Answer:
2081.65 m
Explanation:
We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:
Height (h) = 3000 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
3000 = ½ × 10 × t²
3000 = 5 × t²
Divide both side by 5
t² = 3000 / 5
t² = 600
Take the square root of both side
t = √600
t = 24.49 s
Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:
Horizontal velocity (u) = 85 m/s
Time (t) = 24.49 s
Horizontal distance (s) =?
s = ut
s = 85 × 24.49
s = 2081.65 m
Thus, the load should be released from 2081.65 m.
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.
To solve this problem it is necessary to apply the concepts related to wavelength as a function of frequency and speed, as well as to determine the wavelength as a function of length.
From the harmonic vibration generated we know that the total length of the string will be equivalent to a half of the wavelength, that is

Where,
Wavelength
Therefore the wavelength for us would be,

From the relationship of speed, frequency and wavelength we know that



Therefore the speed of the wave is 232.75m/s
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