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bulgar [2K]
2 years ago
5

Spiders may "tune" strands of their webs to give an enhanced response at frequencies corresponding to the frequencies at which d

esirable prey might struggle. Orb web silk has a typical diameter of 0.0020 mm, and spider silk has a density of 1300 kg/m3. To give a resonance at 200Hz to what tension must a spider adjust a 16cm long strand of silk?
Physics
1 answer:
Bingel [31]2 years ago
3 0

Answer:

0.0000167283525619 N

Explanation:

\rho = Density of silk = 1300 kg/m³

d = Diameter = 0.002 mm

r = Radius = 0.001 mm

l = Length = 16 cm

f = Frequency = 200 Hz

Mass of the string is

m=\rho V\\\Rightarrow m=\rho Al\\\Rightarrow m=1300\times \pi (0.001\times 10^{-3})^2\times 0.16\\\Rightarrow m=6.5345127195\times 10^{-10}\ kg

Frequency is given by

f=\dfrac{1}{2l}\sqrt{\dfrac{T}{m/l}}\\\Rightarrow T=4lf^2m\\\Rightarrow T=4\times 0.16\times 200^2\times 6.5345127195\times 10^{-10}\\\Rightarrow T=0.0000167283525619\ N

The tension on the string is 0.0000167283525619 N

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Why is the entropy change negative for ring closures?
Rom4ik [11]

Answer:ring closure result in fewer molecules

Explanation:

entropy is a measure of the amount of disorderliness in a system.

Therefore, the connection causes orderliness in the ring closure

3 0
2 years ago
A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen
NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
2 years ago
Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc
jarptica [38.1K]
Calculate the weight of the table through the equation,

   W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
   W = (0.44 kg)(9.8 m/s²) 
   <em>W = 4.312 N</em>

The components of this weight can be calculated through the equation,

   Wx = W(sin θ) 

and Wy = W(cos θ)

x - component:
   Wx = W(sin θ)
Substituting,
  Wx = (4.312 N)(sin 150°) = <em>2.156 N</em>

  Wy = (4.312 N)(cos 150°) =<em> -3.734 N</em>
6 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor
mr_godi [17]

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

5 0
2 years ago
Read 2 more answers
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