Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.
to have a fundamental frequency at 150hz , to what tension must a spider adjust a 14cm -long strand of silk?

T=2.94*10^-10  N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

l=length of the spider silk, 14cm

velocity of wave = √(T/μ)          

where T = tension and

μ = mass per unit length)

λ/2=l

for fundamental frequency λ/2 =14cm    

 (λ= wavelength of standing wave;  as there will be no node

   except the endpoints of silk strand)

               λ = 28 cm = 0.28 m

and since frequency * wavelength = speed of wave. we have,

                  150 * 0.28 = √(T/μ)                                        (#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

         = [π * (10 * 10^(-6))²] * 1300  = 13π * 10^(-8).

now putting this in equation (#) we get

    150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)²     =  

2.94*10^-10  N/m.

c, 1a hope this

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer