Answer :
To find the pressure exerted on the floor by the heel, we need to follow a few steps systematically. Let's break it down:
1. Convert the area from square centimeters to square meters.
- The area of the heel is given as [tex]\( 1.00 \, \text{cm}^2 \)[/tex].
- Since [tex]\( 1 \, \text{cm} = 0.01 \, \text{m} \)[/tex], we square this conversion factor to change square centimeters to square meters:
[tex]\[ 1.00 \, \text{cm}^2 = 1.00 \times (0.01 \, \text{m})^2 = 0.0001 \, \text{m}^2. \][/tex]
2. Calculate the force exerted by the woman on the heel.
- The force exerted by the woman's weight is calculated using Newton's second law: [tex]\( F = m \cdot g \)[/tex], where [tex]\( m \)[/tex] is the mass and [tex]\( g \)[/tex] is the acceleration due to gravity.
- The woman's mass [tex]\( m = 52.5 \, \text{kg} \)[/tex].
- The acceleration due to gravity [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]:
[tex]\[ F = 52.5 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 515.025 \, \text{N}. \][/tex]
3. Calculate the pressure exerted on the floor.
- Pressure is defined as force per unit area: [tex]\( P = \frac{F}{A} \)[/tex], where [tex]\( F \)[/tex] is the force and [tex]\( A \)[/tex] is the area.
- Substituting the values we have:
[tex]\[ P = \frac{515.025 \, \text{N}}{0.0001 \, \text{m}^2} = 5150249.999999999 \, \text{Pa}. \][/tex]
Therefore, the pressure exerted on the floor by the heel is approximately [tex]\( 5.15 \times 10^6 \, \text{Pa} \)[/tex].
1. Convert the area from square centimeters to square meters.
- The area of the heel is given as [tex]\( 1.00 \, \text{cm}^2 \)[/tex].
- Since [tex]\( 1 \, \text{cm} = 0.01 \, \text{m} \)[/tex], we square this conversion factor to change square centimeters to square meters:
[tex]\[ 1.00 \, \text{cm}^2 = 1.00 \times (0.01 \, \text{m})^2 = 0.0001 \, \text{m}^2. \][/tex]
2. Calculate the force exerted by the woman on the heel.
- The force exerted by the woman's weight is calculated using Newton's second law: [tex]\( F = m \cdot g \)[/tex], where [tex]\( m \)[/tex] is the mass and [tex]\( g \)[/tex] is the acceleration due to gravity.
- The woman's mass [tex]\( m = 52.5 \, \text{kg} \)[/tex].
- The acceleration due to gravity [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]:
[tex]\[ F = 52.5 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 515.025 \, \text{N}. \][/tex]
3. Calculate the pressure exerted on the floor.
- Pressure is defined as force per unit area: [tex]\( P = \frac{F}{A} \)[/tex], where [tex]\( F \)[/tex] is the force and [tex]\( A \)[/tex] is the area.
- Substituting the values we have:
[tex]\[ P = \frac{515.025 \, \text{N}}{0.0001 \, \text{m}^2} = 5150249.999999999 \, \text{Pa}. \][/tex]
Therefore, the pressure exerted on the floor by the heel is approximately [tex]\( 5.15 \times 10^6 \, \text{Pa} \)[/tex].