Some New Problems 


PHYS 101 
CHAPTER 1 Problem 1.1مسيرة 500 عام A unit of length that was commonly used in the past is the daywalking distance. It is defined as the distance traveled by a caravan in a single day (roughly 20 miles). What would a 500yr walking distance be in meters? Solution: 500yr walking distance = (500 yr) (365 day / 1 yr) (20 mi / 1 day)(1.61 km / 1 mi) = 5.84 x 10^{6} km = 5.84 x 10^{9} m In other words, the distance would be equivalent to approximately 20 times the distance from the earth to the moon. Problem 1.2 Just to practice working with the prefixes of the SI (Table 12/page 3) The femto (f) is equivalent to the factor 10^{15}, while the nano (n) is equivalent to 10^{9}. What is the distance, in fm, that light travels in 10 ns? (speed of light is c = 3.00 x 10^{8} m/s) Solution: distance = speed x time In 10 ns light travels a distance = (3.00 x 10^{8} m/s) (10 ns) (10^{9}s/1 ns) = 3.00 ^{ }m Which is expressed in units of fm as = (3.00 m) (1 fm/ 10^{–15} m) = 3.00 x 10^{15} fm CHAPTER 2 Problem 2.1 سبعون خريف What if a stone takes 70 years to reach the bottom of a canyon! Have you ever thought what the depth of that canyon would be! (Assume the stone was released with zero initial velocity) Solution: Here is an easy way to remember how many seconds, approximately, there are in a year: 1 yr ~ π x 10^{7} s (i.e., 1 yr ~ 3.14 x 10^{7} s) We need to use an equation that describes displacement as a function of time (but not as a function of velocity). From the menu of equations we choose: yy_{0} = v_{0}t (1/2)gt^{2} Therefore –y = (1/2)gt^{2 } , since v_{0} = 0 (the stone was just released) (If you don't see why we put a negative sign in front of y come and see me at the office) Substituting g with 9.8m/s^{2} and t with 3.14 x 10^{7} s gives y = 4.8 x 10^{14} m! The depth of the canyon would be something like 3000 times the distance from the earth to the sun (i.e., 3000 Astronomical Unit).
