Answer
Given:
Wavelength = λ = 18.7 cm
= 0.187 m
Amplitude, A = 2.34 cm
Velocity, v = 0.38 m/s
A) Calculate the angular frequency.
Angular frequency,
ω = 2π f
ω = 2π x 2.03
ω = 12.75 rad/s
B) Calculate the wave number:
C)
Since the wave is traveling in the -x direction, the sign is positive between x and t
y (x, t) = A sin(k x - ω t)
y (x, t) = 2.34 sin(33.59 x - 12.75 t)
Δd = 23 cm. When the eta string of the guitar has nodes at both ends, the resulting waves create a standing wave, which can be expressed with the following formulas: Fundamental: L = ½ λ, 1st harmonic: L = 2 ( λ / 2), 2nd harmonic: L = 3 ( λ / 2), Harmonic n: L = n λ / 2, where n is an integer. The rope's speed can be calculated using the formula v = λ f. This speed remains constant based on the tension and linear density of the rope. Now, let's determine the speed with the provided data: v = 0.69 × 196, yielding v = 135.24 m/s. Next, we will find the wavelengths for the two frequencies: λ₁ = v / f₁, which gives λ₁ = 135.24 / 233.08, equaling λ₁ = 0.58022 m; λ₂ = v / f₂ results in λ₂ = 135.24 / 246.94, consequently λ₂ = 0.54766 m. We'll substitute into the resonance equation Lₙ = n λ/2. At the third fret, m = 3, therefore L₃ = 3 × 0.58022 / 2, resulting in L₃ = 0.87033 m. For the fourth fret, m = 4, which gives L₄ = 4 × 0.54766 / 2, equating to L₄ = 1.09532 m. The distance between the two frets is Δd = L₄ – L₃, so Δd = 1.09532 - 0.87033, leading to Δd = 0.22499 m or 22.5 cm, rounded to 23 cm.
This initial stage represents a right triangle. If the aircraft is positioned 400 km to the east and 300 km to the south of the origin while traveling in a straight path, you can form a right triangle with sides measuring 300, 400, and c. You might notice that these dimensions correspond to multiples of the Pythagorean triple 3, 4, 5, meaning that the length of c is 500 km. Alternatively, you would indicate
.
For the second step, assuming I am correctly understanding "degrees south of east," it involves calculating the angle between the horizontal line indicating east and the trajectory of the aircraft. I created a diagram illustrating this (see attached). You could employ a trigonometric function related to one of the angles to find the solution. I selected
. Therefore, I deduce that the angle is 37° south of east.
Answer:
D, C, B, A
Explanation:
The derivative from a velocity-time graph provides the acceleration value.
Segment A
Segment B
Segment C
Segment D
Sorted from the lowest to the highest acceleration:
D, C, B, A
(6-16)/4.0=-2.5 m/s²
The car's acceleration is -2.5 m/s²
