Ask question

Ask question

All categories

3 years ago

6

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

t which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m^3. To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

1 answer:

Nat2105 [25]3 years ago

8 0

Answer:

Explanation:

Given

diameter $d=20 \mu m$

density $\rho =1300 kg/m^3$

frequency $\nu =150 Hz$

Length of silk strand $L=14 cm$

Velocity in the string is as follows

$\nu =\sqrt{\frac{T}{\mu }}$

The expression for Fundamental Frequency

$f=\frac{\nu }{2l}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}$

$f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}$

Squaring

$f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}$

$T=4\rho \cdot A\cdot l^2\cdot f^2$

$T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2$

$T=7.2\times 10^{-4} N$

You might be interested in

A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference

Andre45 [30]

a) $400 \Omega$

b) 0.43 V

c) 0.44 %

Explanation:

a)

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

$V=E-Ir$ (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when $R_1=550 \Omega$ , $V_1=0.25 V$

Using Ohm's Law, the current is:

$I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A$

2) when $R_2=1000 \Omega$ , $V_2=0.31 V$

Using Ohm's Law, the current is:

$I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A$

Now we can rewrite eq.(1) in two forms:

$V_1 = E-I_1 r$

$V_2=E-I_2 r$

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

$V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega$

b)

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

$V=E-Ir$

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

$V=0.25 V$ is the voltage

$I=4.5\cdot 10^{-4}A$ is the current

$r=400\Omega$ is the internal resistance

Solving for E,

$E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V$

c)

In this part, we are told that the area of the cell is

$A=4.0 cm^2$

While the intensity of incoming radiation (the energy received per unit area) is

$Int.=5.5 mW/cm^2$

This means that the power of the incoming radiation is:

$P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W$

This is the power in input to the resistor.

The power in output to the resistor can be found by using

$P'=I^2R$

where:

$R=1000 \Omega$ is the resistance of the resistor

$I=3.1\cdot 10^{-4} A$ is the current on the resistor (found in part A)

Susbtituting,

$P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W$

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

$\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%$

7 0

3 years ago

The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and

Pie

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:

X = 4t^2

X = 4(2.10)^2

X = 17.64 m

b) The position at t = 2.10 + ∆t s will be:

X = 4(2.10 + ∆t)^2

X = 17.64 + 4∆t^2 + 16.8∆t m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:

∆X/∆t = 4∆t + 16.8

Taking the limit as ∆t approaches to zero we get:

Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

Velocity = 16.8 m/s

8 0

3 years ago

Read 2 more answers

Air at the poles tends to flow close to the surface toward the equator. What can you conclude about the characteristics of this

LenaWriter [7]

Answer:

That the polar air has has more pressure than the air at the equator.

Explanation:

5 0

3 years ago

Why are some consolation visible to New York State observers at midnight during April , but not visible at midnight during Octob

Svetach [21]

"Midnight" means looking away from the Sun. But in 6 months from April to October the earth goes halfway around the Sun. So midnight in April and midnight in October are exactly opposite directions.

3 0

3 years ago

Standards for all managers ethical responsibilities are covered in a company's?

katen-ka-za [31]

Answer:

C. Code of ethics

Explanation:

The code of ethics of a company is the document where all the values that the members of the company should live and obbey, they are often guideliness on how to act and beheave and how to treat customers, providers and co-workers. IT also has all of the standards for all managers ethical responsabilities, and ethical responsabilities of the general of the employees of the company.

5 0

3 years ago

Read 2 more answers