Answer:
The pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²
Explanation:
Given;
mass of the woman, m = 55 kg
diameter of the circular heel, d = 6.0 mm
radius of the heel, r = 3.0 mm = 0.003 m
Cross-sectional area of the heel is given by;
A = πr²
A = π(0.003)²
A = 2.8278 x 10⁻⁵ m²
The weight of the woman is given by;
W = mg
W = 55 x 9.8
W = 539 N
The pressure exerted by the woman on the floor is given by;
P = F / A
P = W / A
P = 539 / (2.8278 x 10⁻⁵ )
P = 1.9061 x 10⁷ N/m²
Therefore, the pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²
Answer:
Explanation:
la frecuencia = ω/2π, nada cambio
v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima
a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima
la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía
Answer:
T = 0.96 seconds
Explanation:
Given that,
The speed of wave, v = 2.6 m/s
The distance between wave crests, 
We need to find the period of the wave motion. Let T be the period. So,
So, the period of the wave motion is equal to 0.96 seconds.
Since the acceleration is uniform, we can calculate it from the data we are given:
a = (vf - vi)/2
where vf=33 m/s and vi=11 m/s
Then use Suvat's equation:
x(t) = vi*t + 0.5 * a * t
where t=20s
