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Solution #3
1. H16.24L
T b. use the equation = $%%%%%%% 2. H16.40L T = 2 = 158 N
a. y = A Sin@k x - tD A is the amplitude A = 2.00 x 10-4 = 2 f = 3140 rad s k = k = 16.0 rad m
a. we have y@x, tD = 0.35 SinA 10 t - 3 x +
b. We can find the energy per cycle by multiplying the power by the period. 1 2 1 = = sec, E = 15.1 5 = 3.02 J f 5
E 4 compare with the standard wave function y = A Sin@k x - t + D rad rad = 10 , A = 0.35 m, k = 3 sec m = = 3.33 m s k 1 Power = 2 A2 = 15.1 W 2
3. H16.68L
average velocity is distance traveled over time, so we have m 4450 km km 1000 m 1 hr 1 min = = 468 = 130 9.5 h hr 1 km 60 min 60 sec s using the equation given !!!!!!!! 2 = gd, d = = 1730 m g
4. H16.56L
solutionset3[1].nb
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The tension in the string is only due to the weight of the block. Ignore the mass of the string for calculating the tension. T 2 T = m g, and = $%%%%%%% . m = g
5. H16.62L a.
is also given by,
m = 2 I M g k
k 0.250 kg m 18 s-1 = J N = 14.7 kg 9.80 m s2 0.750 m-1 =
6. H16.34L
f = 7. H16.63L Y = and
= 15 = 5 m s, traveling in positive x - direction 3 15 b. = = 5 m s, traveling in negative x - direction 3 15 = 7.5 m s, traveling in negative x - direction c. = 2 15 = 24 m s traveling in positive x - direction d. = 0.5
use =
k
= 60.0 Hz, = 2 f = 120 1 0.180 1 N H120 L2 H0.100L2 H30.0L = 1.07 kW J Power = 2 A2 = 3.60 2 2
Young ' s Modulus is a ratio of stress over strain, which can be expressed as L is the strain Hor elongationL. Also note that you can express in terms L of as = A. So the speed of the wire Y is
T L T A L % = $%%%%%% = $%%%%%%%%% = $%%%%%%%%%%%%%% % A T A L L
where T is the tension, A is the cross sectional area of the string
2 L = = 3.86 x 10-4 L Y
solutionset3[1].nb
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8. H16.25L
The tension is the rope is simply, T = M gplanet Have two expressions for of the wave
where L is the length of the wire and t is the time it takes for the wave to reach the bottom. M gplanet L L2 = 2 t mL Lm gplanet = = 1.64 m s2 M t2
M gplanet L = $%%%%%%%%%%%%%%%%%%%%%% = mL t
Written Problem 1. velocity of light is given by c = f f = c c = f
c is the speed of EM waves a. of light waves range @4 x10-7 , 7 x10-7 D c = 7.5 x 1014 Hz high f = 4 x10-7 m c = 4.28 x1014 Hz low f = 7 x10-7 m b. f of shortwaves range @1.5 x106 , 300 x106 D c = 4.28 x1014 Hz high = 1.5 x106 Hz c = 4.28 x1014 Hz low = 300 x106 c. of X - rays range @5 x10-9 , 1 x10-11D c = 3 x 1019 Hz high f = 1 x10-11 m c = 6 x1016 Hz low f = 5 x10-9 m
Written Problem 2.
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