The energy travels in a disturbance, in an ocean that disturbance is a wave, so the wave makes energy and moves it through the water
Answer:
The change in entropy ΔS = 0.0011 kJ/(kg·K)
Explanation:
The given information are;
The mass of water at 20.0°C = 1.0 kg
The mass of water at 80.0°C = 2.0 kg
The heat content per kg of each of the mass of water is given as follows;
The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg
The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg
Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg
The heat energy of the mixture =
1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)
∴ T = 60°C
The heat content, of the water at 60° = 251.154 kJ/kg
Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462
The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).
<u>Given:</u>
Temperature T = 0.20 μK
<u>To determine:</u>
The de Broglie wavelength of Rubidium atoms
<u>Explanation:</u>
The de broglie wavelength (λ) is related to the temperature (T) as:
λ = h/√2πmkT -----(1)
where h = Planck's constant = 6.626*10⁻³⁴ Js
m = mass of Rubidium = 85.47 amu * 1.66*10⁻²⁷ kg/ 1 amu = 1.419*10⁻²⁵ kg
k = Boltzmann constant = 1.38*10⁻²³ J.K⁻¹
T = temperature = 0.2 μK = 0.2 *10⁻⁶ K
Substituting these values in equation (1) we get:
λ = 6.626*10⁻³⁴ Js/√2π * 1.419*10⁻²⁵ kg * 1.38*10⁻²³ J.K⁻¹ * 0.2 *10⁻⁶ K
= 4.224*10⁻⁷ m
Ans: The de Broglie wavelength is 4.224*10⁻⁷ m
O is an element (Oxygen) and H2O2 is a compound (Hydrogen Peroxide)
Answer:
<em>Ceteris Paribus</em>
Explanation:
The process of examining a change in one variable in a model while assuming that all the other variables remain constant is called <em><u>Ceteris Paribus</u></em>.
<em>Ceteris Paribus</em> is a Latin phrase that means "all other things being equal" or "all other things held constant" in English. The phrase has found application in disciplines like Economics and Statistics. This phrase as being adopted as a process of examining a change in one variable in a model while assuming that all the other variables remain constant to ascertain the relationship between the variables or make deductions from an experimental study. An example of <em>Ceteris Paribus</em> application is the law of demand and supply in Economics. The law of demand states, <em>Ceteris Paribus</em>, the higher the price, the lower the quantity demanded and <em>vice versa. </em>Conversely, the law of supply states, <em>Ceteris Paribus</em>, the higher price, the higher the quantity supply and <em>vice versa</em>.