Q#13
(II) Two locomotives approach each other on parallel tracks. Each has a speed 155 km/h of with respect to the ground. If they are initially 8.5 km apart, how long will it be before they reach each other? (See Fig. 2–35.)
$\overline{v}=\frac{d}{\Delta t}$
$\Delta t=\frac{d}{\overline{v}}=\frac{4.25 \ km}{155 \ km/h}$
$\Delta t_1=0.0274 h = 1.645 \ min=99 \ s$
Q#14
(II) Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius $R_1=2.5 \ cm $ and finishes at radius $R_2=5.8 \ cm$. The distance between the centers of neighboring spiral-windings is 1.6 $\mu m$ (=1.6 $\times 10^{-6} \ m$)
(a) Determine the total length of the spiraling path. [Hint: Imagine “unwinding” the spiral into a straight path of width and note that the original spiral and the straight path both occupy the same area.]
(b) To read information, a CD player adjusts the rotation of the CD so that the player’s readout laser moves along the spiral path at a constant speed of about Estimate the maximum playing time of such a CD.
Answer:
Inner radius $R_1$ = 2.5 cm and
outer radius $R_2$ = 5.8 cm.
the width of spiral winding is (d) = $1.6 \times 10^{-6}$ m
(a) The area between the concentric circles is equal to the length times the width of the spiral path.
$Area = \pi(R_2^2-R_1^2)=dL$\
$L = \frac{\pi(R_2^2-R_1^2)}{d}$
$L = \frac{\pi[(0.058 \ m)^2-(0.025 \ m)^2]}{1.6 \times 10^{-6} \ m}$
$L = 5.378 \times 10^3 \ m=5400 \ m$
(b) $t = \frac{L}{v}$
$t=\frac{5.378 \times 10^3 \ m}{1.2 \ m/s}$
$t = 4481.67 \ s \approx 75 \ min$
Q#15
(III) A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound of the ball hitting the pins 2.80 s after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is 340 m/s.
Answer:
The average speed of sound is given by $v_{sound}=\frac{\Delta x}{\Delta t}$ so the time for the sound to travel from the end of the lane back to the bowler is
$\Delta t = \frac{\Delta x}{v_{sound}}$
$\Delta t = \frac{16.5 \ m}{340 \ m/s}=4.85 \times 10^{-2} \ s$
Thus the time for the ball to travel from the bowler to the end of the lane is given by
$\Delta t_{ball}=\Delta t_{total}-\Delta t_{sound}$
$\Delta t_{ball}=2.80 \ s-4.85 \times 10^{-2} \ s=2.7515 \ s$
The speed of the ball is as follows:
$v_{ball}=\frac{\Delta x}{\Delta t_{ball}}$
$v_{ball}=\frac{16.5 \ m}{2.7515 \ s}$
$v_{ball}=5.9967 \ m/s \approx 6.00 \ m/s$
Q#16
(III) An automobile traveling 95 km/h overtakes a 1.30-km long train traveling in the same direction on a track parallel to the road. If the train’s speed is 75 km/h how long does it take the car to pass it, and how far will the car have traveled in this time? See Fig. 2–36. What are the results if the car and train are traveling in opposite directions?
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