2. A string with a length of 0.9m that is fixed at both ends
1 answer:
Answer:
2a.) Wavelength = 1.8 m
2b.) F = 66.67 Hz
3a.) Find the attached file
3b.) Wavelength = 0.6 m
Explanation:
Given that the
Length L = 0.9m
Wavelength (λ) = 2L/n
Where n = number of harmonic
If n = 1, then
Wavelength (λ) = 2L = 2 × 0.9 = 1.8 m
b.)
If waves travel at a speed of 120m/s on this string, what is the frequency
associated with the longest wave (first harmonic)?
Given that V = 120 m/s
V = Fλ
But λ = 2L, therefore,
F = V/2L
F = 120/1.8
F = 66.67 Hz
3. b.) If there are two node, the position will be in 3rd position which is 3rd harmonic
Using the same formula,
Wavelength (λ) = 2L/n
Where n = 3
Wavelength (λ) = 2 × 0.9/3
Wavelength (λ) = 0.6 m
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Answer:
32.3 m/s
Explanation:
The ball follows a projectile motion, where:
- The horizontal motion is a uniform motion at costant speed
- The vertical motion is a free fall motion (constant acceleration)
We start by analyzing the horizontal motion. The ball travels horizontally at constant speed of
and it covers a distance of
d = 165 m
So, the total time of flight of the ball is
In order to find the vertical velocity of the ball, we have now to analyze its vertical motion.
The vertical motion is a free-fall motion, so the ball is falling at constant acceleration; therefore we can use the following suvat equation:
where
is the vertical velocity at time t
is the initial vertical velocity
is the acceleration of gravity (taking downward as positive direction)
Substituting t = 3.3 s (the time of flight), we find the final vertical velocity of the ball: