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3 years ago

Relationships between tables can be illustrated through a(n) ____ relationship diagram.

1 answer:

4 0

I believe the correct term to fill in the blank is "entity". Relationships between tables can be illustrated through an entity relationship diagram. This diagram shows the relationships of entity sets stored in a database. Hope this answers the question. Have a nice day.

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Which of Jafar's statements are correct regarding distance and displacement?

shepuryov [24]

Answer:The distance and magnitude of displacement are sometimes equal." Jafar is correct. The distance traveled and the magnitude of displacement are equal if and only if the path is a straight line in one direction.

Explanation:

5 0

4 years ago

Some ideal gas is constrained in the V"part of anadiabatic container of volume V" V$. The rest of the container is vacuum. When

solmaris [256]

Answer:

Using the log combination rules to reduce the famous Sakur-Tetrode equation, The change in entropy is given as:

∆S = NK*ln(V"V$/V").

Where V"V$ is final Volume (Vf) after constraint's removal,

V" is Initial Volume (Vi) before constraint's removal.

Temperature (T) is constant, Internal Energy, U is constant, N and K have their usual notations

Explanation:

Given in the question, the container is an adiabatic container.

For an adiabatic contain, it does not permit heat to the environment due to its stiff walls. This implies that the Internal Energy, U is kept constant(Q = U). The temperature is also constant (Isothermal). Thus, the famous Sakur-Tetrode equation will reduce to ∆S = NK* In(Vf/Vi).

Vf is the volume after the constraint is removed(Vf = V"V$). Vi is the volume occupied before the constraint is removed (Vi = V")

5 0

3 years ago

How to find the acceleration of a system?

natta225 [31]

The acceleration can be found by:
$\frac{v}{t}$
Where v= velocity and t= time
or
$a = \frac{f}{m}$

6 0

3 years ago

Read 2 more answers

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

7 0

4 years ago

The process of evaluating evidence to form an attitude is called

tatyana61 [14]

The process of evaluating evidence to form an attitude should be called Cognitive evaluation.

5 0

3 years ago