Silica aerogel, an extremely porous, thermally insulating material made of silica, has a density of \(1.00 \mathrm{mg} / \mathrm{cm}^{3}\). A thin circular slice of aerogel has a diameter of \(2.00 \mathrm{~mm}\) and a thickness of \(0.10 \mathrm{~mm}\). a) What is the weight of the aerogel slice (in newtons)? b) What is the intensity and radiation pressure of a \(5.00-\mathrm{mW}\) laser beam of diameter \(2.00 \mathrm{~mm}\) on the sample? c) How many \(5.00-\mathrm{mW}\) lasers with a beam diameter of \(2.00 \mathrm{~mm}\) would be needed to make the slice float in the Earth's gravitational field? Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
Answer: 10 lasers are needed to make the aerogel slice float in Earth's gravitational field.

Step by step solution

01

a) Weight of the aerogel slice

First, we need to find the volume of the aerogel slice. The volume (V) of a cylinder with height (h) and radius (r) is: V = πr^2h The radius is half the diameter, so r = 1 mm. The thickness is the height, so h = 0.1 mm. V = π (1 mm)^2 (0.1 mm) = 0.1π mm^3 Now we'll convert the volume to cubic centimeters, since the given density is in mg/cm³: V = 0.1π (1 cm³ / 1000 mm³) ≈ 3.14 x 10⁻⁴ cm³ Now we can find the mass (m) of the aerogel slice using its density (ρ): m = ρV = (1.00 mg/cm³)(3.14 x 10⁻⁴ cm³) ≈ 3.14 x 10⁻⁴ mg We'll convert this mass to kilograms: m ≈ 3.14 x 10⁻⁴ mg (1 kg / 10⁶ mg) ≈ 3.14 x 10⁻¹⁰ kg Now we can find the weight (W) of the aerogel slice by multiplying its mass by Earth's gravitational acceleration (g): W = mg = (3.14 x 10⁻¹⁰ kg)(9.81 m/s²) ≈ 3.08 x 10⁻⁹ N
02

b) Intensity and radiation pressure of the laser beam

The intensity (I) of a laser beam is the power (P) divided by the area (A) it covers. The area covered by the laser can be found using the formula for the area of a circle (πr²). First, convert the diameter to radius: r = (2.00 mm) / 2 = 1.00 mm = 0.1 cm Now, calculate the area: A = πr² = π (0.1 cm)² = 0.01π cm² Find the intensity: I = P / A = (5 mW) / (0.01π cm²) Remember to convert the power from milliwatts to watts: I = (5 x 10⁻³ W) / (0.01π cm²) ≈ 15.92 W/cm² Now we need to find the radiation pressure (RP) of the laser beam. The formula is: RP = 2I / c where c is the speed of light in vacuum (c ≈ 3.00 × 10^8 m/s). First, convert the intensity (I) to SI units (W/m²): I = 15.92 W/cm² (10^4 cm² / 1 m²) = 1.592 x 10⁵ W/m² Now calculate the radiation pressure: RP = (2)(1.592 x 10⁵ W/m²) / (3.00 × 10^8 m/s) ≈ 1.061 × 10⁻³ N/m²
03

c) Number of lasers to make the slice float

First, let's find the area (A) of the front surface of the aerogel slice exposed to the laser beams. Since the slice is circular and has a diameter of 2.00 mm, we can find the area as: A = πr² = π (1 mm)² = π mm² Convert this area to square meters: A = π mm² (1 m² / 10⁶ mm²) ≈ 3.14 x 10⁻⁶ m² We need to balance the force due to radiation pressure (F_r) with the gravitational force acting on the slice (F_g). The number of lasers (n) needed can be found using the equation: F_r = F_g n(RP)(A) = W n = W / (RP)(A) Substitute the values we found earlier: n ≈ (3.08 x 10⁻⁹ N) / ((1.061 × 10⁻³ N/m²)(3.14 x 10⁻⁶ m²)) ≈ 9.15 Since we cannot have a fraction of a laser, we must round up to the nearest whole number: n = 10 Therefore, 10 lasers are needed to make the slice float in Earth's gravitational field.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free