Answer: How to solve for FX and FY?
to find fx(x, y): keeping y constant, take x derivative; • to find fy(x, y): keeping x constant, take y derivative. f(x1,...,xi−1,xi + h, xi+1,...,xn) − f(x) h . ∂y2 (x, y) ≡ ∂ ∂y ( ∂f ∂y ) ≡ (fy)y ≡ f22. similar notation for functions with > 2 variables.
Explanation:
Answer:
The rotation of a planet around it's sun
Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.
The energy of the emitted photon is given by:

where
h is the Planck constant
f is the photon frequency
Substituting

, we find

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
Answer:
Explanation:
a is the acceleration
μ is the coefficient of friction
Acceleration of the object is given by

Velocity at the bottom

after travelling 4m , its velocity becomes 0



Coefficient of kinetic friction
μ = F/N

Therefore, the Coefficient of kinetic friction is 0.31
Answer:
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