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3 years ago

The moons light-colored highlights are

Engineering

1 answer:

gulaghasi [49]3 years ago

3 0

The light colored material in areas of the mood is the earliest crust on the moon.

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Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r

irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area = 1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000 × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0

3 years ago

In the idealized Otto cycle, heat is added during: a. Isentropic Compression b. Constant (minimum) volume c. Constant (maximum)

docker41 [41]

Answer:

(b) Constant (minimum) volume

Explanation:

In the idealized Otto cycle there are 4 process that are

Reversible adiabatic compression
Addition of heat at constant volume
Reversible adiabatic expansion
Rejection of constant volume

So from above discussion we can see that heat is added when there is constant (minimum) volume which is given in option (b) so option (b) will be the correct answer

3 0

3 years ago

Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical

Olin [163]

Answer:

The theoretical density for Niobium is $1.87 g/cm^3$ .

Explanation:

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density of the unit cell

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

$a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm$

$1 nm = 10^{-7} cm$

On substituting all the given values , we will get the value of 'a'.

$\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}$

$\rho =1.87 g/cm^3$

The theoretical density for Niobium is $1.87 g/cm^3$ .

6 0

3 years ago

Read 2 more answers

Air enters the compressor of a cold air-standard Brayton cycle with regeneration at 100 kPa, 300 K with a mass flow rate of 6 kg

Marta_Voda [28]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

7 0

3 years ago

WIL MARK U AS BRAINLIEST HELP PLZ

ss7ja [257]

Answer:

The 1st one:Your natural ability

4 0

3 years ago