Download Quest Chapter 29

Quest Chapter 29 # Problem Hint 1 “turned back” think “bounce”
2 3 4 A wave is turned back when it meets the boundary of the medium in which it is traveling. The wave is said to have undergone 1. interference. 2. diffraction. 3. reflection. 4. refraction. What is the law of reflection? 1. The angle of refraction equals the angle of reflection. 2. Reflection is not consistent from an irregular surface. 3. The critical angle is 90 degrees. 4. The angle of reflection equals the angle of incidence. 5. A faster light in the media results in a larger reflective angle. 6. The angle of incidence equals the angle of refraction. An object, slanted at an angle of 45o, is placed in front of a vertical plane mirror, as shown. Which of the following shows the apparent position and orientation of the object’s image? Hold a pocket mirror at almost an arm’s length from your face and note the amount of your face you can see. To see more of your face, should you hold the mirror closer or farther, or would you have to have a larger mirror? 1. You have to have a larger mirror. 2. You can hold the mirror closer. 3. You can hold the mirror farther. Reread the definition of the law of reflection on page 444. You are looking for line symmetry. Which one matches? Draw a ray diagram for an object at two different distances from a mirror. Mark the location of the “eye” on the farthest object and then draw the axis from that point through the near object to the mirror. As you draw the rays, you might remember something about isosceles triangles… # Problem Hint 5 Draw a diagram on which you label the distances. Include the image and its distance. A butterfly at eye level is 23 cm in front of a plane mirror. You are behind the butterfly, 77 cm from the mirror. What is the distance between your eye and the image of the butterfly in the mirror? Answer in units of cm. 6 If you take a photograph of your image in a plane mirror, how many meters away should you set your focus if you are 2.5 m in front of the mirror? Answer in units of m. 7 Suppose you walk toward a mirror at 4 m/s. How fast do you and your image approach each other? Answer in units of m/s. 8 Hint: A ray diagram would be helpful. Determine the minimum height of a vertical flat mirror in which a person 71 in. in height can see his or her full image. Answer in units of in. 9 Why is it difficult to see the roadway in front of you when driving on a rainy night? 1. There is no moon to light the roadway. 2. It is much darker outside than on a clear night. 3. The road is covered with water which acts like a plane mirror. 4. Lightning may prevent you from seeing the roadway clearly. 10 What causes refraction? 1. None of these 2. the existence of different colors of light 3. the difference in the speed of light in different transparent media 4. the speed of light being higher than c in some media 5. reflection 11 If the speed of light through material Z is 2.5×108 m/s, what is this material’s index of refraction? The speed of light in a vacuum is 3 × 108 m/s. 12 Diamond has a high index of refraction at about 2.4, which helps account for its sparkle. How fast does light travel through a diamond? The speed of light in a vacuum is 3 × 108 m/s. Answer in units of m/s. How far does the light have to travel? How far does the light have to travel? Use the hint.
Hmmm…What does water do to a surface? See page 446. Reread the definition of refraction on page 448 and the diagram on page 449. Reread the definition of index of refraction on the bottom of page 451. Reread the definition of index of refraction on the bottom of page 451. # Problem Hint 13 (part 1 of 2) A beam of light with a wavelength 810 nm in air travels in water. The index of refraction of water is 1.33. What is the wavelength in water? Answer in units of nm. 14 (part 2 of 2) Does a swimmer underwater observe the same color or a different color for this light? 1. A swimmer underwater observes a different color because the frequency changes. 2. A swimmer underwater observes the same color because the frequency does not change. 3. It is not possible to know what a swimmer underwater will observe. 4. A swimmer underwater observes a the same color because the wavelength does not change. 5. A swimmer underwater observes a different color because the wavelength changes. 15 Light rays bend as they pass from air into water at an angle (not 90 degrees). This is refraction. Which quantity doesn’t change when light refracts? 1. frequency of light 2. index of refraction of the material 3. wavelength of light 4. average speed of light 16 When light travels from medium “X” to medium “Y” as shown in the figure: 1. both the speed and the frequency decrease 2. both the wavelength and the frequency are unchanged 3. both the speed and the frequency increase 4. both the speed and the wavelength decrease 5. both the speed and the wavelength increase The index of refraction in air is about 1.0003. Remember: f doesn’t change. So, n1/n2 = λ2/ λ 1. Observed color depends on the frequency. What happens to the frequency? Check out the diagram on page 449. See the hint for 14. Remember the diagram on page 449? What is changing? We know the speed changes, but what else changes: f or λ? # Problem Hint 17 A light wave passes from air into water. Which of the following is true? 1. Its frequency remains unchanged, but its wavelength decreases. 2. Both its wavelength and its frequency remain unchanged. 3. Its wavelength remains unchanged, but its frequency increases. 4. Its frequency remains unchanged, but its wavelength increases. 5. None of these. 6. Its wavelength remains unchanged, but its frequency decreases. 18 (part 1 of 2) Given: A ray approaching an interface. What is the approximate refracted ray? Remember the diagram on page 449? What is changing? We know the speed changes, but what else changes: f or λ? 19 (part 2 of 2) What is the approximate refracted ray? 20 (part 1 of 2) In movies that involve a character lost in the wilderness somewhere, you often see the hero vainly try to spear a fish in a river or tidal basin. Even if his aim is good, how should he aim to spear the fish? 1. below the fish 2. at the tail 3. on the heart 4. above the fish 21 (part 2 of 2) Due to refraction, how deep is the fish in relation to its image? 1. shallower 2. at the point of the image 3. deeper Given the different media, which will have the greater angle, the incident ray or the refracted ray? Given the different media, which
will have the greater angle, the incident ray or the refracted ray? Draw the ray diagram. Extend the ray from the fisherman into the water. It will show the fisherman’s direct line of sight. Now, figure out where he should aim. Use the ray diagram from 20. # Problem Hint 22 During a lunar eclipse, the moon is not completely dark, but is often a deep red in color. Why? 1. High frequencies pass more easily through the long grazing path through the Earth’s atmosphere to be refracted finally onto the moon. 2. Low frequencies pass more easily through the long grazing path through the Earth’s atmosphere to be refracted finally onto the moon. 3. During the sunset, the Sun transfers some red light to the moon. 23 The separation of white light into its component colors is called 1. refraction. 2. dispersion. 3. reflection. 4. diffraction. 5. polarization. 24 Of the following statement about the speeds of the various colors of light in glass, which is true? 1. Red has the highest speed; violet the lowest. 2. Violet has the highest speed; red the lowest. 3. Red and violet have the highest speed; green the lowest. 4. Green has the highest speed; red and violet the lowest. 5. All colors of light have the same speed in glass. 25 If you left a glass fiber‐optic cable unshielded by any plastic covering, should the light still be able to travel through the cable? 1. No 2. Yes 26 If you swim just below the surface of a pool and look straight up, you will see the sky. However, if you look at a glancing angle to the surface of the water, you will see a reflection of the bottom of the pool. What caused this to happen? 1. to reflection 2. total internal reflection 3. None of these 4. refraction Draw a ray diagram using the Earth’s atmosphere as the other medium. Which will be bent more: blues or reds? Read pages 454 & 455. Read pages 454 & 455. Read 459. What is happening that causes you to see down when you look up? Read page 457. # Problem Hint 27 A(n) 6.8 kg object moving at 4.8 m/s eastward has a one‐dimensional elastic collision with a(n) 5.2 kg object moving at 3.9 m/s in the opposite direction. What is the total kinetic energy of the two‐ mass system after the collision? Answer in units of J Think. What is conserved?
How much KE before? How much is lost in an elastic bump? How much should remain? Look through chapters 7 & 8. O boy! A projectile motion problem. Look at the diagram. What are you to find? The vertical position? No. Δh doesn’t go to the ground. So, what are you to find? Do you need the mother or equation or grandma? Absolute pressure included both the pressure from the water AND the pressure from the air. Remember: Pfluid = dweighth Think. Draw your energy diagram: Heatin, Workout, Heatout.
Efficiency = Useful Work Total Energy Used Think? How much energy was thrown away? Check your energy diagram. 28 A cannon fires a 0.276 kg shell with initial
◦
velocity vi= 9.5 m/s in the direction θ= 53
above the horizontal.
The shell’s trajectory curves downward
because of gravity, so at the time t= 0.235 s
the shell is below the straight line by some
vertical distance ∆h.
Find this distance ∆h in the absence of air
resistance. The acceleration of gravity is
9.8 m/s2.
Answer in units of m
29 Determine the absolute pressure at the bottom
of a lake that is 22.4 m deep. The acceleration
of gravity is 9.8 m/s2 and atmospheric pressure
is 1.01×105Pa.
Answer in units of Pa
30 (part 1 of 2)
A heat engine absorbs 388 J of thermal energy
and performs 39.6 J of work in each cycle.
Find the efficiency of the engine.
31 (part 2 of 2) Find the thermal energy expelled in each
cycle.
Answer in units of J
# Problem Hint 20 Why does the sun look red at sunrise and sunset but not at noon? 1. Red light is easily scattered through the air since its frequency is the lowest. At noon, sunlight travels through the least amount of atmosphere with little transmission of high frequency light to the surface. 2. The sun is colder at sunrise and sunset than at noon. 3. Red light is easily transmitted through the air since its frequency is the lowest. At noon, sunlight travels through the least amount of atmosphere with little scattering of high frequency light to the surface. 4. The temperature at noon is warmer than at sunrise and sunset. 5. Our eyes are more sensitive to the red light at sunrise and sunset than at noon. 6. The sun itself is red at sunrise and sunset, but is yellow at noon; it has nothing to do with the effect of the atmosphere. Think of the whole blue sky problem. Why is it scattered first? What is different about sunset/sunrise that would make red scatter at those times?