Various Math #28 - Q&A

Question 1 Light of a certain frequency has a wavelength of 526 nm in water. What is the wavelength of this light in benzene? Answer Step 1 Wavelength of light in a material: λ = λ₀/n λ ⇒ wavelength in material, λ₀⇒ wavelength of light in vacuum, n ⇒ index of refraction of the material proof: v = λf, n = c/v = λ₀f/ λf = λ₀/λ → f is constant as the number of wave cycles arriving per unit must equal the number leaving per unit time. When light passes from one material into another, the frequency of the wave doesn't change. The boundary surface cannot create or destroy waves; the number arriving per unit time must equal the number leaving per unit time; otherwise, incident and transmitted waves couldn't have a definite phase relationship. In any material, (V = fλ) because f is the same in any material as in vacuum and (v) is always less than the wave speed c in vacuum by the factor n,(λ) is also correspondingly reduced. Thus, the wavelength of the light in a material is less than its wavelength in vacuum by the factor n. Apply: this law can provide us with the index of refraction for each wavelength for the same material by knowing the wavelength in the material. In other problems, we are given the index of refraction for certain material and certain wavelength of light in vacuum and we are asked to get the wavelength in the material. Step 2 From table 33.1 → n₁ = 1.333 → for water From table 33.1 → n₂ = 1.501 → for benzene λ₁ = λ₀/n₁ λ₀ = λ₁n₁ = (526nm)(1.333) = 701.158nm λ₂ = λ₀/n₂ = 701.158nm/1.501 = 467.13nm Result 467.13nm Question 2 A medical technician is trying to determine what percentage of a patient's artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20 × 10^4 Pa, while in the region of blockage it is 1.15 × 10^4 Pa. . Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s, and the specific gravity of this patient's blood is 1.06. What percentage of the cross-sectional area of the patient's artery is blocked by the plaque? Answer Step 1 The plaque presence will cause the speed to change because of the changing in the cross-sectional area of the artery. To grasp the change we employ the Bernoulli's equation P₁ + ρgy₁ + (1/2)ρv₁² = P₂ + ρgy₂ + (1/2)ρv₂² and we will combine it with the continuity equation v₁A₁ = v₂A₂ from which we get A₁/A₂ = v₂/v₁ Step 2 Since two points are close to each other y₁ ≈ y₂ we can rewrite Bernoulli's equation as Step 3 Now, we can express the velocity when the plaque is present, v₂ as follows (1/2)ρv₂² = P₁ - P₂ + (1/2)ρv₁² v₂² = (2p₁ - 2p₂ + ρv₁²)/ρ After inserting the given values we get that v₂² = (2 × 1.2 × 10^4 - 2 × 1.15 × 10^4 + 1.06 × 10³ × 0.3² )/1.06 × 10³ = 1(m/s)² v₂ = 1 m/s Step 4 Now, we recall the continuity equation and use A₁/A₂ = v₂/v₁ to write A₂ = 0.3A₁ So we get that the 70% of the artery is blocked by the plaque Question 3 a. An FM radio station broadcasts at a frequency of 101.3 MHz. What is the wavelength? b. What is the frequency of a sound source that produces the same wavelength in air at 20° Answer For part a) The wavelength is calculated as: λ = c/f = (3.0 × 10^8 m/s) / (101.3 × 10^6 Hz) = 2.96m ≈ 3.0m For part b) The speed of sound in air at 20° given as 343 m/s. Hence, the frequency is calculated as: f = v_sound/λ = 343 m/s / 2.96m = 120Hz Result: α) λ = 2.96m b) f = 1.6Hz Question 4 The electric potential along the x-axis is V = 100e^(-2x) V, where x is in meters. A) What IS E(x) at x=1.0 m? B) What is E(x) at x=1.6m? Express your answer to two significant figures and include the appropriate units. Answer Part A E(x) = -dV(x)/dx = -d/dx (100e^(-2x)) = -100e^(-2x)(-2) = 200e^(-2x) At x=1 E(1) = 200e^(-2(1)) = 27 V/m Part B At x=1.6 E(1.6) = 200e^(-2(1.6)) = 8.2 V/m Question 5 Eliminate the parameter to find a Cartesian equation of the curve x = sin t, y = csc t, 0