Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
= K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
Answer:
a = 2.94 m/s²
Explanation:
In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:
Unbalanced Force = Frictional Force
ma = μR = μW
ma = μmg
a = μg
where,
a = maximum acceleration for the cup not to slip = ?
μ = coefficient of static friction = 0.3
g = acceleration due to gravity = 9.8 m/s²
Therefore,
a = (0.3)(9.8 m/s²)
<u>a = 2.94 m/s²</u>
Answer:
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon)
A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by. E=hf=hcλ(energy of a photon) E = h f = h c λ (energy of a photon) , where E is the energy of a single photon and c is the speed of light.
The answere is No pain, no gain
