Answer:
v = 17.71 m / s
Explanation:
We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity
v² = v₀² - 2 g (y -y₀)
v² = 0 - 2g (y -y₀)
when it hits the stone the height is zero and part of the height of the seagull I
v² = 2g y₀
v = Ra (2g i)
let's calculate
v =√ (2 9.8 16)
v = 17.71 m / s
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Answer:
Explanation:
Q1 = 35 nC = 35 x 10^-9 C
m = 3.5 micro gram = 3.5 x 10^-9 Kg
d = 35 cm = 0.35 m
(a) The electrostatic force between the two charges is balanced by the weight of another charge.
F = m g
(b) By substituting the values
Q2 = 13.34 x 10^-12 C
Q2 = 0.0134 nC
Referring to Compton scattering
Δλ = h/m₀c (I- cos Ф)
λ' =λ = (0,0242×10⁻¹⁰) (1- cos 60°)
λ= λ' -(0.0242 × 10⁻¹⁰) (1- cos 60°)
7.19 ˣ 10⁻¹²m
The increased potential is given by
Vₐc = hc/eλ = 6.625 × 10 ⁻³⁴ J,s) ( 3× 10⁸ m/s ( 1.6 ˣ 10 ⁻¹⁰C)
(7.19 ˣ 10⁻¹²m)
173kV.
