Answer:
Explanation:
m₂ is hanging vertically and m₁ is placed on inclined plane . Both are in limiting equilibrium so on m₁ , limiting friction will act in upward direction as it will tend to slip in downward direct . Tension in cord connecting the masses be T .
For equilibrium of m₁
m₁ g sinα= T + f where f is force of friction
m₁ g sinα= T + μsx m₁ g cosα
m₁ g sinα - μs x m₁ g cosα = T
For equilibrium of m₂
T = m₂g
Putting this value in equation above
m₁ g sinα - μs x m₁ g cosα = m₂g
m₂ = m₁ sinα - μs x m₁ cosα
Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer,
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :
Time period of oscillation measured by the observer is :
So, the time period of oscillation measured by the observer is 5.79 seconds.
Answer:
Explanation:
Given that:
p = magnitude of charge on a proton =
k = Boltzmann constant =
r = distance between the two carbon nuclei = 1.00 nm =
Since a carbon nucleus contains 6 protons.
So, charge on a carbon nucleus is
We know that the electric potential energy between two charges q and Q separated by a distance r is given by:
So, the potential energy between the two nuclei of carbon is as below:
Hence, the energy stored between two nuclei of carbon is .