Well, the density of the water is
so i believe that is what the question is asking for :)
False, according to Boyle's law it's pressure increases, volume decreases
The fraction of the water must evaporate to remove precisely enough energy to keep the temperature constant when water at 37°c has a latent heat of vaporization of lv = 580 kcal/kg is 2.58 times 10 to the minus 3.
Vaporization is the process by which a substance is transformed from its liquid or solid state into its gaseous (vapour) state. Boiling is the term for the vaporization process when conditions permit the creation of vapour bubbles within a liquid. Sublimation is the process of directly converting a solid to a liquid.
Boiling and evaporation are the two processes that cause vaporization. Evaporation is the process by which a liquid body's surface changes from a liquid to a gas, as in the case of a drop of water on hot concrete evaporating into a gas. A liquid is said to be boiling when it is heated to the point at which it begins to give off steam, as when you boil water on a stove. The process of converting a substance from its liquid or solid state into its gaseous (vapour) state is known as vaporization.
To learn more about vaporization please visit - brainly.com/question/12625048
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Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
