Question

A spider of mass 6.9 x 104 kg hangs from a 78.1-cm-long thread whose diameter is 2.0 x 10 mm. Youngs modulus of thread is 3

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Given: m = 6.9x10-4 kg L = 78.1 cm = 0.781 m d = 2x103 mm= 2x106m so radius (%) = 1x106m Y=3x109 Namiz Since; Y= (F/A) - F.L
Alright Dude, If that worked for you... don't forget to give THUMBS UP.(that will work for me!)
Please Vote...
If I missed something feel free to leave a comment.
atleast before giving down vote.
and, Thanks for using Chegg- Smarter way to study.

Add a comment
Know the answer?
Add Answer to:
A spider of mass 6.9 x 104 kg hangs from a 78.1-cm-long thread whose diameter is...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A diamond rod and a tungsten rod have the same length and diameter and are subjected...

    A diamond rod and a tungsten rod have the same length and diameter and are subjected to the same force. If the diamond rod stretches by 2.30 times 10^-6 m, by what amount will the tungsten rod stretch? Young's modulus for diamond = 1.22 times 10^12 N/m^2; for tungsten = 3.60 times 10^11 N/m^2. A spider of mass 7.0 times 10^-4 kg hangs from a 75.4-cm-long thread whose diameter is 2.0 times 10^-3 mm. Young's modulus of thread is 3....

  • An orb weaver spider with a mass of 0.23 grams hangs vertically by one of its...

    An orb weaver spider with a mass of 0.23 grams hangs vertically by one of its threads. The thread has a Young's modulus of 4.7 x 109 N/m2 and a radius of 9.8 x 10-6 m. 1) What is the fractional increase in the thread's length caused by the spider? fs = 2) Suppose a 73 kg person hangs vertically from a nylon rope. What radius must the rope have if its fractional increase in length is to be the...

  • An orb weaver spider with a mass of 0.23 grams hangs vertically by one of its...

    An orb weaver spider with a mass of 0.23 grams hangs vertically by one of its threads. The thread has a Young's modulus of 4.7 x 109 N/m2 and a radius of 9.8 x 10-6 m. 1. What is the fractional increase in the thread's length caused by the spider? fs = 2. Suppose a 73 kg person hangs vertically from a nylon rope. What radius must the rope have if its fractional increase in length is to be the...

  • Orb spiders make silk with a typical diameter of 0.15 mm. The Young's modulus of spider silk is 0.2×1010N/m2 and its te...

    Orb spiders make silk with a typical diameter of 0.15 mm. The Young's modulus of spider silk is 0.2×1010N/m2 and its tensile strength is 1000×106N/m2. Part A: A typical large orb spider has a mass of 0.50 g. If this spider suspends itself from a single 12-cm-long strand of silk, by how much will the silk stretch? (answer in mm) Part B: What is the maximum weight that a single thread of this silk could support? (answer in N)

  • How much force does it take to stretch a 10-m-long, 1.0 cm -diameter steel cable by...

    How much force does it take to stretch a 10-m-long, 1.0 cm -diameter steel cable by 3.0 mm ? The Young's modulus of steel is 20×1010N/m2.

  • A person whose mass is 80 kg is being raised up the side of a cliff...

    A person whose mass is 80 kg is being raised up the side of a cliff by a rope with a diameter of 2 cm. They are moving upward with an acceleration of 1 m/s2. The unstretched length of the rope from the top of the cliff to the person is 10 m. The Young's modulus of the rope is 5 X 107 N/m2. What distance does the rope stretch? Ignore the weight of the rope.

  • Problem 3: Suppose a 65-kg mountain climber has a 0.72 cm diameter nylon rope. Randomized Variables...

    Problem 3: Suppose a 65-kg mountain climber has a 0.72 cm diameter nylon rope. Randomized Variables m = 65 kg d = 0.72 cm l = 33 m By how much does the mountain climber stretch her rope, in centimeters, when she hangs 33 m below a rock outcropping? Assume the Young's modulus of the rope is 5 × 109 N/m2.

  • A sign (mass 2100 kg k g ) hangs from the end of a vertical steel...

    A sign (mass 2100 kg k g ) hangs from the end of a vertical steel girder with a cross-sectional area of 0.015 m2 m 2 . Ignore the mass of the girder itself. The elastic modulus for steel is 2.0×1011N/m2 2.0 × 10 11 N / m 2 . A. What is the stress within the girder? B. What is the strain on the girder? C. If the girder is 7.20 mm long, how much is it lengthened?

  • A mass of 50. kg is suspended from a steel wire of diameter 1.0 mm and...

    A mass of 50. kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch? (The Young's modulus for steel is 20 × 10^10 N/m^2) The answer is 34.9 mm Showing all work please explain how this answer came about.

  • Many caterpillars construct cocoons from silk, one of the strongest naturally occurring materials known. Each thread...

    Many caterpillars construct cocoons from silk, one of the strongest naturally occurring materials known. Each thread is typically 2.0 um in diameter, and the silk has a Young's modulus of 4.0 x 109 N/m2 Assuming that there is no appreciable space between the N =1087222 strands parallel strands, how many strands N would be needed to make a rope 8.0 m long that would stretch only 1.06 cm when supporting a pair of 87-kg mountain climbers? Again assuming that there...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT