Respuesta :
Answer: 2.35*10^-4 N
Explanation:
The linear mass density is
μ = m / L
Recall, ρ = m / V, so that, m = ρV
Substitute this in the equation
μ = ρV / L
Also, recall, V = AL. Substituting this in the equation, we have
μ = ρAL / L = ρA
Again, recall, A = πr², so that
μ = ρπr²
μ = 1300 * 3.142 * [10*10^-6]²
μ = 4.085*10^-7 kg/m
the wavelength of the wave in the silk, λ = 2L
λ = 2 * 0.12 = 0.24 m
The wave speed, v = fλ
v = 100 * 0.24 = 24 m/s
v = √(T / μ)
v² = T / μ
T = v²μ
T = 24² * 4.085*10^-7
T = 2.35*10^-4 N
Answer:
The tension is 2.35x10⁻⁴N
Explanation:
Given:
d = 20 μm = 20x10⁻⁶ m
L = 12 cm = 0.12 m
f = 100 Hz
The transverse velocity is:
[tex]v=\sqrt{\frac{T_{s} }{\mu } }[/tex] (eq. 1)
Where
Ts = web´s tension
μ = linear mass density
The linear mass density is:
[tex]\mu =\frac{\pi d^{2}\rho }{4}[/tex] (eq. 2)
Where
ρ = density of the spider silk = 1300 kg/m³
The fundamental frequency is:
[tex]f=\frac{v}{2L}[/tex]
Replacing eq. 1 and 2 and clearing the tension Ts:
[tex]f=\frac{1}{2L} \sqrt{\frac{4T_{s} }{\pi d^{2}\rho } } \\T_{s} =\pi d^{2} f^{2} L^{2} \rho[/tex]
[tex]T_{s} =\pi *(20x10^{-6} )^{2}*(100 )^{2} *(0.12 )^{2}*1300=2.35x10^{-4} N[/tex]
