The two different isotopes have weights :
w1 = 78.918 amu
w2 = 80.916 amu
average weight w3 = 79.903 amu
The mixing of two components can be modeled as
let the fraction of w1 be 'x'
hence
now this is a linear equation in 'x'. Substituting the values we get
x = 0.507
hence the percentage of Br79 = 50.7% and the percentage of BR81 = 49.3%
Answer:
the resulting temperature is 23.37 ⁰C
Explanation:
Given;
mass of the iron, m₁ = 80 g = 0.08 kg
mass of the water, m₂ = 200 g = 0.2 kg
mass of the iron vessel, m₃ = 50 g = 0.05 kg
initial temperature of the iron, t₁ = 100 ⁰C
initial temperature of the water, t₂ = 20 ⁰C
specific heat capacity of iron, c₁ = 462 J/kg⁰C
specific heat capacity of water, c₂ = 4,200 J/kg⁰C
let the temperature of the resulting mixture = T
Apply the principle of conservation of energy;
heat lost by the hot iron = heat gained by the water
Therefore, the resulting temperature is 23.37 ⁰C
Explanation:
It is given that,
Mass of the rim of wheel, m₁ = 7 kg
Mass of one spoke, m₂ = 1.2 kg
Diameter of the wagon, d = 0.5 m
Radius of the wagon, r = 0.25 m
Let I is the the moment of inertia of the wagon wheel for rotation about its axis.
We know that the moment of inertia of the ring is given by :
The moment of inertia of the rod about one end is given by :
l = r
For 6 spokes,
So, the net moment of inertia of the wagon is :
So, the moment of inertia of the wagon wheel for rotation about its axis is . Hence, this is the required solution.