University:
University of North Georgia
Course:
MATH 1450 | Calculus I
Academic year:

2024
Views:

50

Pages:

3
Author:

Autumn Cox

Question 1 Calculate the theoretical fracture strength of a brittle material if it is known that fracture happens as a result of the spread of a 0.5 mm long elliptical surface crack (0.02 in.) and a tip radius of curvature of 5 × 10-3mm (2×10−4),, when a stress of 1035 MPa (150,000 psi) is applied. Answer We are given following data for brittle material: L = 0.5mm r = 5.10-3mm σο = 1035MPa Calculating the theoretical fracture strength: L 0.5 σ= σο· = 1035. r 5.10 0-3 = 20700MPa σ = 20700MPa Question 2 What is the energy of an electron in a hydrogen atom in its ground state? Now consider an excited- state hydrogen atom, what is the energy of the electron in the n=4 level? Answer The equation for energy is given by: Eo E = n2 At its ground state n=1 E= -2.18 × 10-18 12 At n=4 E= -2, 18 × 10-18 42 = -2.18 × 10-18J = -1.4 × 10-19 J Question 3 An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ = 1.8 × 10-7C/m² and the plates are separated by a distance of 1.5 x 10-2m. How fast is the electron moving just before it reaches the positive plate? Answer Step 1 Concept The Electric field due to capacitor having surface charge density a is The acceleration of electron is given by σ E= €0 F ả m qE a= m We have to find the speed of electron as it reaches the plate Step 2 The acceleration ασ ả meo a= (1.6 × 10-19C) (1.8×10-7) (9.11 × 10-31kg) (8.85 × 10-12N) a = 3.572 × 1015 m Step 3 The velocity $2 The velocity can be found from the equation of kinematics v² = v + 2(a)s v2 = 0 0+2(3.572 02 v = X = 1.0716 × 1014 m² $2 V1.0716 × 1014 m2 $2 1015 m )(1.5 × 10-2m) v = 1.035 × 107m/s Result v = 1.035 × 107m Question 4 s LASIK eye surgery uses pulses of laser light to shave off tissue from the cornea, reshaping it. A typical LASIK laser emits a 1.0-mm-diameter laser beam with a wavelength of 193 nm. Each laser pulse lasts 15 ns and contains 1.0 mJ of light energy. a. What is the power of one laser pulse? b. During the very brief time of the pulse, what is the intensity of the light wave? Answer a) The power will be equal to the ratio of the energy emitted over the time during which it was emitted. So, we will have: E P = t 1.10-3 = 66.7kW 1.5.10-8 b) The intensity will be equal to the power over area; that is, P 4P 4.66700 I= = 8.5-1010 10 W A πα2 π. 0.0012 m2 Result: a) 66.7 kW b) 8.5-1010 10 W m² Question 5 Gas Y effuses half as fast as 02. What is the molar mass of gas Y? Answer Given, gas Y effuses half as fast as 02. The molar mass of gas Y is calculated by using Graham's law. R1 R2 where, 2 M2 M1 R₁ and R2 =rates of diffusion M₁ = molar mass of O2 M2 = = molar mass of gas Y The molar mass of gas Y is 2 ()= 4 M2 1 32 M2 32 M2 = 32 × 4 = 128 9 mol Molar mass of gas Y = 128 9 mol