Chapter 1 Problems. 1micron m = microns. = cm. 1micron m = microns. 1 ft


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1 Chapter 1 Problems 1.3 The micrometer is often called the micron. (a) How man microns make up 1 km? (b) What fraction of a centimeter equals 1µm? (c) How many microns are in 1.0 yard We begin by calculating 1 km in microns d =1.0km 1000m 1km Now calculate a distance of 1 micron in cm. 1micron m = microns d =1micron m 1micron 100cm 1m = cm Finally we compute a distance of 1 yard in microns. d =1yd 3 ft 1yd m 1 ft 1micron m = microns 1.5 The Earth is approximately a sphere of radius 6.37 x 10 6 m. (a) What is is its circumference in kilometers? (b) What is its surface area in square kilometers? (c) What is its volume in cubic kilometers? To do all three sections of this problem, we can first convert the radius to kilometers. r = km m 1000m = km (a) The formula for circumference can be found in most calculus books (and in Appendix E of your Physics text). We will assume that we are finding the circumference of the equator. (b) The surface area of a sphere is: (c) The volume of a sphere is: c = π r = π km = km a = 4π r = 4π ( km) = km v = 4 3 π r3 = 4 3 π ( km) 3 = km 3
2 1.7 Antarctica is roughly semicircular, with a radius of 000 km. The average thickness of its ice cover is 3000m. How many cubic centimeters of ice does Antarctica contain? (Ignore the curvature of the earth) (Corrected) r=000m We find the volume in cubic meters first and then convert. Volume = (Area of half circle) (depth) = π r d 3000m = π ( m) 3000m = m 3 = m 3 ( 100cm 1m )3 = cm Hydraulic engineers often use, as a unit of volume of water, the acrefoot, defined a the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumps.0 inches of rain in 30 min. on a town of area 6 km. What volume of water in acrefeet, fell on the town? We first convert the depth to feet and the area to acres. 1ft d =.0in = ft 1in a = 6km ( 1000m 1km ).471acres 10 4 m = acres Volume = a d = acres ft = acre ft 1.10 The fastest growing plan on record is Hesperoyucca whipplei that grew 3.7 m in 14 days. What was its growth rate in micrometers/second r = 3.7m 14days 1micron m 1day 4hrs 1hr microns = s s
3 1.11 A fortnight is a charming English measure of time equal to.0 weeks (the word is a contraction of fourteen nights ). That is a nice amount of time in pleasant company but perhaps a painful string of microseconds in unpleasant company. How many microseconds are in a fortnight. 14days # micros =1 fortnight 1 fortnight 4h 1d 3600s 1µs 1h 10 6 s = Enrico Fermi once pointed out that that standard lecture period (50 min.) is close to 1 microcentury. How long is a microcentury in minutes and what is the percent difference from Fermi s approximation? # s = yr century 1century 365.5days 1yr %error = = 5.596min 5.596min 50min 5.596min 100 = 4.9% 4hrs 1day 60min 1hr 1.14 Time standards are now based on atomic clocks. A promising second standard is based on pulsar, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a light house beacon. Pulsar PSR is an example; it rotates once every ± 3 ms, where the trailing ±3 indicates the uncertainty in the last decimal place (it does not mean ±3 ms). (a) How many times does PSR rotate in 7.00 days? (b) How much time does the pulsar to rotate 1.0x10 6 times and (c) what is the associated uncertainty? (a) We compute the number of rotations No.of rotations = (7days 4hrs 1day 3600s 1hr (b) We can compute the time: = rotations 1000ms 1s t = rotations ms 1rotation = s 1rotation ) ms (c) A quick and dirty computation of the uncertainty os just to take the uncertainty per rotation and multiply by the number of rotations Δt = ms = ms = s This might not be quite right. For a product of numbers with uncertainties, like r is the product of two numbers, a and b, each with uncertainties..
4 r = ab Δ r r = Δ a a + Δ b b If we apply this equation here, where the total time T for N turns is the product of N turns and the time per turn. T = N t ΔT T = Δ N N + Δt t ΔT s = ms ms ΔT = s = s ms ms The uncertainty IS our quick calculationbecause there was no uncertainty in the number of turns Because Earth s rotation is gradually slowing, the length of each day increases: The day at the end of the 1.0 century is 1.0 ms longer than the day at the start of the century. In 0 centuries, what is the total of the daily increases in time (that is, the sum of the gain on the first day, the gain on the second day) See class notes.
5 1.19 Suppose that, while lying on the beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height h=1.7m, and stop the watch when the top of the Sun again disappears If the elapsed time is 11.1s, what is the radius of the earth. Getting the picture is critical for doing this problem. To proceed, we find the angle that the earth turns through in 11.1 s. re θ re h This is the angle in the picture. θ 11.1s = 360 4hrs 60min/ hr 60sec/min θ = We now use some geometry to find h cosθ = r e r e + h (r e + h)cosθ = r e hcosθ = r e r e cosθ r e = hcosθ 1 cosθ = 1.7m cos( ) 1 cos( ) = m This isn t a great value, but not bad for lying on the beach Gold, which as a mass of 19.3 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If oz of gold, with a mass of 7.63 g is pressed into a leaf of microns thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius of.5 microns, what is the length of the fiber. In this problem, the volume of the gold remains constant. It simply gets reshaped into different
6 shapes. We compute the volume first... a) In the case of a thin sheet of leaf, the volume is 1cm 3 V = 7.63g 19.3g =1.43cm 3 V = Area thickness Area = (b) In the case of a cylinder, the volume is V = π r l l = V thickness = 1.43cm cm = cm V π r = 1.43cm 3 π ( cm) = cm 1.4 One cubic centimeter of a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 microns. For that range, give the lower value and the higher values, respectively, for the following. (as) How many cubic meters of water are in a cylindrical cumulus cloud of height 3.0 km and radius 1.0 km. (b) How many 1liter pop bottles would that water fill? (c) Water has a mass of 1000 kg per cubic meter of volume. How much mass does the water in the cloud have? We will calculate the lower number first. The higher values will be 10 times larger since the number of drops is 10 times higher. a) We are told that the cloud is a cylinder. We begin by calculating the volume of the cloud in cubic cm. This will require us to convert the cloud dimensions to cm. r cloud = 1km 1000m 1km h cloud = 3km 1000m 1km 100cm 1m = cm 100 cm 1m = cm V cloud = π r h = π ( cm) cm = cm 3 # drops V water in cloud = V cloud cm 3 of cloud V drop V water in cloud = cm 3 50drops cm π ( m) 3 = m 3 The upper limit would be 1979 m 3 b. We can find the number of 1 L popbottles by converting the volume of water to L. The
7 number of L will be the number of bottles. 1m 3 =1000L V water in cloud = 197.9m L 1m 3 The upper limit is L. = L c We can now compute the mass of water in the cloud. The upper limit is kg. m = V water incloud ρ water = 197.9m kg 1m 3 = kg 1.6 A mole of atoms is atoms to the nearest order of magnitued, how many moles of atoms are in a large domestic cat. The masses of a hydrogen atom, an oxyten atom and a cargon atom are 1.0 u, 16 u, and 1 u respectively. (Hint: Cats are sometimes know to kill a mole). We begin by estimating the mass of a large cat to be 10 kg. We also take the cat to be made entirely out of water (we ll ignore the carbon). Since each mole of water has a mass of 18 g (16 for O, 1g for each H), we can compute m cat = # moles 18g mole m # moles = cat 18g / mole = 10, 000g 18g / mole = 555 moles 1 kilomole 1.30 Water is poured inot a container that has a leak. The mass m of the ater is given as a function of time t by m = 5.00t t with t 0, m in grams, and t in seconds. (a) AT what time is the water mass greatest, and (b)what is that greatest mass? In kilograms per minute, what is the rate of mass change at (c) t=.00s and (d) t=5.00s.
8 If we plot this equation We can see that the maximum mass occurs at t 4.5s. The maximum mass can also be read off the graph. The maximum mass is m 3.15g We can use calculus to do this as well. The rate of change of the mass is given by the derivative. dm dt = d dt (5.00t t ) = t = 4.00t The maximum mass occurs in this problem when the rate goes to zero (the slope of the mass curve goes to zero. dm = 4.00t dt 0 = 4.00t t 0. = t 0. = t = = 4.14s Our estimate from the graph was pretty good. Now we can find the mass by substituting in this time. 5
9 m = 5.00t t m(4.14s) = 3.16g Again, our estimate from the graph was very good. We can find the rate of change at s and 5s from the derivative that we have computed. dm = 4.00t dt At t =.0s dm = 4.00t = 0.48g / s dt At t = 5.0s dm = 4.00t = g / s dt 1.46 What mass of water fell on the town in problem 9 Let s calculate the volume of water in MKS units and then find mass. d = in.54cm 1in A = 6 km 1m 100cm = m 1000m 1km = m V = A d = m m = m 3 m = V ρ = m kg 1m 3 = kg
Chapter 1 Problems. To do all three sections of this problem, we can first convert the radius to kilometers. r = 6.37 10 6 1km 1000m = 6.
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