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Which of the following describes a polar orbit?

Low altitude is the answer

4 0

3 years ago

A binary liquid system exhibits LLE at 25°C. Determine from each of the following sets of miscibility data estimates for paramet

Phoenix [80]

Answer:

(a) - A12 = A21 = 2.747

(b) - A12 = 2.148; A21 = 2.781

(c)- A12 = 2.781; A21 = 2.148

Explanation:

(a) - x1(a) = 0.1 | x2(a) = 0.9 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = A21 = 2.747

(b) - x1(a) = 0.2 | x2(a) = 0.8 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = 2.148; A21 = 2.781

(c) - x1(a) = 0.1 | x2(a) = 0.9 | x1(b) = 0.8 | x2(b) = 0.2

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = 2.781; A21 = 2.148

7 0

3 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

Define Plastic vs elastic deformation.

Snowcat [4.5K]

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

3 0

3 years ago

The tolerance for a geometric can be easily identified by reading the _________.

baherus [9]

Answer:

C

Explanation:

4 0

3 years ago

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