In a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.
To find the answer, we have to know more about the transformer.
<h3>
How transformer works?</h3>
- An item utilized in the transfer of electric energy is a transformer.
- AC current is used for transmission.
- It is frequently used to modify the supply voltage between circuits without altering the AC frequency.
- The fundamentals of mutual and electromagnetic induction govern how the transformer operates.
- Magnetic field through the primary coil changes when primary coil current varies. the iron core of the secondary coil likewise has a magnetic field.
- EMF is therefore generated in the secondary coil.
Thus, we can conclude that, in a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.
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Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
<h3>How to solve for the time interval</h3>
We have y = 0.175
y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.5
99.62 = pi/6
t1 = 5.257 x 10⁻³
99.6t = pi/6 + 2pi
= 0.0683
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
b. we have k = 1.25, w = 99.6t
v = w/k
99.6/1.25 = 79.68
s = vt
= 79.68 * 0.0683
= 5.02
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complete question
A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?
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.
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Answer:
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