Work, Energy and Power
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New answer posted
a month agoContributor-Level 10
First, find the initial velocity (v) of the ball as it leaves the machine, using the kinematic equation for maximum height:
v² = u² + 2as ⇒ 0 = v² - 2gh_max
v² = 2gh_max = 2 * 10 * 20 = 400
v = 20 m/s.
Now, apply the work-energy theorem to the ball while it is being pushed by the machine. The work done by the constant force F equals the change in kinetic energy of the ball.
Work done W = F * d
Change in K.E. = (1/2)mv² - 0
F * 0.2 = (1/2) * 0.15 * (20)²
F * 0.2 = (1/2) * 0.15 * 400 = 0.15 * 200 = 30
F = 30 / 0.2 = 150 N.
New answer posted
a month agoContributor-Level 10
ΔE = have = 13.6Z² (1/n? ² - 1/n? ²)eV.
hv = 13.6 (1²) (1/n² - 1/ (n+1)²) = 13.6 (n+1)²-n²)/ (n² (n+1)²)
v = (13.6/h) * (2n+1)/ (n² (n+1)²). For n>>1, v ≈ (13.6/h) (2n/n? ) ∝ 1/n³.
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