All categories

3 years ago

What's the opposite of vaporization

Physics

2 answers:

Temka [501]3 years ago

8 0

Condensation is the opposite

Nesterboy [21]3 years ago

7 0

Condensation. Remember, Vaporization happens when energy is taken in (enfothermic) the opposite will be the process that releases energy ( exothermic) which will be condensation. Put ice in a glass of water. the ice melts, taking in energy from the water in the glass, which in turn takes heat energy away from the vapor in the surrounding air, thus causing the water vapor in the air to condense.

You might be interested in

Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.

Vesnalui [34]

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

$ma = \text{Mass}\times \text{Acceleration}$

$ma = kg \cdot \frac{m}{s^2}$

At the same time we have that

$\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2$

$\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2$

$\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}$

Therefore there is not have same units and both are not consistent and the correct answer is B.

5 0

3 years ago

The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc

prohojiy [21]

Answer:

The answer is 218

Explanation:

Weight = mass * gravitational acceleration

weight is represented by F

F = 25kg (8.7)

(I'm pretty sure that you don't have to include the meters per second/per second thing)

4 0

3 years ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

4 0

3 years ago

Solve the equation below for vi.<br> d=vit +.5at2

Andrej [43]

Answer: $\dfrac{d-0.5at^2}{t}=v_i$

<u>Explanation:</u>

$d=v_it+0.5at^2\\\\\\\text{Subtract}\ 0.5at^2\ \text {from both sides:}\\d-0.5at^2=v_it\\\\\\\text{Divide both sides by t:}\\\dfrac{d-0.5at^2}{t}=v_i$

8 0

3 years ago

If a person can't accompdate and the glasses is +2.50D. At which distance will the person see cleary

snow_tiger [21]

I believe the correct answer is 40 cm.

5 0

3 years ago