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
Answer:
(a): a = 0.4m/s²
(b): α = 8 radians/s²
Explanation:
First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):
a= (Vf-Vi)/t = (2m/s)/t
a: linear acceleration.
Vf: speed at the end of the ramp.
Vi: speed at the beginning of the ramp (zero).
d= (1/2)×a×t² = 5m
d: distance of the ramp (5m).
We replace the first equation in the second to determine the travel time on the ramp:
d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s
And the linear acceleration will be:
a = (2m/s)/5s = 0.4m/s²
Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:
perimeter = π×diameter = π×0.1m = 0.3142m
To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:
α = (0.4m/s²)/(0.05m) = 8 radians/s²
α: angular aceleration.
The sound wave will have traveled 2565 m farther in water than in air.
Answer:
Explanation:
It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.
Distance = Velocity × Time.
So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.
As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.
Distance = 343 × 2.25 =771.75 m
And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.
Distance = 1483×2.25=3337 m.
Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.
Difference in distance covered in water and air = 3337-772 m = 2565 m
So the sound wave will have traveled 2565 m farther in water than in air.
Answer: A
Explanation: How large a parachute is (in other words, the parachute's surface area) affects its air resistance, or drag force. ... In the case of these parachutes, the drag force is opposite to the force of gravity, so the drag force slows the parachutes down as they fall.
