All categories

3 years ago

Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?

Engineering

1 answer:

lianna [129]3 years ago

5 0

Answer:

Both B and G ( Hexagonal and Tetragonal )

Explanation:

The crystals system listed below has the following relationship for the unit cell edge lengths; a = b ≠ c ( hexagonal and Tetragonal )

hexagonal ; represents a crystal system which has three equal axes that have an angle of 60⁰ between them while Tetragonal denotes crystals that have three axes which have only two of its axes equal in length.

You might be interested in

Visual aids are useful for all of the following reasons except

Ksivusya [100]

Answer:

Explanation:

their presence allows the speaker to take a rest from talking

I couldnt see the question

4 0

2 years ago

A rigid, sealed tank initially contains 2000 kg of water at 30 °C and atmospheric pressure. Determine: a) the volume of the tank

Bad White [126]

Given:

mass of water, m = 2000 kg

temperature, T = $30^{\circ}C$ = 303 K

extacted mass of water = 100 kg

Atmospheric pressure, P = 101.325 kPa

Solution:

a) Using Ideal gas equation:

PV = m $\bar{R}$ T (1)

where,

V = volume

m = mass of water

P = atmospheric pressure

$\bar{R} = \frac{R}{M}$

R= Rydberg's constant = 8.314 KJ/K

M = molar mass of water = 18 g/ mol

Now, using eqn (1):

$V = \frac{m\bar{R}T}{P}$

$V = \frac{2000\times \frac{8.314}{18}\times 303}{101.325}$

$V = 2762.44 m^{3}$

Therefore, the volume of the tank is $V = 2762.44 m^{3}$

b) After extracting 100 kg of water, amount of water left, m' = m - 100

m' = 2000 - 100 = 1900 kg

The remaining water reaches thermal equilibrium with surrounding temperature at T' = $30^{\circ}C$ = 303 K

At equilibrium, volume remain same

So,

P'V = m' $\bar{R}$ T'

$P' = \frac{1900\times \frac{8.314}{18}\times 303}{2762.44}$

Therefore, the final pressure is P' = 96.258 kPa

4 0

3 years ago

typedef struct bitNode { int data; struct bitNode *left; struct bstNode *right; } bstNode; int solve(bstNode* root) { if (root =

Sergio039 [100]

Answer:

The ten numbers to be filled in the blanks are: 18, 7, 7, 11, 18, 36, 3, 8, 13, 50.

Explanation:

keeps on going to the left node until node->left == NULL;

now at Node 18;

left = right = 0; hence condition is not satisfied

18 is printed first.

the value 18 is returned .

Then we reach at 4;

from there we move to 7;

just like 18, similar things happen with 7 and 7 is printed, the value 7 is returned.

Now coming to Node 4,

left = 0, right = 7 ; hence the condition is satisfied & res = 7; 7 is printed.

For Node 16, left = 7 ; right = 11(but for this we visit 11 first and 11 is printed)

for 16; condition is satisfied; res = 7 + 11 = 18 ; 18 is printed

Now for 5; left = right = 18; the condition is satisfied; so res = 18 + 18 = 36; 36 is printed

Next we visit Node 3; 3 is printed & 3 is returned

Then Node 8 ; 8 is printed & 8 is returned

for Node 13; left = 3, right = 8 ; condition is not satisfied, 13 is printed.

For Node 50; left = 36 right = 13 ; condition is not satisfied hence 50 is printed.

So the order of printing is 18 7 7 11 18 36 3 8 13 50.

4 0

3 years ago

A column carries 5400 pounds of load and is supported on a spread footing. The footing rests on coarse sand. Design the smallest

Citrus2011 [14]

Answer:

Following are the responses to the given question:

Explanation:

7 0

2 years ago

Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph

kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

$S.G=\frac{\rho_{sw}}{\rho_{w}}$