Answer:
a=2.378 m/s^2
Explanation:
a=Δv/Δt------eq(1)
Δv=Vf-Vi=120 km/h-0 km/h=120 km/h
or Δv=33.3 m/sec
or time=t=14s
putting values in eq(1)
a=33.3/14
a=2.378 m/s^2
Based on the calculation of the resultant of vector forces:
- the resultant force due to the quadriceps is 1795 N
- 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:
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
Explanation: The first one
Source: it literally has fusion in the name
Answer:
the final velocity of the car is 59.33 m/s [N]
Explanation:
Given;
acceleration of the car, a = 13 m/s²
initial velocity of the car, u = 120 km/h = 33.33 m/s
duration of the car motion, t = 2 s
The final velocity of the car in the same direction is calculated as follows;
v = u + at
where;
v is the final velocity of the car
v = 33.33 + (13 x 2)
v = 59.33 m/s [N]
Therefore, the final velocity of the car is 59.33 m/s [N]
Answer:
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