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:3.56 nanosecond
In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:
time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft= 3.56
second or 3.56 nanosecond
Answer: 13.2 seconds.
Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0
S= distance travelled
a = acceleration due gravity
t= time.
1 foot = 0.305m so,
S= 2860 feet =872.3m
S= ut+1/2 at²
872.3 = 0×t + 1/2×10 × t²
872.3 =0 + 5t²
T²= 872.3/5
T²= 174.46
Take the square root of T we then have;
t = 13.2 seconds to one decimal place.
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
Answer:
6.5454 m
Explanation:
Let the distance from the wire carrying 3 A current is x
Then the distance from the the carrying current 8 A is 24-x
We know that magnetic field due to long wire is given by 
It is given that magnetic field is zero at some distance so
Here 
So 
