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lakkis [162]
3 years ago
6

Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det

ect vibrations of their web to determine what has landed in their web, and where. In fact, spiders carefully adjust the tension of strands to "tune" their web. Suppose an insect lands and is trapped in a web. The silk of the web serves as the spring in a spring-mass system while the body of the insect is the mass. The frequency of oscillation depends on the restoring force of the web and the mass of the insect. Spiders respond more quickly to larger - and therefore more valuable - prey, which they can distinguish by the web's oscillation frequency. Suppose a 15 mg fly lands in the center of a horizontal spider's web, causing the web to sag by 4.0 mm .
Assuming that the web acts like a spring, what is the spring constant of the web?
Physics
1 answer:
Leokris [45]3 years ago
3 0

Answer:

0.037 N/m

Explanation:

The web acts as a spring, so it obeys Hook's law:

F=kx (1)

where

F is the force exerted on the web

k is the spring constant

x is the stretching/compression of the web

In this problem, we have:

- The mass of the fly is m=15 mg=15\cdot 10^{-6} kg

- The force exerted on the web is the weight of the fly, so:

F=mg=(15\cdot 10^{-6}kg)(9.81 m/s^2)=1.47\cdot 10^{-4}N

- The stretching of the web is

x=4.0 mm=0.004 m

So if we solve eq.(1) for k, we find the spring constant:

k=\frac{F}{x}=\frac{1.47\cdot 10^{-4} N}{0.004 m}=0.037 N/m

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sergij07 [2.7K]

Answer:

9.60 m/s

Explanation:

The escape speed of an object from the surface of a planet/asteroid is given by:

v=\sqrt{\frac{2GM}{R}}

where

G is the gravitational constant

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R is the radius of the planet/asteroid

In this problem we have

\rho = 2.02\cdot 10^6 g/m^3 is the density of the asteroid

V=3.09\cdot 10^{12}m^3 is the volume

So the mass of the asteroid is

M=\rho V=(2.02\cdot 10^6 g/m^3)(3.09\cdot 10^{12} m^3)=6.24\cdot 10^{18} g=6.24\cdot 10^{15} kg

The asteroid is approximately spherical, so its volume can be written as

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v=\sqrt{\frac{2(6.67\cdot 10^{-11})(6.24\cdot 10^{15})}{(9036)}}=9.60 m/s

7 0
4 years ago
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Arte-miy333 [17]

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Explanation:

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Explanation:

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Answer:

Natural frequencies of oscillation and typical earthquake frequencies should be different.

Damping on the structure should be large.

Explanation:

The natural frequency of the structure must be different from the typical earthquake frequency, the more different the better. This is because if both frequencies were the same or similar there is a risk that the building will <u>resonate </u>and collapse.

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