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	<title>Physics Tuition Singapore by The Physics Coach</title>
	
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		<title>Gravitation Quiz</title>
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		<pubDate>Mon, 30 Jul 2012 05:22:35 +0000</pubDate>
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				<category><![CDATA[Gravitation]]></category>

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		<description><![CDATA[These are 10 multiple choice questions on the topic of Gravitation according to the Singapore-Cambridge A-level Physics exams (syllabus 9646). Try it out to test your understanding of the topic. Worked solutions are provided at the end of the quiz.]]></description>
			<content:encoded><![CDATA[<p>These are 10 multiple choice questions on the topic of Gravitation according to the Singapore-Cambridge A-level Physics exams (syllabus 9646). Try it out to test your understanding of the topic. Worked solutions are provided at the end of the quiz.</p>
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  <h2>Gravitation</h2>
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            <div class='mtq_question mtq_scroll_item-1' id='mtq_question-1-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 1</div><div id='mtq_stamp-1-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-1-1' class='mtq_question_text'>A boy can jump to a maximum height of h on the surface of Earth. If he were on another planet whose radius is four times that of Earth’s, and density is only half that of Earth’s, determine the maximum height the boy can attain when he jumps.</div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-1-1-1' onclick='mtq_button_click(1,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-1-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 1, Choice 1'>A</div><div id='mtq_marker-1-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-1-1-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_2a11b0ccdeff1400499ce04e15193bd5.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{h}{4}" /></span><script type='math/tex'>\frac{h}{4}</script></div></td></tr><tr id='mtq_row-1-2-1' onclick='mtq_button_click(1,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-1-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 1, Choice 2'>B</div><div id='mtq_marker-1-2-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-1-2-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_8b35f2ed497b02eb80db965263552ec4.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{h}{2}" /></span><script type='math/tex'>\frac{h}{2}</script></div></td></tr><tr id='mtq_row-1-3-1' onclick='mtq_button_click(1,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-1-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 1, Choice 3'>C</div><div id='mtq_marker-1-3-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-1-3-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_2bdde03f6b9c93b8f56a9f77a1a5136c.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="2h" /></span><script type='math/tex'>2h</script></div></td></tr><tr id='mtq_row-1-4-1' onclick='mtq_button_click(1,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-1-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 1, Choice 4'>D</div><div id='mtq_marker-1-4-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-1-4-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_936446dfd48b33e71d8cd7619cc72256.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="4h" /></span><script type='math/tex'>4h</script></div></td></tr></table><div id='mtq_question_explanation-1-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 1 Explanation:&nbsp;</div><div class='mtq_explanation-text'> <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_8b35f2ed497b02eb80db965263552ec4.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{h}{2}" /></span><script type='math/tex'>\frac{h}{2}</script> is the answer as <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_f1c50764fc65fecd82a5c28ae1688eb5.gif' style='vertical-align: middle; border: none; ' class='tex' alt="g=\frac{GM}{r^2} \\= \frac{G\rho{\frac{4}{3}}\pi r^3}{r^2} \\=  G\rho{\frac{4}{3}}\pi r " /></span><script type='math/tex'>g=\frac{GM}{r^2} \\= \frac{G\rho{\frac{4}{3}}\pi r^3}{r^2} \\=  G\rho{\frac{4}{3}}\pi r </script>
<p>
As <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_d2606be4e0cd2c9a6179c8f2e3547a85.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="\rho" /></span><script type='math/tex'>\rho</script> decreases by half and <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_4b43b0aee35624cd95b910189b3dc231.gif' style='vertical-align: middle; border: none; padding-bottom:2px;' class='tex' alt="r" /></span><script type='math/tex'>r</script> increases to 4 times, the overall effect is that <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_b2f5ff47436671b6e533d8dc3614845d.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="g" /></span><script type='math/tex'>g</script> is doubled.</p> <p>Hence, if we approximate the change in gravitational potential energy as the boy jumps to be <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_d769f3ab30112939f2b4469f8a92804e.gif' style='vertical-align: middle; border: none; ' class='tex' alt="mgh" /></span><script type='math/tex'>mgh</script>, his maximum height is halved.</p></div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-2-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 2</div><div id='mtq_stamp-2-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-2-1' class='mtq_question_text'>An Earth-orbiting satellite experiences a continuous small resistance from external influences and gradually transits into lower orbits. Which of the following correctly gives the changes in the kinetic energy and the gravitational potential energy of the satellite?</div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-2-1-1' onclick='mtq_button_click(2,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-2-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 2, Choice 1'>A</div><div id='mtq_marker-2-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-2-1-1' class='mtq_answer_text'>Kinetic energy increases and gravitational potential energy increases</div></td></tr><tr id='mtq_row-2-2-1' onclick='mtq_button_click(2,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-2-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 2, Choice 2'>B</div><div id='mtq_marker-2-2-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-2-2-1' class='mtq_answer_text'>Kinetic energy increases and gravitational potential energy decreases</div></td></tr><tr id='mtq_row-2-3-1' onclick='mtq_button_click(2,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-2-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 2, Choice 3'>C</div><div id='mtq_marker-2-3-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-2-3-1' class='mtq_answer_text'>Kinetic energy decreases and gravitational potential energy increases</div></td></tr><tr id='mtq_row-2-4-1' onclick='mtq_button_click(2,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-2-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 2, Choice 4'>D</div><div id='mtq_marker-2-4-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-2-4-1' class='mtq_answer_text'>Kinetic energy decreases and gravitational potential energy decreases</div></td></tr></table><div id='mtq_question_explanation-2-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 2 Explanation:&nbsp;</div><div class='mtq_explanation-text'> From the graph below, for an object in orbit (where gravitational force provides the centripetal force), when distance from the centre of the orbit <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_4b43b0aee35624cd95b910189b3dc231.gif' style='vertical-align: middle; border: none; padding-bottom:2px;' class='tex' alt="r" /></span><script type='math/tex'>r</script> decreases, kinetic energy <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_08f5af0d3874b9d9388108ed70fa0102.gif' style='vertical-align: middle; border: none; ' class='tex' alt="E_k" /></span><script type='math/tex'>E_k</script> increases and potential energy <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_4c614360da93c0a041b22e537de151eb.gif' style='vertical-align: middle; border: none; ' class='tex' alt="U" /></span><script type='math/tex'>U</script> decreases. <p><img src="http://www.thephysicscoach.com/wp-content/uploads/2012/07/energygraphsgravitation.png"></p></div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-3-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 3</div><div id='mtq_stamp-3-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-3-1' class='mtq_question_text'>Star X of mass 2M and Star Y of mass M perform circular motion about their common centre of mass under their gravitational attraction. What is the ratio of the force acting on X to the force acting on Y, ignoring the effects of any other bodies?</div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-3-1-1' onclick='mtq_button_click(3,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-3-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 3, Choice 1'>A</div><div id='mtq_marker-3-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-3-1-1' class='mtq_answer_text'>0.5</div></td></tr><tr id='mtq_row-3-2-1' onclick='mtq_button_click(3,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-3-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 3, Choice 2'>B</div><div id='mtq_marker-3-2-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-3-2-1' class='mtq_answer_text'>1</div></td></tr><tr id='mtq_row-3-3-1' onclick='mtq_button_click(3,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-3-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 3, Choice 3'>C</div><div id='mtq_marker-3-3-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-3-3-1' class='mtq_answer_text'>2</div></td></tr><tr id='mtq_row-3-4-1' onclick='mtq_button_click(3,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-3-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 3, Choice 4'>D</div><div id='mtq_marker-3-4-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-3-4-1' class='mtq_answer_text'>4</div></td></tr></table><div id='mtq_question_explanation-3-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 3 Explanation:&nbsp;</div><div class='mtq_explanation-text'> There is no need to do any calculations as the forces are action-reaction pairs. This is a tricky one.</div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-4-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 4</div><div id='mtq_stamp-4-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-4-1' class='mtq_question_text'>The diagram below shows how the gravitational potential varies between the moon and Earth. At which position will a particle experience zero net force?

<a href="http://www.thephysicscoach.com/wp-content/uploads/2012/07/potential.png"><img src="http://www.thephysicscoach.com/wp-content/uploads/2012/07/potential.png" alt="" title="potential" width="295" height="195" class="aligncenter size-full wp-image-511" /></a></div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-4-1-1' onclick='mtq_button_click(4,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-4-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 4, Choice 1'>A</div><div id='mtq_marker-4-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-4-1-1' class='mtq_answer_text'>A</div></td></tr><tr id='mtq_row-4-2-1' onclick='mtq_button_click(4,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-4-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 4, Choice 2'>B</div><div id='mtq_marker-4-2-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-4-2-1' class='mtq_answer_text'>B</div></td></tr><tr id='mtq_row-4-3-1' onclick='mtq_button_click(4,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-4-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 4, Choice 3'>C</div><div id='mtq_marker-4-3-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-4-3-1' class='mtq_answer_text'>C</div></td></tr><tr id='mtq_row-4-4-1' onclick='mtq_button_click(4,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-4-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 4, Choice 4'>D</div><div id='mtq_marker-4-4-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-4-4-1' class='mtq_answer_text'>D</div></td></tr></table><div id='mtq_question_explanation-4-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 4 Explanation:&nbsp;</div><div class='mtq_explanation-text'> Since <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_837b2300bdb35805e5b4ec6df07439b4.gif' style='vertical-align: middle; border: none; ' class='tex' alt="F = mg" /></span><script type='math/tex'>F = mg</script> and <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_4dda5920e870b7f9fbc682f26a08674a.gif' style='vertical-align: middle; border: none; ' class='tex' alt="g=-\frac{\text{d}\phi}{\text{d}r}" /></span><script type='math/tex'>g=-\frac{\text{d}\phi}{\text{d}r}</script>, the position with zero net force is when the gradient of the <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_aae480686b7cb1f3f07c7fe4bc8333d9.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\phi-r" /></span><script type='math/tex'>\phi-r</script> graph is zero.</div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-5-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 5</div><div id='mtq_stamp-5-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-5-1' class='mtq_question_text'>Two stars of equal mass M move with constant speed v in a circular orbit of radius R about their common centre of mass as shown in the diagram below.

<p><div id="attachment_512" class="wp-caption aligncenter" style="width: 158px"><a href="http://www.thephysicscoach.com/wp-content/uploads/2012/07/binarystars.png"><img class="size-full wp-image-512" title="binarystars" src="http://www.thephysicscoach.com/wp-content/uploads/2012/07/binarystars.png" alt="" width="148" height="137" /></a><p class="wp-caption-text">Binary Stars</p></div>
</p><p>
What is the net force on each star?</p></div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-5-1-1' onclick='mtq_button_click(5,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-5-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 5, Choice 1'>A</div><div id='mtq_marker-5-1-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-5-1-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_eb1c87ef46a7826cb2b5794e8d05ee8e.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{GM^2}{4R^2}" /></span><script type='math/tex'>\frac{GM^2}{4R^2}</script></div></td></tr><tr id='mtq_row-5-2-1' onclick='mtq_button_click(5,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-5-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 5, Choice 2'>B</div><div id='mtq_marker-5-2-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-5-2-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_8ea35f3e1d87e4eeaa0170ff8e6ec409.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{Mv^2}{2R}" /></span><script type='math/tex'>\frac{Mv^2}{2R}</script></div></td></tr><tr id='mtq_row-5-3-1' onclick='mtq_button_click(5,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-5-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 5, Choice 3'>C</div><div id='mtq_marker-5-3-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-5-3-1' class='mtq_answer_text'>zero</div></td></tr><tr id='mtq_row-5-4-1' onclick='mtq_button_click(5,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-5-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 5, Choice 4'>D</div><div id='mtq_marker-5-4-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-5-4-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_eec814f61a5b9eb31e9083c6a1aee50f.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{2Mv^2}{R}" /></span><script type='math/tex'>\frac{2Mv^2}{R}</script></div></td></tr></table><div id='mtq_question_explanation-5-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 5 Explanation:&nbsp;</div><div class='mtq_explanation-text'> The distance apart is 2R. Using Newton's law of gravitation, <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_d1ddca7acd7598d3697b1c26e093278e.gif' style='vertical-align: middle; border: none; ' class='tex' alt="F=\frac{GM^2}{(2R)^2}" /></span><script type='math/tex'>F=\frac{GM^2}{(2R)^2}</script>. Alternatively, <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_37af79061214a733a6be2c20a819410a.gif' style='vertical-align: middle; border: none; ' class='tex' alt="F=\frac{Mv^2}{R}" /></span><script type='math/tex'>F=\frac{Mv^2}{R}</script> could have been an option too as the gravitational force provides the centripetal force.</div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-6-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 6</div><div id='mtq_stamp-6-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-6-1' class='mtq_question_text'>The acceleration of free-fall at the Earth’s surface is <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_b2f5ff47436671b6e533d8dc3614845d.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="g" /></span><script type='math/tex'>g</script> and the radius of the Earth is <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_e1e1d3d40573127e9ee0480caf1283d6.gif' style='vertical-align: middle; border: none; ' class='tex' alt="R" /></span><script type='math/tex'>R</script>.
What will be the acceleration of free fall at a height <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_2510c39011c5be704182423e3a695e91.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="h" /></span><script type='math/tex'>h</script> above the surface?
</div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-6-1-1' onclick='mtq_button_click(6,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-6-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 6, Choice 1'>A</div><div id='mtq_marker-6-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-6-1-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_ce180e064169f2c2f03089e7c675788a.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{gR^2}{h^2}" /></span><script type='math/tex'>\frac{gR^2}{h^2}</script></div></td></tr><tr id='mtq_row-6-2-1' onclick='mtq_button_click(6,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-6-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 6, Choice 2'>B</div><div id='mtq_marker-6-2-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-6-2-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_1ea913112d21f4c9bc33044f21577013.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{gh^2}{R^2}" /></span><script type='math/tex'>\frac{gh^2}{R^2}</script></div></td></tr><tr id='mtq_row-6-3-1' onclick='mtq_button_click(6,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-6-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 6, Choice 3'>C</div><div id='mtq_marker-6-3-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-6-3-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_10a50aa5bf37251fbcc86a0d82f68a95.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{gh}{R+h}" /></span><script type='math/tex'>\frac{gh}{R+h}</script></div></td></tr><tr id='mtq_row-6-4-1' onclick='mtq_button_click(6,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-6-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 6, Choice 4'>D</div><div id='mtq_marker-6-4-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-6-4-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_ab24fdefa15c55c6f8b9bac6ce5d8fd5.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{gR^2}{(R+h)^2}" /></span><script type='math/tex'>\frac{gR^2}{(R+h)^2}</script></div></td></tr></table><div id='mtq_question_explanation-6-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 6 Explanation:&nbsp;</div><div class='mtq_explanation-text'> <p><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_c870edc4db1934de50f585020f41b1e5.gif' style='vertical-align: middle; border: none; ' class='tex' alt="g\propto{\frac{1}{R^2}}" /></span><script type='math/tex'>g\propto{\frac{1}{R^2}}</script></p>
<p><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_8f94755c202756c2598ba88b39ac7a7b.gif' style='vertical-align: middle; border: none; ' class='tex' alt="g'(R+h)^2=gR^2" /></span><script type='math/tex'>g'(R+h)^2=gR^2</script></p>
</div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-7-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 7</div><div id='mtq_stamp-7-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-7-1' class='mtq_question_text'>Two satellites A and B orbit the Earth in circular orbits, the radius of satellite A’s orbit being 4 times that of satellite B. If the orbital period and tangential velocity of satellite A are T and v respectively, what are the corresponding values for satellite B?
<table border="black" cellpadding=5>
<tbody>
<tr>
<td></td>
<td>Period</td>
<td>Tangential velocity</td>
</tr>
<tr>
<td>A</td>
<td><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_c7c57fc1c2cf59f2817e40115476010f.gif' style='vertical-align: middle; border: none; ' class='tex' alt="8T" /></span><script type='math/tex'>8T</script></td>
<td><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_a0d428d574a42aad6669cb42b01ae234.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="2v" /></span><script type='math/tex'>2v</script></td>
</tr>
<tr>
<td>B</td>
<td><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_bb7cd4cbdf02cae2a65da705ae63cedb.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{1}{8}T" /></span><script type='math/tex'>\frac{1}{8}T</script></td>
<td><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_a0d428d574a42aad6669cb42b01ae234.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="2v" /></span><script type='math/tex'>2v</script></td>
</tr>
<tr>
<td>C</td>
<td><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_bb7cd4cbdf02cae2a65da705ae63cedb.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{1}{8}T" /></span><script type='math/tex'>\frac{1}{8}T</script></td>
<td><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_38cfe3d42b14530aed539fd49088172f.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{1}{2}v" /></span><script type='math/tex'>\frac{1}{2}v</script></td>
</tr>
<tr>
<td>D</td>
<td><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_c7c57fc1c2cf59f2817e40115476010f.gif' style='vertical-align: middle; border: none; ' class='tex' alt="8T" /></span><script type='math/tex'>8T</script></td>
<td><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_38cfe3d42b14530aed539fd49088172f.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\frac{1}{2}v" /></span><script type='math/tex'>\frac{1}{2}v</script></td>
</tr>
</tbody>
</table></div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-7-1-1' onclick='mtq_button_click(7,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-7-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 7, Choice 1'>A</div><div id='mtq_marker-7-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-7-1-1' class='mtq_answer_text'>A</div></td></tr><tr id='mtq_row-7-2-1' onclick='mtq_button_click(7,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-7-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 7, Choice 2'>B</div><div id='mtq_marker-7-2-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-7-2-1' class='mtq_answer_text'>B</div></td></tr><tr id='mtq_row-7-3-1' onclick='mtq_button_click(7,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-7-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 7, Choice 3'>C</div><div id='mtq_marker-7-3-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-7-3-1' class='mtq_answer_text'>C</div></td></tr><tr id='mtq_row-7-4-1' onclick='mtq_button_click(7,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-7-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 7, Choice 4'>D</div><div id='mtq_marker-7-4-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-7-4-1' class='mtq_answer_text'>D</div></td></tr></table><div id='mtq_question_explanation-7-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 7 Explanation:&nbsp;</div><div class='mtq_explanation-text'> <p><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_152262e9beb2310031660e49d98314d0.gif' style='vertical-align: middle; border: none; ' class='tex' alt="F=\frac{GMm}{r^2}=mr{\omega}^2 \\
\frac{GM}{r^3}=({\frac{2\pi}{T}})^2 \\
r^3 \propto T^2" /></span><script type='math/tex'>F=\frac{GMm}{r^2}=mr{\omega}^2 \\
\frac{GM}{r^3}=({\frac{2\pi}{T}})^2 \\
r^3 \propto T^2</script></p>
<p>When <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_a4bfc0ca9a5edaf346271b2777fbb32c.gif' style='vertical-align: middle; border: none; ' class='tex' alt="r_B = \frac{1}{4}r_A" /></span><script type='math/tex'>r_B = \frac{1}{4}r_A</script>, <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_568be88dc709293281c02c7b399efee2.gif' style='vertical-align: middle; border: none; ' class='tex' alt="T_B = \frac{1}{8}T_A" /></span><script type='math/tex'>T_B = \frac{1}{8}T_A</script><p>
<p><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_269c4f1e39ef7405cdb4fcda07c328e1.gif' style='vertical-align: middle; border: none; ' class='tex' alt="F=\frac{GMm}{r^2}=\frac{mv^2}{r} \\
r \propto \frac{1}{v^2}" /></span><script type='math/tex'>F=\frac{GMm}{r^2}=\frac{mv^2}{r} \\
r \propto \frac{1}{v^2}</script></p>
<p>When <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_a4bfc0ca9a5edaf346271b2777fbb32c.gif' style='vertical-align: middle; border: none; ' class='tex' alt="r_B = \frac{1}{4}r_A" /></span><script type='math/tex'>r_B = \frac{1}{4}r_A</script>, <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_5c19972643ff35b045ef4c89867bfaf2.gif' style='vertical-align: middle; border: none; ' class='tex' alt="v_B = 2v_A" /></span><script type='math/tex'>v_B = 2v_A</script><p>
</div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-8-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 8</div><div id='mtq_stamp-8-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-8-1' class='mtq_question_text'>The diagram shows two points A and B which are at distances X and 2X from the centre of the Earth respectively. At point A, the gravitational potential is -8 kJ kg<sup>-1</sup>.

<p><a href="http://www.thephysicscoach.com/wp-content/uploads/2012/08/potentialdiff.png"><img class="aligncenter size-full wp-image-513" title="potentialdiff" src="http://www.thephysicscoach.com/wp-content/uploads/2012/08/potentialdiff.png" alt="" width="256" height="155" /></a></p>

What is the change in potential energy when a satellite of mass 2.0 kg is moved from A to B?</div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-8-1-1' onclick='mtq_button_click(8,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-8-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 8, Choice 1'>A</div><div id='mtq_marker-8-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-8-1-1' class='mtq_answer_text'>- 4 kJ</div></td></tr><tr id='mtq_row-8-2-1' onclick='mtq_button_click(8,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-8-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 8, Choice 2'>B</div><div id='mtq_marker-8-2-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-8-2-1' class='mtq_answer_text'>- 8 kJ</div></td></tr><tr id='mtq_row-8-3-1' onclick='mtq_button_click(8,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-8-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 8, Choice 3'>C</div><div id='mtq_marker-8-3-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-8-3-1' class='mtq_answer_text'>+ 4 kJ</div></td></tr><tr id='mtq_row-8-4-1' onclick='mtq_button_click(8,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-8-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 8, Choice 4'>D</div><div id='mtq_marker-8-4-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-8-4-1' class='mtq_answer_text'>+ 8 kJ</div></td></tr></table><div id='mtq_question_explanation-8-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 8 Explanation:&nbsp;</div><div class='mtq_explanation-text'> <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_4c3f4253002d03547c494747d2ce7461.gif' style='vertical-align: middle; border: none; ' class='tex' alt="\phi_B = - 4 \text{ kJ kg}^{-1} \\
\Delta \phi= \phi_B - \phi_A \\
= +4 \text{ kJ kg}^{-1} \\
\Delta U = m \Delta \phi \\
= +8 \text{ kJ}" /></span><script type='math/tex'>\phi_B = - 4 \text{ kJ kg}^{-1} \\
\Delta \phi= \phi_B - \phi_A \\
= +4 \text{ kJ kg}^{-1} \\
\Delta U = m \Delta \phi \\
= +8 \text{ kJ}</script>
</div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-9-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 9</div><div id='mtq_stamp-9-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-9-1' class='mtq_question_text'>A satellite is at a height h above the surface of the Earth. If the radius of the Earth is r, and the acceleration due to gravity at the Earth’s surface is g, the period of orbit of the satellite will be </div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-9-1-1' onclick='mtq_button_click(9,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-9-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 9, Choice 1'>A</div><div id='mtq_marker-9-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-9-1-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_b6bef551502ffb2d856992e296f314f6.gif' style='vertical-align: middle; border: none; ' class='tex' alt="2\pi\sqrt{\frac{r+h}{g}}" /></span><script type='math/tex'>2\pi\sqrt{\frac{r+h}{g}}</script></div></td></tr><tr id='mtq_row-9-2-1' onclick='mtq_button_click(9,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-9-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 9, Choice 2'>B</div><div id='mtq_marker-9-2-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-9-2-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_9914f596803f0c1013f62ef7acd10d7d.gif' style='vertical-align: middle; border: none; ' class='tex' alt="2\pi\sqrt{\frac{(r+h)^2}{gr}}" /></span><script type='math/tex'>2\pi\sqrt{\frac{(r+h)^2}{gr}}</script></div></td></tr><tr id='mtq_row-9-3-1' onclick='mtq_button_click(9,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-9-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 9, Choice 3'>C</div><div id='mtq_marker-9-3-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-9-3-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_c61e2a496ecc382e4654df5dc09c4d2b.gif' style='vertical-align: middle; border: none; ' class='tex' alt="2\pi\sqrt{\frac{r^2}{g(r+h)}}" /></span><script type='math/tex'>2\pi\sqrt{\frac{r^2}{g(r+h)}}</script></div></td></tr><tr id='mtq_row-9-4-1' onclick='mtq_button_click(9,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-9-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 9, Choice 4'>D</div><div id='mtq_marker-9-4-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-9-4-1' class='mtq_answer_text'><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_40de8885896f13c9eb2d5b0973330f2a.gif' style='vertical-align: middle; border: none; ' class='tex' alt="2\pi\sqrt{\frac{(r+h)^3}{gr^2}}" /></span><script type='math/tex'>2\pi\sqrt{\frac{(r+h)^3}{gr^2}}</script></div></td></tr></table><div id='mtq_question_explanation-9-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 9 Explanation:&nbsp;</div><div class='mtq_explanation-text'> <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_d07db2218e58b4175292cb0064f905eb.gif' style='vertical-align: middle; border: none; ' class='tex' alt="g=\frac{GM}{r^2} \\
g'=(\frac{r}{r+h})^2 g \\
g' = (r+h)\omega^2=(r+h)(\frac{2\pi}{T})^2 \\
T =2\pi\sqrt{\frac{(r+h)^3}{gr^2}}" /></span><script type='math/tex'>g=\frac{GM}{r^2} \\
g'=(\frac{r}{r+h})^2 g \\
g' = (r+h)\omega^2=(r+h)(\frac{2\pi}{T})^2 \\
T =2\pi\sqrt{\frac{(r+h)^3}{gr^2}}</script>
</div></div></div><div class='mtq_question mtq_scroll_item-1' id='mtq_question-10-1'><table class='mtq_question_heading_table'><tr><td><div class='mtq_question_label '>Question 10</div><div id='mtq_stamp-10-1' class='mtq_stamp'></div></td></tr></table><div id='mtq_question_text-10-1' class='mtq_question_text'>An artificial satellite is orbiting around the Earth at a steady speed. The astronaut in the satellite can be regarded as ‘weightless’ because</div><table class='mtq_answer_table'><colgroup><col class='mtq_oce_first'/></colgroup><tr id='mtq_row-10-1-1' onclick='mtq_button_click(10,1,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-10-1-1' class='mtq_css_letter_button mtq_letter_button_0'  alt='Question 10, Choice 1'>A</div><div id='mtq_marker-10-1-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-10-1-1' class='mtq_answer_text'>the gravitational force acting on him is zero.</div></td></tr><tr id='mtq_row-10-2-1' onclick='mtq_button_click(10,2,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-10-2-1' class='mtq_css_letter_button mtq_letter_button_1'  alt='Question 10, Choice 2'>B</div><div id='mtq_marker-10-2-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-10-2-1' class='mtq_answer_text'>the centripetal force he experienced is zero. </div></td></tr><tr id='mtq_row-10-3-1' onclick='mtq_button_click(10,3,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-10-3-1' class='mtq_css_letter_button mtq_letter_button_2'  alt='Question 10, Choice 3'>C</div><div id='mtq_marker-10-3-1' class='mtq_marker mtq_correct_marker' alt='Correct'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-10-3-1' class='mtq_answer_text'>the satellite’s acceleration is the same as his.</div></td></tr><tr id='mtq_row-10-4-1' onclick='mtq_button_click(10,4,1)' class='mtq_clickable'><td class='mtq_letter_button_td'><div id='mtq_button-10-4-1' class='mtq_css_letter_button mtq_letter_button_3'  alt='Question 10, Choice 4'>D</div><div id='mtq_marker-10-4-1' class='mtq_marker mtq_wrong_marker' alt='Wrong'></div></td><td class='mtq_answer_td'><div id='mtq_answer_text-10-4-1' class='mtq_answer_text'>the acceleration he experienced is zero.</div></td></tr></table><div id='mtq_question_explanation-10-1' class='mtq_explanation'><div class='mtq_explanation-label'>Question 10 Explanation:&nbsp;</div><div class='mtq_explanation-text'> When the satellite's acceleration is the same as his, all of the gravitational force acting on him is exactly providing for the centripetal force, hence there is no contact between him and the floor of the satellite.</div></div></div>            <div id="mtq_results_request-1" class="mtq_results_request mtq_scroll_item-1">
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]]></content:encoded>
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		<title>Special group for Dunman High School (DHS) students</title>
		<link>http://www.thephysicscoach.com/special-group-for-dunman-high-school-dhs-students/</link>
		<comments>http://www.thephysicscoach.com/special-group-for-dunman-high-school-dhs-students/#comments</comments>
		<pubDate>Wed, 23 May 2012 03:44:30 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[Blog]]></category>
		<category><![CDATA[Education]]></category>

		<guid isPermaLink="false">http://www.thephysicscoach.com/?p=502</guid>
		<description><![CDATA[I have now started a dedicated group for Dunman High School (DHS) senior high students for JC1 physics because their syllabus is far ahead of the other JCs, since they do not have a two-month wait for the O-level intake to start JC. DHS students who are interested in signing up for my special group]]></description>
			<content:encoded><![CDATA[<p>I have now started a dedicated group for Dunman High School (DHS) senior high students for JC1 physics because their syllabus is far ahead of the other JCs, since they do not have a two-month wait for the O-level intake to start JC.</p>
<p>DHS students who are interested in signing up for my special group can email me at spencer@neulearning.com or call me at the number shown on the left.</p>
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		<item>
		<title>Definition of the Tesla</title>
		<link>http://www.thephysicscoach.com/definition-of-the-tesla/</link>
		<comments>http://www.thephysicscoach.com/definition-of-the-tesla/#comments</comments>
		<pubDate>Mon, 07 May 2012 04:35:38 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[Electromagnetism]]></category>

		<guid isPermaLink="false">http://www.thephysicscoach.com/?p=476</guid>
		<description><![CDATA[I was just discussing the different definitions that some JCs give for the definition of the Tesla (the SI unit for magnetic flux density) with my JC2 students last week. According to the 'A' level examiners, they are looking for something more like: The magnetic flux density of a magnetic field is said to be]]></description>
			<content:encoded><![CDATA[<p>I was just discussing the different definitions that some JCs give for the definition of the Tesla (the SI unit for magnetic flux density) with my JC2 students last week. According to the 'A' level examiners, they are looking for something more like:</p>
<blockquote><p>The magnetic flux density of a magnetic field is said to be 1 Tesla, if the force acting per unit length on an <strong>infinitely long conductor</strong> carrying a current of 1 A and placed perpendicularly to the magnetic field is <strong>1 N m<sup>-1</sup></strong>.</p></blockquote>
<p>Rather than what is found in some A-level guide books:</p>
<blockquote><p>A tesla is the magnetic flux density if a force of 1 N acts on a wire of length 1m, carrying a current of 1A placed perpendicular to the magnetic field.</p></blockquote>
<p>Some JCs still teach this erroneous definition in their lectures.</p>
<p>So, why is the second definition wrong? In the measurement of a tesla, if we refer only to a wire of finite length, due to the end effects, the magnetic flux density at both ends of the wire will not be the same as that in the middle.</p>
<p>I will be addressing this error and more in my upcoming <a href="http://www.thephysicscoach.com/june-holiday-physics-workshop-for-jc1-and-jc2s/" title="June holiday A-level Physics intensive workshop">June holiday Physics intensive workshop</a> meant for JC1 and JC2 students. Do sign up early to avoid disappointment!</p>
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		<title>Is Singapore considering Thorium as Nuclear Fuel?</title>
		<link>http://www.thephysicscoach.com/is-singapore-considering-thorium-as-nuclear-fuel/</link>
		<comments>http://www.thephysicscoach.com/is-singapore-considering-thorium-as-nuclear-fuel/#comments</comments>
		<pubDate>Wed, 04 Apr 2012 16:12:16 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[A-level Units]]></category>
		<category><![CDATA[Nuclear Physics]]></category>

		<guid isPermaLink="false">http://www.thephysicscoach.com/?p=424</guid>
		<description><![CDATA[It's been more than a year since Fukushima's nuclear disaster and it's time to take a new look at the possibility of finding a safe way to handle nuclear power. In Guardian.co.uk, thorium has been hailed has a nuclear fuel that is safer to use than uranium. It many advantages include: greater abundance, less likelihood]]></description>
			<content:encoded><![CDATA[<p>It's been more than a year since Fukushima's nuclear disaster and it's time to take a new look at the possibility of finding a safe way to handle nuclear power. In <a href="http://www.guardian.co.uk/environment/blog/2012/mar/09/fukushima-thorium-nuclear-power-uranium">Guardian.co.uk</a>, thorium has been hailed has a nuclear fuel that is safer to use than uranium. It many advantages include:</p>
<ol>
<li>greater abundance,</li>
<li>less likelihood of runaway chain reactions that can lead to nuclear disasters</li>
<li>much shorter half-life for its waste products (therefore minimising waste storage problems)</li>
<li>byproducts that are not useful for making nuclear weapons</li>
</ol>
<p>Research has begun to design new reactors for thorium in India, China, USA and Europe. Perhaps one day, there can be a design for a reactor so small and so safe that it can find its way to Singapore's shores and replace our heavy dependence on fossil fuels.</p>
<p>We might even set a test question on Thorium to celebrate that when it happens!</p>
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		<title>Ideal Gas</title>
		<link>http://www.thephysicscoach.com/ideal-gas/</link>
		<comments>http://www.thephysicscoach.com/ideal-gas/#comments</comments>
		<pubDate>Sun, 11 Mar 2012 14:32:19 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[A-level Units]]></category>
		<category><![CDATA[Thermal Physics]]></category>

		<guid isPermaLink="false">http://www.thephysicscoach.com/?p=348</guid>
		<description><![CDATA[An ideal gas is a theoretical gas composed of a set of randomly-moving, non-interacting point particles. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state: where is the pressure of the gas, is its volume, is the amount of substance (in moles), is the Gas Constant]]></description>
			<content:encoded><![CDATA[<p>An ideal gas is a theoretical gas composed of a set of randomly-moving, non-interacting point particles. The ideal gas concept is useful because <strong>it obeys the ideal gas law</strong>, a simplified equation of state:</p>
<p style="text-align: center;"><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_39259e73816a953048a079ba4ea1454f.gif' style='vertical-align: middle; border: none; ' class='tex' alt="pV = nRT" /></span><script type='math/tex'>pV = nRT</script></p>
<p>where <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_83878c91171338902e0fe0fb97a8c47a.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="p" /></span><script type='math/tex'>p</script> is the pressure of the gas, <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_5206560a306a2e085a437fd258eb57ce.gif' style='vertical-align: middle; border: none; ' class='tex' alt="V" /></span><script type='math/tex'>V</script> is its volume, <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_7b8b965ad4bca0e41ab51de7b31363a1.gif' style='vertical-align: middle; border: none; padding-bottom:2px;' class='tex' alt="n" /></span><script type='math/tex'>n</script> is the amount of substance (in moles), <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_e1e1d3d40573127e9ee0480caf1283d6.gif' style='vertical-align: middle; border: none; ' class='tex' alt="R" /></span><script type='math/tex'>R</script> is the Gas Constant <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_d23fbd5347db08def2e81bc3014650e0.gif' style='vertical-align: middle; border: none; ' class='tex' alt="8.31 J K^{-1}mol^{-1}" /></span><script type='math/tex'>8.31 J K^{-1}mol^{-1}</script> and <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_b9ece18c950afbfa6b0fdbfa4ff731d3.gif' style='vertical-align: middle; border: none; padding-bottom:1px;' class='tex' alt="T" /></span><script type='math/tex'>T</script> is the absolute temperature.</p>
<p>At normal conditions such as standard temperature and pressure, most real gases behave qualitatively like an ideal gas. Generally, a gas behaves even more like an ideal gas at higher temperature and lower density (i.e. lower pressure), as the work performed by intermolecular forces (i.e. potential energy) becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them.</p>
<p>For a monoatomic ideal gas, internal energy <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_4c614360da93c0a041b22e537de151eb.gif' style='vertical-align: middle; border: none; ' class='tex' alt="U" /></span><script type='math/tex'>U</script> is simply the sum of the random kinetic energies of the constituent molecules and <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_729bd5c1cd3e1bd4dd62df7640221474.gif' style='vertical-align: middle; border: none; ' class='tex' alt="U=\frac{3}{2}nRT" /></span><script type='math/tex'>U=\frac{3}{2}nRT</script>.</p>
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		<title>Institute of Physics Singapore (IPS) Gold Medal</title>
		<link>http://www.thephysicscoach.com/institute-of-physics-gold-medal-ips/</link>
		<comments>http://www.thephysicscoach.com/institute-of-physics-gold-medal-ips/#comments</comments>
		<pubDate>Fri, 02 Mar 2012 08:37:46 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[Blog]]></category>
		<category><![CDATA[Education]]></category>

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		<description><![CDATA[Every year, after the A level results are released, I'm keen to find out which schools produce the top students in Physics in the nation. This year, I've gathered early intel to come up with the list below: Hwa Chong Institution: 5 (no names given) Raffles Institution: 2 Yuen Wing Yan Kang Zi Yang Temasek]]></description>
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<ol>Every year, after the A level results are released, I'm keen to find out which schools produce the top students in Physics in the nation. This year, I've gathered early intel to come up with the list below:</ol>
</ol>
<p>Hwa Chong Institution: 5<br />
(no names given)</p>
<p>Raffles Institution: 2<br />
Yuen Wing Yan<br />
Kang Zi Yang</p>
<p>Temasek Junior College: 2<br />
Chua Chin Wei<br />
Liang Liyuan</p>
<p>National Junior College: 1<br />
Yuan Bo</p>
<p>Anglo-Chinese Junior College:1<br />
Hans Adrian</p>
<p>Congratulations to these 11 students!</p>
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		<title>Line of Best Fit</title>
		<link>http://www.thephysicscoach.com/line-of-best-fit/</link>
		<comments>http://www.thephysicscoach.com/line-of-best-fit/#comments</comments>
		<pubDate>Wed, 29 Feb 2012 01:59:22 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[SPA]]></category>
		<category><![CDATA[A level Physics]]></category>
		<category><![CDATA[best fit line]]></category>

		<guid isPermaLink="false">http://thephysicscoach.com/?p=322</guid>
		<description><![CDATA[Here is a chapter from a laboratory guide book by the College of the Redwoods on the plotting of best-fit line using graph paper and ruler, as well as the graphing calculator. Comes with a set of practice questions with solutions. This work (including all text, photographs, Portable Document Format files, and any other original works),]]></description>
			<content:encoded><![CDATA[<p>Here is a chapter from a laboratory guide book by the College of the Redwoods on the plotting of best-fit line using graph paper and ruler, as well as the graphing calculator.</p>
<p><a title="Line of Best Fit" href="http://thephysicscoach.com/wp-content/uploads/2012/02/Line-of-Best-Fit.pdf"><img title="download" src="http://thephysicscoach.com/wp-content/uploads/2012/02/download.png" alt="" width="120" height="27" /></a></p>
<p>Comes with a set of practice questions with solutions.</p>
<p><a href="http://thephysicscoach.com/wp-content/uploads/2012/02/Line-of-Best-Fit-Practice-Questions.pdf"><img title="download" src="http://thephysicscoach.com/wp-content/uploads/2012/02/download.png" alt="" width="120" height="27" /></a></p>
<div><small>This work (including all text, photographs, Portable Document Format files, and any other original works), except where otherwise noted, is licensed under a <a href="http://creativecommons.org/licenses/by-nc-sa/2.5/" rel="license">Creative Commons Attribution-NonCommercial-ShareAlike 2.5 License</a>, and is copyrighted © 2006, Department of Mathematics, College of the Redwoods.</small></div>
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		<title>Monopolar Motor</title>
		<link>http://www.thephysicscoach.com/monopolar-motor/</link>
		<comments>http://www.thephysicscoach.com/monopolar-motor/#comments</comments>
		<pubDate>Tue, 28 Feb 2012 13:15:46 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[Electromagnetism]]></category>
		<category><![CDATA[motor]]></category>

		<guid isPermaLink="false">http://thephysicscoach.com/?p=326</guid>
		<description><![CDATA[This is an interesting application of electromagnetism that a physics teacher can show his student: a monopolar motor. The simplest type of DC motor there is.]]></description>
			<content:encoded><![CDATA[<p>This is an interesting application of electromagnetism that a physics teacher can show his student: a monopolar motor. The simplest type of DC motor there is.</p>
<p><iframe width="420" height="315" src="http://www.youtube.com/embed/w2f6RD1hT6Q" frameborder="0" allowfullscreen></iframe></p>
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		<title>Reaction Meter</title>
		<link>http://www.thephysicscoach.com/reaction-meter/</link>
		<comments>http://www.thephysicscoach.com/reaction-meter/#comments</comments>
		<pubDate>Thu, 23 Feb 2012 04:44:13 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[A-level Units]]></category>
		<category><![CDATA[Fun Science]]></category>
		<category><![CDATA[Kinematics]]></category>

		<guid isPermaLink="false">http://thephysicscoach.com/?p=297</guid>
		<description><![CDATA[This is a simple tool created for my JC1 students as they begin to learn about kinematics. It's designed based on the equation . Feel free to print it out (on A4 sized paper that is at least 80 gsm). What I tell my students to do is this: 1. Get a partner. One will]]></description>
			<content:encoded><![CDATA[<p>This is a simple tool created for my JC1 students as they begin to learn about kinematics. It's designed based on the equation <span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_cbea8d956824a00ef8ef45444d314bc1.gif' style='vertical-align: middle; border: none; ' class='tex' alt="s = ut +\frac{1}{2}at^2" /></span><script type='math/tex'>s = ut +\frac{1}{2}at^2</script>. Feel free to print it out (on A4 sized paper that is at least 80 gsm).</p>
<p>What I tell my students to do is this:</p>
<p>1. Get a partner. One will release the card while the other will try to catch it with his fingers.<br />
2. The releaser will place the card betweeen the catcher's thumb and index finger which should leave a small gap of about 2 cm.<br />
3. Without any advance notic, the one holding the card will let go of the card.<br />
4. The position where the catcher grips the card will indicate his reaction time. </p>
<p><a href="http://thephysicscoach.com/wp-content/uploads/2012/02/Reaction-Meter.pdf"><img class="aligncenter size-full wp-image-247" title="download" src="http://thephysicscoach.com/wp-content/uploads/2012/02/download.png" alt="" width="120" height="27" /></a></p>
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		<title>Fermi Questions</title>
		<link>http://www.thephysicscoach.com/fermi-questions/</link>
		<comments>http://www.thephysicscoach.com/fermi-questions/#comments</comments>
		<pubDate>Fri, 17 Feb 2012 06:22:23 +0000</pubDate>
		<dc:creator />
				<category><![CDATA[A-level Units]]></category>
		<category><![CDATA[Measurement]]></category>

		<guid isPermaLink="false">http://thephysicscoach.com/?p=259</guid>
		<description><![CDATA[2008 GCE A Level Physics Paper 1 Question 2 Which estimate is realistic? A. The kinetic energy of a bus travelling on an expressway is 30 000 J. B. The power of a domestic light is 300 W. C. The temperature of a hot oven is 300 K. D. The volume of air in a]]></description>
			<content:encoded><![CDATA[<p style="text-align: center;"><a href="http://thephysicscoach.com/wp-content/uploads/2012/02/tyres1.jpg"><img class="aligncenter size-full wp-image-276" title="tyres" src="http://thephysicscoach.com/wp-content/uploads/2012/02/tyres1.jpg" alt="GCE A level physics question: volume of a tyre" width="400" height="300" /></a></p>
<h4>2008 GCE A Level Physics Paper 1 Question 2</h4>
<p>Which estimate is realistic?</p>
<p>A. The kinetic energy of a bus travelling on an expressway is 30 000 J.<br />
B. The power of a domestic light is 300 W.<br />
C. The temperature of a hot oven is 300 K.<br />
D. The volume of air in a car tyre is 0.03 m<sup>3</sup>.<br />
<span id="more-259"></span><br />
The answer to the above question is D.</p>
<p>Based on the assumption that the car type takes the shape of a cylinder with a circular hollow in the middle, and estimating the diameter of the wheel to be 60 cm, the diameter of the hub to be 40 cm and the width to be 20 cm, the volume can be calculated as below:</p>
<p><span class='MathJax_Preview'><img src='http://www.thephysicscoach.com/wp-content/plugins/latex/cache/tex_7b6bc2f1f3ed955efd77ef786320f327.gif' style='vertical-align: middle; border: none; ' class='tex' alt="V =\pi (r_2^2 - r_1^2)\times w= \pi (0.30^2 - 0.20^2)\times 0.20 = 0.031 m^3" /></span><script type='math/tex'>V =\pi (r_2^2 - r_1^2)\times w= \pi (0.30^2 - 0.20^2)\times 0.20 = 0.031 m^3</script></p>
<p><strong>About Fermi Questions</strong></p>
<p>Almost every year, in the A level Physics paper 1, there will be a Fermi Question. A "Fermi question" is a question in physics which seeks a fast, rough estimate of quantity which is either difficult or impossible to measure directly. Also know as "back-of-the-envelope" questions, such questions could potentially stump the brightest students as well as the most hardworking ones.</p>
<p>Such questions are named after the American physicist Enrico Fermi. One example of him solving such a question was his estimate of the strength of the atomic bomb based on the distance travelled by bits of paper dropped from his hand at the test blasting.</p>
<p>Estimation in Physics uses simple numbers (usually 1, sometimes 2 or 5) with the correct order of magnitude (e.g. 10<sup>3</sup> or 10<sup>-4</sup>).</p>
<p>The answer calculated for the above MCQ does not have to be exactly the same as that in the option, just close enough.</p>
<p>The tip I would give my student for such questions will be to pay attention to the physical quantities around them. Values like the power rating of a fluorescent bulb (~10-20W), the mass of a bus (~10 tons), the speed limit on the expressway for a bus (60 km/h), the temperature of an oven (~200 degrees Celsius or ~473 K).</p>
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