When light strikes an object, what happens to some of the light?

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:

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:

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

Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?

Ganezh [65]

As we know that the formula of kinetic energy will be

$KE = \frac{1}{2} mv^2$

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

$KE = \frac{1}{2}(2)(1^2)$

$KE = 1 J$

7 0

3 years ago

An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination o

Ann [662]

Answer:

θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

v_fly² = v_nort² + v_air²

v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

sin θ = v_air / v_fly

θ = sin⁻¹ (v_air / v_fly)

let's calculate

θ = sin⁻¹ (10/120)

θ = 4.78º

with respect to the vertical or 4.78 to the north-east

5 0

3 years ago

A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude

marusya05 [52]

Answer:

a) 8.83*10⁵ m/s b) 2.80*10⁶ m/s

Explanation:

a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.

By definition, the force due to the electric field is as follows:

F = q*E = e*E (1)

where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.

According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:

F = mp*a (2)

From (1) and(2), being left sides equal, right sides must be equal too:

$F = e*E = mp*a$

Solving for a:

$a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2$

⇒ a = 1.3*10¹³ m/s²

As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$