRR = 125 cm³/min

4 0
3 years ago
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