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

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.