A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
1 answer:
Answer:
a. Speed = 1.6 m/s
b. Amplitude = 0.3 m
c. Speed = 1.6 m/s
Amplitude = 0.15 m
Explanation:
a.
The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:
frequency = = 0.25 Hz
The wavelength of the wave is the distance between consecutive crests of wave. Therefore,
Wavelength = 6.4 m
Now, the speed of the wave is given as:
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
<u>Speed = 1.6 m/s</u>
<u></u>
b.
Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:
Amplitude = (0.5)(0.6 m)
<u>Amplitude = 0.3 m</u>
<u></u>
c.
frequency = = 0.25 Hz
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
<u>Speed = 1.6 m/s</u>
<u></u>
Amplitude = (0.5)(0.3 m)
<u>Amplitude = 0.15 m</u>
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Answer:
Number of electron wavelength will Double
Explanation:
let the radius = r and wavelength = λ
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Using Bohr's angular momentum quantization to show this
attached below
Answer:
a) I = 1,944 10⁻⁴⁶ Kg m²
, b) I = 7,915 10⁻⁵¹ Kg m²
Explanation:
a) The moment of inertia of point masses is
I = m r²
The nuclei have a very small size (10-15 m) so we can consider them punctual.
The distance to the center of mass passing through the middle of the two nuclei of equal mass
r = 0.6 10⁻¹⁰ m
The moment of total inertia
I = I₁ + I₂ = 2 I₀
I = 2 2.7 10⁻²⁶ (0.6 10⁻¹⁰)²
I = 1,944 10⁻⁴⁶ Kg m²
The kinetic energy of the rotation is
w = h / 2π
K = ½ I w²
K = ½ 1,944 10⁻⁴⁶ (h / 2π)² = ½ 1.944 10⁻⁴⁶ (6.63 10⁻³⁴ / 2π)²
K = 2.16 10⁻¹¹⁴ J (1eV / 1.6 10⁻¹⁹ J)
K = 2.16 10⁻⁹⁵ eV
B) the moment of inertia of the electron in the orbit, we can calculate it with the parallel axis theorem
I = + m R²
I = r² +
R²
Where R we calculate it by Pythagoras
R² = (0.6 10⁻¹⁰)² + (0.5 10⁻¹⁰) 2
R = √ (0.61 10⁻²⁰)
R = 0.78 10⁻¹⁰ m
I = 9.1 10⁻³¹ (0.5 10⁻¹⁰)² + 9.1 10⁻³¹ (0.78 10⁻¹⁰)²
I = (2,275 +5.54) 10⁻⁵¹
I = 7,915 10⁻⁵¹ Kg m²
The rotation energy of the electron
K = ½ I w²
If the angular velocity is the electrons outside the core, its kinetic energy is much lower by an order 10⁵, but the angular velocity of the electrons is much higher.
Explanation:
Below is an attachment containing the solution
