What is the compressive stress f/a on a woman's feet, in units of kilopascals(kpa) , if the woman's weight is spread out evenly over both soles of her shoes? assume her mass is 59.0 kg and each shoe has an area of 11.7 cm2 in contact with the ground (as might be the case for high‑heeled shoes).

247.09 kPa

Explanation:

given,

mass of the woman = 59 kg

area of each shoes = 11.7 cm² = 0.00117 m²

acceleration due to gravity (g) = 9.8 m/s²

weight of the woman = m g

                                    = 59 × 9.8

                                    = 578.2 N

weight is evenly distributed hence the weight on each foot will be equal to

                           =

                           = 289.1 N

compressive stress F/A on the woman's feet

                        =  

                        =  

                        =247094.4 Pa = 247.09 kPa

Hence, compressive stress under women's feet is 247.09 kPa

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach