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Contact [7]
3 years ago
10

All spiders are very sensitive to vibrations. An orb spider sits at the center of its web and monitors radial threads for vibrat

ions created when an insect lands. Assume that these threads are made of silk with a linear density of 1.0 x 10-5 kg/m under a tension of 0.40 N. If an insect lands in the web 30 cm from the spider, how long will it take for the spider to find out
Physics
1 answer:
Roman55 [17]3 years ago
3 0

Answer:

The time taken by spider to catch insect 1.5 ms

Explanation:

Given:

Linear mass density \mu = 1 \times 10^{-5} \frac{kg}{m}

Tension on the string T = 0.40 N

Distance between web and insect lands x = 30 \times 10^{-2} m

From the formula of velocity in string,

   v = \sqrt{\frac{T}{\mu} }

First find the velocity and then find time,

   v = \sqrt{\frac{0.40}{1 \times 10^{-5} } }

   v = 200 \frac{m}{s}

The time taken is given by,

   v = \frac{x}{t}

   t = \frac{x}{v}

   t = \frac{30 \times 10^{-2} }{200}

   t = 1.5 \times 10^{-3} sec

   t = 1.5 ms

Therefore, the time taken by spider to catch insect 1.5 ms

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Answer:

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In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. the sharp upward
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<u>Answer:</u>

 The skier’s launch speed = 17.08 m/s

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The motion of skiing is projectile motion.

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In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.  

0 = u sin θ - gt  

t = u sin θ/g  

Total time for vertical motion is two times time taken for upward vertical motion of projectile.  

So total travel time of projectile = 2u sin θ/g  

Horizontal motion:  

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In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g  

So range of projectile, R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}  

Vertical motion (Maximum height reached, H) :

We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.  

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

We have θ = 63° and H = 11.8 meter.

Substituting

   11.8=\frac{u^2sin^263}{2*9.81}\\ \\ u^2=\frac{11.8*2*9.81}{sin^263}=\frac{231.516}{0.794} =291.58\\ \\ u=17.08m/s

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In Part I, the independent variable, the one that is intentionally manipulated, is blank.
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<h3>What is volume?</h3>

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<h3>Volume of the tire</h3>

The volume of the tire is the measure of the product of area and thickness of the tire.

The volume of the tire is calculated as follows;

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Thus, the volume of the tire at the given diameter and thickness of tube is determined as  1,128.2 cubic inch.

Learn more about Volume of tire here: brainly.com/question/1972490

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