Explanation:
The observation from the gold foil experiment by Rutherford was that, while most of the alpha particles went through the foil almost unaffected, a small fraction of the particles were deflected in directions with large angles away from the original path (including bouncing straight back). This was unexpected and led to a complete revision of the model (as relates to your part 1).
The nuclear model proposes a spacially small but massively charged nucleus. Due to its small size, most alpha particles will pass through the atoms of the gold foil unaffected (they "miss" the small nucleus). But some of them will come very close to the nucleus and those will be deflected strongly by its charge. This is one of the main arguments explaining the observation and speaking for the validity of the nuclear model.
Answer:
31.92 h
Explanation:
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Original amount (N₀) = 1
Amount remaining (N) = ⅛
Number of half-lives (n) =?
N = 1/2ⁿ × N₀
⅛ = 1/2ⁿ × 1
Cross multiply
2ⁿ = 8
Express 8 in index form with 2 as the base.
2ⁿ = 2³
n = 3
Thus, 3 half-lives has elapsed.
Finally, we shall determine the time. This can be obtained as follow:
Half-life (t½) = 10.64 h
Number of half-lives (n) = 3
Time (t) =?
n = t / t½
3 = t / 10.64
Cross multiply
t = 3 × 10.64
t = 31.92 h
Therefore, it will take 31.92 h for lead-212 to decay to one-eighth its original strength.
Answer:
(a) 1.11sec
(b) 14.37m/s
(c) 31.78m
Explanation:
U = 18m/s, A = 37°, g = 9.8m/s^2
(a) t = UsinA/g = 18sin37°/9.8 = 18×0.6018/9.8 = 1.11sec
(b) Ux = UcosA = 18cos37° = 18×0.7986 = 14.37m/s
(c) R = U^2sin2A/g = 18^2sin2(37°)/9.8 = 324sin74°/9.8 = 324×0.9613/9.8 = 31.78m
Answer:
Work done, W =1520 J
Explanation:
We have,
The brakes on a bicycle apply 95 N of force to the wheels. When the brakes are applied, the bicycle comes to a stop in 16 m.
It is required to find the work done by the brakes on the wheels. We know that the product of force and displacement is equal to the work done. It is given by :
So, the work done by the brakes is 1520 J.