Question 1The heels on a pair of women's shoes have radii of 0.50 cm at the bottom. If 30% of the weight of a woman weighing 480 Nis supported by each heel, find the stress on each heel.Question 2For Satety in climbing., a mountaineer uses a nylon rope that is 50 m long and 0 cm in diameter when supporting a 90-kgclimber the rope elongates 1.6m. Find Young modulusQuestion 3Bone has a Young's modulus of about 18 × 109 P a Under compression, it can withstand a stress of about 160 x 106 Pabefore breaking Assume that a femur (thighbone) is 0.50 m long, and calculate the amount of compression this bone canwithstand before breaking.

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Question 1The heels on a pair of women&#039;s shoes have radii of 0.50 cm at the bottom. If 30% of the weight of a woman weighing 480 Nis supported by each heel, find the stress on each heel.Question 2For Satety in climbing., a mountaineer uses a nylon rope that is 50 m long and 0 cm in diameter when supporting a 90-kgclimber the rope elongates 1.6m. Find Young modulusQuestion 3Bone has a Young&#039;s modulus of about   18  ×  109  P  a Under compression, it can withstand a stress of about 160 x 106 Pabefore breaking Assume that a femur (thighbone) is 0.50 m long, and calculate the amount of compression this bone canwithstand before breaking.

Kamila Donovan · Accepted Answer

Question 1 )given  r  =  0.5  c  m    =  0.005  m30 % of the woman of 480 Nso   F  =  480  ×  0.3  F  =  144  N    A  =  π      r    2      =  3.14  ×      0.005    2      A  =  7.85  ×      10          −      5            m    2  stress is defined as force for unit areastress = F / A  =      144    7.85    ×      10          −      5      stress   =  1.834394  ×      10    6    N      /        m    2  so the stress on each wheel is   1.834394  ×      10    6    N      /        m    2  Question 2 )given  L  =  50  m    D  =  0.01  m    r  =  D      /    2    A  =  3.14  ×  (  D      /    2      )    2      F  =  900  N    d  L  =  1.6  m    Y  =  F  L      /    A  d  L    =  900  ×  50      /    7.85  ×      10          −      5        ×  0.01    Y  =  3.58  ×      10    8    N      /        m    2  so the Young’s Modulus (Y) for the rope is   Y  =  3.58  ×      10    8    N      /        m    2  Question 3 )  Y  =  18  ×      10    9    P  a   or   N      /        m    2  stress   =  160  ×      10    6    N      /        m    2    L  =  0.5  m  d  L  =  stress   ×  (      L    Y    )  =  160  ×      10    6    ×  (  0.5      /    18  ×      10    9    )  d  L  =  4.44  ×      10          −      3        mor  d  L  =  0.444  c  mor  d  L  =  4.44  m  m

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