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grin007 [14]
3 years ago
7

What is the pressure on a horizontal surface with an area of 8 msquared that is 4 m​ underwater? Use 1000 kg divided by m cubed

for the density of water and 9.8 m divided by s squared for the acceleration due to gravity. What is the correct relationship to use to find the pressure on the horizontal​ surface?
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
7 0

Answer:

140525 pa

Explanation:

h = depth of water = 4 m

A = Area of surface = 8 m²

\rho = Density of water = 1000 kg m⁻³

tex]g[/tex] = acceleration due to gravity = 9.8 ms⁻²

P_{o} = Atmospheric pressure = 101325 Pa

P = Pressure at the given depth "h"

Pressure at the given depth is given as

P = P_{o} + \rho g h\\P = 101325 + (1000) (9.8) (4)\\P = 140525 pa

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The quadriceps muscles pull on the patella simultaneously. Below are the forces from each
Nostrana [21]

Based on the calculation of the resultant of vector forces:

  1. the resultant force due to the quadriceps is 1795 N
  2. the resultant force due to the quadriceps is 1975 N. Training and strengthening the vastus medialis results in a greater force of muscle contraction.

<h3>What is the resultant force due to the quadriceps?</h3>

The resultant of more than two vector forces is given by:

  • F = √Fₓ² + Fₙ²

where:

  • Fₓ is the sum of the horizontal components of the forces
  • Fₙ is the sum of the vertical components of the forces
  • Fx = F₁cosθ + F₂cosθ + F₃cosθ + F₄cosθ
  • Fₙ = F₁sinθ + F₂sinθ + F₃sinθ + F₄sinθ
  • F₁ = 680N, θ = 90 = 30 = 120°
  • F₂ = 220 N, θ = 90 + 16 = 106°
  • F₃ = 600 N, θ = 90 + 15 = 105°
  • F₄ = 480 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 480 * cos 55

Fx = -280.6 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 480 * sin 55

Fₙ = 1773.1 N

then:

F = √(-280.6)² + ( 1773.1)²

F = 1795.16 N

F ≈ 1795 N

Therefore, the resultant force due to the quadriceps is 1795 N

<h3>What would happen if the vastus medialis was trained and strengthened to contract with 720N of force?</h3>

From the new information provided:

  • F₁ = 680N, θ = 90 = 30 = 120°
  • F₂ = 220 N, θ = 90 + 16 = 106°
  • F₃ = 600 N, θ = 90 + 15 = 105°
  • F₄ = 720 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 720 * cos 55

Fx = -142.95 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 720 * sin 55

Fₙ = 1969.72 N

then:

F = √(-142.95)² + ( 1969.72)²

F = 1974.9 N

F ≈ 1975 N

Therefore, the resultant force due to the quadriceps is 1975 N.

Training and strengthening the vastus medialis results in a greater force of muscle contraction.

Learn more about resultant of forces at: brainly.com/question/25239010

3 0
2 years ago
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