Answer:
3.6 x 10⁶ Pa
Explanation:
A = Area of the heel = 1.50 cm² = 1.50 x 10⁻⁴ m²
m = mass of the woman = 55.0 kg
g = acceleration due to gravity = 9.8 m/s²
Force of gravity on the heel is given as
F = mg
Inserting the values
F = (55) (9.8)
F = 539 N
Pressure exerted on the floor is given as
P = 3.6 x 10⁶ Pa
Answer:
Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).
Explanation:
The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):
José
Txomin
Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).
Answer:
its 1/2 the mass of the object times by its velocity ^ 2
Answer:
153.6 kN
Explanation:
The elastic constant k of the block is
k = E * A/l
k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m
0.12% of the original length is:
0.0012 * 0.25 m = 0.0003 m
Hooke's law:
F = x * k
Where x is the change in length
F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)
The compressive load will generate a stress of
σ = F / A
F = σ * A
F = 80*10^6 * 0.048 * 0.04 = 153.6 kN
The smallest admisible load is 153.6 kN