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:
= 0.331 J / g ° C
Explanation:
We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.
Heat ceded Qh = m1 ce1 ( -
)
Heat absorbed Qc = m2 ce2 ( - T₀)
Body 1 is metal and body 2 is water . Where m are the masses of the two bodies, ce their specific heat and T the temperatures
Qh = Qc
m₁ (
-
) = m₂
(
- T₀)
we clear the specific heat of the metal
= m₂
(
- T₀) / (m₁ (
-
))
= 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))
= 209.2 (9.36) / (75 78.85)
= 1958.11 / 5913.75
= 0.331 J / g ° C