No, they have different definitions
C, electrons. Ion<span> of an </span>element has<span> the </span>same<span> nucleus, the </span>same number<span> of protons and neutrons, with a </span>different number<span> of electrons.</span>
<h2>K.E/P.E = m/k tan²φ x ω²</h2>
Explanation:
The given position of block x = x₀ cos(ωt + φ)
The velocity of block v = dx/dt = - x₀ sin(ωt + φ) x ω
The kinetic energy = 1/2 mv² = 1/2 m x₀² sin²(ωt + φ) x ω²
The potential energy of spring = 1/2 k x² , where k is the spring constant
Thus P.E = 1/2 x k x x₀² cos²(ωt + φ)
When t = 0
K.E = 1/2 m x₀²sin²φ x ω²
P.E = 1/2 k x₀² cos²φ
Dividing these , we have
K.E/P.E = m/k tan²φ x ω²
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
