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vesna_86 [32]
3 years ago
13

Which statement about the physical change of liquid water boiling into steam is true

Physics
2 answers:
ioda3 years ago
6 0

Answer:

The heat added represents an energy change.

Explanation:

Whenever the heat is supplied to the water at normal atmospheric pressure, its molecules gain the thermal energy and their kinetic energy increases, the inter-molecular bonding decreases due to which its molecules leave the surface of water and convert into the vapor form.

  • The action can be reversed by condensation.
  • The mass of the matter remains conserved irrespective of its phase.
  • There is a loss of weight in the water phase because some of the molecules are now in the gaseous state but if we calculate the total mass of all the water particles before and after the process it remains constant.
sp2606 [1]3 years ago
3 0
Option 1 would be right one ! as the heat added represent itself into the energy change
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Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.
MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>
  • It is given that, the infinite well having a width of 0.050 mm.
  • We have the expression for energy of electron in an infinite well as,

                  E_n=\frac{n^2h^2}{8mL^2}

  • where;

                m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1

  • Thus, the lowest energy of electron in an infinite well is,

                E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

brainly.com/question/20317353

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1 year ago
The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla
kondor19780726 [428]

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A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

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Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]

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