Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 2 01

... at one end A person at the other end hears two sounds as a result of the propagation of two longitudinal waves, one that traveled through the rod and the other that traveled through the air at ... at constant pressure to the heat capacity at constant volume (a) Show that Bad = γ P for an ideal gas (b) Show that the speed of a longitudinal wave √ in the adiabatic process of an ideal gas ... sinusoidal wave and has the same frequency f and wavelength λ as any one of the contributing waves y1 and y2 , (2) It has an amplitude of 2A cos(φ /2) , (3) It has a phase of φ /2 Now, let us consider...

... diffraction, and polarization of light and need to get satisfactory descriptions of these phenomena, we treat light as waves Such a study is referred to as physical optics H A Radi and J O Rasmussen, ... position of the nodes and antinodes (c) What is the maximum amplitude of an element at an antinode? (9) The two sources shown in the evacuated vessel of Fig 16 .22 , are 1 .2 m apart, and send sound ... wall at this end does not allow longitudinal motion of the air and acts like a pressure antinode (point of maximum pressure variation) The open end of a pipe acts like the end of a string that...

... sin[ (A + δm ) /2] sin (A/ 2) −−−−−−−−−−→ When A is small n= A + δm A (17. 12) The most charming example of color dispersion is that of a rainbow To understand the formation of a rainbow we consider a ... An anti-shoplifter convex spherical mirror has a radius of curvature of 0.4 m Locate and describe the image formed by a man standing 3.8 m away from the mirror Solution: The focal length of a ... at a real focal point after reflecting from a concave mirror (b) The same rays will diverge from a convex mirror and appear to come from a virtual focal point When parallel rays reach the surface...

... in air falls on water at an angle of incidence of 45◦ and then passes through a glass block before it emerges out to air again The surfaces of water and glass are parallel and their indexes of ... the air-alcohol boundary at an angle of incidence of 60◦ ? (6) A beam of light in air falls on a liquid surface at an angle of incidence of 55◦ The liquid has an unknown index of refraction (a) ... Eq 18 .23 Part (a) of Fig 18. 12 displays the variation of I as a function of δ /2 Part (b) is a representation of the obtained photograph 18.4 Diffraction of Light Waves 619 Fig 18. 12 (a) A sketch...

... defined as the direction of polarization of the wave However, an ordinary beam of light contains a large number of electromagnetic waves emitted by atoms having random vibrational orientations, → and ... the Coulomb constant Coulomb found that each charged particle (also called a point charge) exerts a force of this magnitude on the other particle, and the two forces form an action–reaction pair ... materials nonconductors or insulators Spotlight Insulators are materials that contain electrons that are bound to their atoms and cannot move freely through the material Semiconductors are materials...

... Gravitational attraction A classical model of the hydrogen atom Solution: (a) From Coulomb’s law, the magnitude of the electrostatic force of → attraction Fe between the electron and the proton ... 20 Electric Fields When x a we can take out a factor of x from each denominator of the brackets of the last formula for x > +a and then expand each of these terms by binomial expansion Therefore, ... equal point charges of magnitude q lie on a semicircle of radius R as √ shown in Fig 19 .21 Show that the net force on q2 has a magnitude kq2 / 2R2 and points downward away from the center C of...

... that: (x dx x = 3 /2 +a ) a2 (x + a2 ) (20 .33) Thus: L /2 E = 2kλ a = 2kλ a x dx = kλ a √ 2 3 /2 (x + a ) a2 x + a2 L /2 a2 (L /2) 2 + a2 −0 = L /2 kλ L a (L /2) 2 + a2 (20 .34) 20 .4 Electric Field of a ... Example 20 .4 Figure 20 .14 shows a non-conducting rod that has a uniform positive charge density +λ and a total charge Q along its right half, and a uniform negative charge density −λ and a total ... terms of x From the geometry of Fig 20 .13, we have: x + a2 and sin θ = r= a a =√ r x + a2 (20 .31) Therefore, Eq 20 .30 becomes: L /2 E = kλ a dx (x + a2 )3 /2 (20 . 32) From the table of integrals in Appendix...

... central axes along a common horizontal line and separated by a distance of R, see Fig 20 .36 Ring has a uniform positive charge q1 , while ring has a uniform positive charge q2 Given that the ... results of Exercise 21 to the near-field approximation E = σ /2 ◦ as well as to the point charge approximation E = k(π R2 σ ) /a2 , and find which result(s) of Exercise 21 match the two approximations ... when a formula: 2kQ a 1− √ 2 R R + a2 R, the electric field approaches that of a point charge E≈k Q (a a2 R) You may use the binomial expansion (1 + δ)p ≈ + pδ when δ (23 ) Compare the obtained...

... such an electron has to be accelerated from rest through a potential difference of 4.55 kV to reach a speed of × 107 m/s Equations 22 .3 and 22 .6 hold true for both discrete and continuous source ... Potential 735 Suppose we move a particle of an arbitrary charge q from point A to point B by applying a force on it and changing its kinetic energy The applied force must perform work WAB (app), ... identical Therefore, the general features of the gravitational force introduced in Chap apply to electrostatic forces In particular, electrostatic forces are conservative Consequently, it is more convenient...

... 22 Electric Potential For a Point Above One End of the Rod When a point P is located at a distance a from one of the rod’s ends, see Fig 22 .17, we can perform similar calculations to find that: ... Therefore: V = 2kλ ln x + a2 ) (22 .48) L /2 (L /2) 2 + a2 ) − ln (a) L /2 + (L /2) 2 + a2 a (22 .49) When we use the fact that the total charge Q = λL, we get: V = L /2 + 2kQ ln L (L /2) 2 + a2 a (22 .50) 7 52 ... potential energy of each pair and add them algebraically For instance, the total potential energy of three charges q1 , q2 , and q3 is: q1 q2 q1 q3 q2 q3 + + r 12 r13 r23 U=k (22 .31) Example 22 .5 Two...

... inverse of the capacitance of an air capacitor of separation d − a, and a/ κ ◦ A is the inverse of the capacitance of a capacitor of separation a but filled with a dielectric 23 .4 Capacitors in Parallel ... two basic combinations of capacitors that allow such a replacement Capacitors in a Parallel Combination Figure 23 .1 1a shows two capacitors of capacitances C1 and C2 , that are connected in parallel ... capacitance for the capacitor under consideration A Parallel-Plate Capacitor Figure 23 . 2a shows an uncharged parallel-plate capacitor of equal area A separated by a distance d The capacitor is...

... Section 23 .2 Calculating Capacitance (4) A computer memory chip contains a large number of capacitors, each of which has a plate area A = 20 × 10− 12 m2 and a capacitance of 50 f F (50 femtofarads) ... (a) and (b) of Fig 23 . 32 Show that the capacitances of the two capacitors of parts (a) and (b) are: C= A d κ + 2 ◦ A κ1 2 and C = respectively, d κ1 + 2 A/ 2 d A/ 2 2 κ1 (a) d κ1 d /2 2 d /2 ... the maximum magnitude of the charge on the plates of an air-filled parallelplate capacitor of area A = 30 cm2 such that a spark is avoided (7) A parallel-plate capacitor has circular plates, each...

... to move In fact, it is a device that works as an energy converter A battery is often called a source of electromotive force or, a source of emf (this unfortunate historical name describes a potential ... Parallel Combination Figure 24 .1 5a shows two resistors of resistances R1 and R2 that are connected in parallel with a battery B Figure 24 .15b shows a circuit diagram for this combination of resistors ... )2 R (24 .24 ) Because P = I V , the same amount of power P can be transported either at high I and low V , or at low I and high V Example 24 .6 A 22 0 V potential difference is maintained across...