{"id":1494,"date":"2020-07-03T19:17:20","date_gmt":"2020-07-03T19:17:20","guid":{"rendered":"https:\/\/murreeroad.org\/physics10\/?p=1494"},"modified":"2020-10-26T00:03:50","modified_gmt":"2020-10-25T19:03:50","slug":"physics-10-chapter-10-examples","status":"publish","type":"post","link":"https:\/\/murreeroad.org\/physics10\/physics-10-chapter-10-examples\/","title":{"rendered":"Physics 10 Chapter 10 Examples"},"content":{"rendered":"
<\/a>Chapter 10 Numerical Problems Examples<\/span><\/strong><\/p>\n <\/p>\n Example 10.2:<\/span> A wave moves on a slinky with frequency of 4Hz and wavelength of 0.4 m. What is the speed of the wave?<\/strong> Given that, f = 4 Hz, \u03bb = 0.4 m v = (4 Hz) (0.4 m)<\/p>\n v = 1.6 m s-1<\/sup><\/p>\n Example 10.3:<\/span> A student performs an experiment with waves in water. The student measures the wavelength of a wave to be 10 cm. By using a stopwatch and observing the oscillations <\/strong>of a floating ball, the student measures a frequency of 2 Hz. If the student starts a wave in one part of a tank of water, how long will it take the wave to reach the opposite side of the <\/strong>tank 2 m away?<\/strong> v = f \u03bb = (2 Hz)(0.1 m) = 0.2 m s-1<\/sup> . t = 2 m\/0.2 m s-1<\/sup> = 10s<\/p>\n <—————————–><\/p>\n <\/p>\n <\/p>\n Goto Top<\/a><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Chapter 10 Numerical Problems Examples Example 10.2: A wave moves on a slinky with frequency of 4Hz and wavelength of 0.4 m. What is the speed of the wave? Solution: Given that, f = 4 Hz, \u03bb = 0.4 m Wave speed\u00a0 v = f \u03bb v = (4 Hz) (0.4 m) v = 1.6… Read More »Physics 10 Chapter 10 Examples<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"neve_meta_sidebar":"","neve_meta_container":"","neve_meta_enable_content_width":"off","neve_meta_content_width":0,"neve_meta_title_alignment":"","neve_meta_author_avatar":"","neve_post_elements_order":"","neve_meta_disable_header":"","neve_meta_disable_footer":"","neve_meta_disable_title":""},"categories":[9],"tags":[],"_links":{"self":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/1494"}],"collection":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/comments?post=1494"}],"version-history":[{"count":6,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/1494\/revisions"}],"predecessor-version":[{"id":2095,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/posts\/1494\/revisions\/2095"}],"wp:attachment":[{"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/media?parent=1494"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/categories?post=1494"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/murreeroad.org\/physics10\/wp-json\/wp\/v2\/tags?post=1494"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nSolution:<\/strong><\/span><\/p>\n
\nWave speed\u00a0 v = f \u03bb<\/p>\n
\nSolution:<\/strong><\/span>
\n(1) We are given the frequency, wavelength, and distance.
\n(2) We have to calculate the time, the wave takes to move a
\ndistance of 2 m.
\n(3) The relationship between frequency, wavelength, and speed is
\nv = f \u03bb. The relationship between time, speed, and distance is
\nv = d \/t
\n(4) Rearrange the speed formula to solve for the time: t = d \/ v
\nThe speed of the wave is the frequency times the wavelength.<\/p>\n
\nUse this value to calculate the time:<\/p>\n