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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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Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?
jasenka [17]

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

8 0
2 years ago
Which body is in equilibrium?
sergey [27]
"(1) a satellite moving around Earth in a circular <span>orbit" is the only option from the list that describes an object in equilibrium, since velocity and gravity are working together to keep the orbit constant. </span>
6 0
3 years ago
Read 2 more answers
A family took a trip in a car traveling East from Greensboro to Wilmington, NC. Use the Graph to answer the questions below.
kotykmax [81]

Answer:

1). Average speed = 1.5 m per second

2). Average velocity = 1.5 m per second

Explanation:

1). Since, speed is a scalar quantity

   Therefore, average speed of the trip = \frac{\text{Total distance covered}}{\text{Total time taken}}

    From the graph attached,

   Total distance covered = 10 + 10 + 20 + 0 + 20 + 30

                                           = 90 meters

   Total time taken = 60 seconds

    Average speed = \frac{90}{60}

                               = 1.5 meter per second

2). Velocity is a vector quantity.

    Therefore, average velocity = \frac{\triangle d}{\triangle t}

                                                   = \frac{d_{60}-d_0}{60-0}

                                                   = \frac{90-0}{60-0}

                                                   = 1.5 meter per second                        

7 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across
NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

A=\frac{\pi }{4}d^{2}  \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper  </em>

<em>    wire: </em>

  L= 2.00 m

From Table  Copper Resistivity p= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

R=\frac{pL}{A}

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

R=\frac{pL}{A}

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

4 0
3 years ago
In creating his definition of horsepower, James Watt, the inventor of the steam engine, calculated the power output of a horse o
Studentka2010 [4]

Answer:

Part a)

F = 182.3 Lb

Part b)

P = 0.55 HP

Explanation:

Diameter of the circle = 24 ft

Diameter = 731.52 cm = 7.3152 m

now the horse complete 144 trips in one hour

so time to complete one trip is given as

t = \frac{3600}{144} s

t = 25 s

now the speed of the horse is given as

v = \frac{2\pi r}{t}

v = \frac{\pi(7.3152)}{25}

v = 0.92 m/s

Part a)

Now we know that the power is defined as rate of work done

it is given as

P = F v

746 = F(0.92)

F = 810.9 N

F = 182.3 Lb

Part b)

Work done to climb up to 3 m height is given by

W = mgh

now we have

Power = \frac{Work}{time}

P = \frac{mgh}{t}

P = \frac{(70kg)(9.81)(3)}{5.0s}

P = 412.02 Watt

now we know that 1 HP = 746 Watt

so we have

P = \frac{412}{746} = 0.55 HP

8 0
3 years ago
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