suppose the spring in the sample problem is replaced with a spring that stretches 36 cm from its equilibrium position
1 answer:
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Answer:
L=0.654 m
Explanation:
<u>Concepts and Principles </u>
1- The speed of sound in air is expressed as a function of the temperature of air as follows:
v=(331 m/s)√(1+T_C/273°C) (1)
where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.
<u>Standing Wave Patterns in Pipes: </u>
A pipe open at both ends can have standing wave patterns with resonant frequencies:
f=v/λ=nv/2L n=1,2,3.........
where v is the speed of sound in air.
<u>Given Data </u>
f_1 (fundamental frequency of the flute) = 262 Hz
T (temperature of the air) = 20°C
The flute is open at both ends.
<u>Required Data </u>
We are asked to determine the length of the tube.
<u>Solution</u><u> </u>
The speed of sound in air at temperature T = 20°C is found from Equation (1):
v=(331 m/s)√(1+T_C/273°C)
=342.91 m/s
The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):
f=v/2L
Solve for L:
L=v/2f_1
L=0.654 m
Answer:
a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m - m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N