標題:
force and wave
發問:
http://i409.photobucket.com/albums/pp172/momoko_deedee/P3.jpg http://i409.photobucket.com/albums/pp172/momoko_deedee/P4.jpg i apologize for the low quality....><
最佳解答:
1. Use: work done by friction = change of kinetic energy Ff x 120 = (1/2)m.(20)^2 where m is the mass of the puck, Ff is the frictional force and v is its initial speed Since Ff = u.(mg), where g is the acceleration due to gravity (taken to be 10 m/s2), and u is the coefficient of friction hence, umg x 120 = (1/2)m.(20)^2 u = 20^2/(2x120g) = 0.17 2. Since the box is moving with constant speed, the pushing force equals to the frictional force hence, pushing force = 0.2 x 11.2g N = 22.4 N where g is the acceleration due to gravity 3. The figures are not clearly seen, I just give you the methods. Y = A.sin(kx).sin(wt) The waveform at an instant of time when the vibration is at its maximum amplitude, where sin(wt) =1, is thus given by y = A.sin(kx) Nodes are palces with displacement always zero thus sin(kx) =0 i.e. kx = 0, pi, 2 x pi, 3 x pi, 4 x pi ...etc where pi = 3.14159... thus, x = 0, pi/k, 2.pi/k, 3.pi/k ... Antinodes are just mid-points between two nodes. after you have found the nodes, you can calculate the antinodes accordingly. 4. When the wire is vibrating at fundamental frequency, wavelength = 2 x lingth of wire. hence wavelength = 2 x 80 cm = 160 cm = 1.6 m speed of wave = frequency x wavelength = 60 x 1.6 m/s = 96 m/s since speeed of wave v = square-root[T/m] where T is the wire tension m is the mass per unit length of the wire, i.e. m = 40/80 g/cm = 0.5 g/cm = 0.05 kg/m Thus, 96 = square-root[T/0.05] solve for T gives T = 461 N
其他解答:
force and wave
發問:
http://i409.photobucket.com/albums/pp172/momoko_deedee/P3.jpg http://i409.photobucket.com/albums/pp172/momoko_deedee/P4.jpg i apologize for the low quality....><
最佳解答:
1. Use: work done by friction = change of kinetic energy Ff x 120 = (1/2)m.(20)^2 where m is the mass of the puck, Ff is the frictional force and v is its initial speed Since Ff = u.(mg), where g is the acceleration due to gravity (taken to be 10 m/s2), and u is the coefficient of friction hence, umg x 120 = (1/2)m.(20)^2 u = 20^2/(2x120g) = 0.17 2. Since the box is moving with constant speed, the pushing force equals to the frictional force hence, pushing force = 0.2 x 11.2g N = 22.4 N where g is the acceleration due to gravity 3. The figures are not clearly seen, I just give you the methods. Y = A.sin(kx).sin(wt) The waveform at an instant of time when the vibration is at its maximum amplitude, where sin(wt) =1, is thus given by y = A.sin(kx) Nodes are palces with displacement always zero thus sin(kx) =0 i.e. kx = 0, pi, 2 x pi, 3 x pi, 4 x pi ...etc where pi = 3.14159... thus, x = 0, pi/k, 2.pi/k, 3.pi/k ... Antinodes are just mid-points between two nodes. after you have found the nodes, you can calculate the antinodes accordingly. 4. When the wire is vibrating at fundamental frequency, wavelength = 2 x lingth of wire. hence wavelength = 2 x 80 cm = 160 cm = 1.6 m speed of wave = frequency x wavelength = 60 x 1.6 m/s = 96 m/s since speeed of wave v = square-root[T/m] where T is the wire tension m is the mass per unit length of the wire, i.e. m = 40/80 g/cm = 0.5 g/cm = 0.05 kg/m Thus, 96 = square-root[T/0.05] solve for T gives T = 461 N
其他解答:
此文章來自奇摩知識+如有不便請留言告知
1. By equation of motion: v^2 = u^2 + 2as 0 = (20)^2 + 2a(120) Acceleration, a = -1.67ms^-2 By Newton's 2nd law of motion, f = ma Friction, f = -1.67m Taking the magnitude, f = 1.67m By the normal reaction, R = mg = 9.8m So, the coefficient of kinetic friction, u = f/R = 1.67/9.8 = 0.170 2. Frictional force, f = umg = (0.20)(11.2)(9.8) = 21.952 N As the box is undergoing constant speed, so the horizontal applied force = friction = 22.0 N 1. At the nodes, sinkx = 0 sin0.75pix = 0 0.75pix = npi, where n is an integer, x = 4n/3 m For the antinodes, sinkx = 1 sin0.75pix = 1 0.75pix = npi +- pi/2 x = (4n/3 +- 2/3) m 2. At fundamental mode, wavelength, 入 = 2l = 1.6 m By v = f入 Speed of propagation, v = (60.0)(1.6) = 96 ms^-1 Density, p = m/l = (0.04)/(0.8) = 0.05 kgm^-1 By v = sqrt(T/p) Tension, T = pv^2 = (0.05)(96)^2 = 460.8 N
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