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	<title>Physics Tuition</title>
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		<title>Kinematics Examples</title>
		<link>http://www.physicsandcalculus.com/2012/kinematics-examples</link>
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		<pubDate>Tue, 21 Feb 2012 09:45:04 +0000</pubDate>
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		<description><![CDATA[Symmetry in Kinematics Problems For those in Sec 3 studying Kinematics now and for those in Sec 4 taking the O-Level this year, remember a couple of quick points about the symmetry of kinematics problems: 1. If you throw a ball up in the air vertically with a speed of 10 m/s, it will come [...]]]></description>
			<content:encoded><![CDATA[<p>Symmetry in Kinematics Problems<br />
<br />
For those in Sec 3 studying Kinematics now and for those in Sec 4 taking the O-Level this year, remember a couple of quick points about the symmetry of kinematics problems:<br />
1.	If you throw a ball up in the air vertically with a speed of 10 m/s, it will come back down with the same speed in the opposite direction.<br />
2.	If you throw a ball up in the air and it takes 2 seconds to reach the highest point, then it takes 2 seconds from the highest point to fall back to the ground.<br />
This all assumes no air friction. With air friction the returning speed is slower.</p>
<p>See if you can also follow these two examples:<br />
<br />
1.	A ball thrown vertically down toward the ground from a 50 meter tall building with an initial speed of 20 m/s. Assuming that it falls without air friction, what is the velocity when it hits the ground?<br />
<br />
<a href="http://www.physicsandcalculus.com/wp-content/uploads/2012/02/kinematics-pic-1.jpg"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2012/02/kinematics-pic-1.jpg" alt="" title="kinematics pic 1" width="329" height="111" class="aligncenter size-full wp-image-1637" /></a><br />
<br />
It should make sense that we now have a faster velocity since we threw the ball downward to begin with.<br />
<br />
2.	A ball is thrown directly upward from a 50 meter tall building with an initial speed of 20 m/s. Assuming that it falls without air friction, what is the velocity when it hits the ground?<br />
<br />
<a href="http://www.physicsandcalculus.com/wp-content/uploads/2012/02/kinematics-pic-2.jpg"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2012/02/kinematics-pic-2.jpg" alt="" title="kinematics pic 2" width="591" height="106" class="aligncenter size-full wp-image-1638" /></a><br />
<br />
We took the initial velocity here as a negative number because the ball was thrown upward and we are calling downward positive. If a ball is thrown up at 20 m/s by the time it gets back down to the top of the roof, it is going downward at 20 m/s. So the answer comes out to the same number!</p>
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		<title>Speed of the Earth in Space</title>
		<link>http://www.physicsandcalculus.com/2012/speed-of-the-earth-in-space</link>
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		<pubDate>Sun, 12 Feb 2012 13:06:07 +0000</pubDate>
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		<description><![CDATA[In Lower Secondary class we are covering the basic formula Speed = distance /time. Lower Secondary Question: Given that the earth is 150 million km from the sun, find the orbital speed of the earth around the sun in km/hr. Assume that the earth’s orbit is circular and take the year to be 365 days. [...]]]></description>
			<content:encoded><![CDATA[<p><a href="http://www.physicsandcalculus.com/wp-content/uploads/2012/02/earth-sun21.jpg"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2012/02/earth-sun21.jpg" alt="" title="earth-sun2" width="325" height="251" class="aligncenter size-full wp-image-1629" /></a></p>
<p>In Lower Secondary class we are covering the basic formula Speed = distance /time.<br />
<br />
<strong>Lower Secondary Question:</strong> Given that the earth is 150 million km from the sun, find the orbital speed of the earth around the sun in km/hr. Assume that the earth’s orbit is circular and take the year to be 365 days.</p>
<p><strong>Solution:</strong> The circumference of a circle is given by 2(pi)r = 2×3.14×150×10^6km = 9.42 ×10^8 km. This is the distance that the earth travels in one year.  Taking speed = distance/time<br />
<br />
<a href="http://www.physicsandcalculus.com/wp-content/uploads/2012/02/formula1.jpg"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2012/02/formula1.jpg" alt="" title="formula" width="328" height="51" class="aligncenter size-full wp-image-1630" /></a><br />
<br />
As a comparison the escape velocity of a rocket traveling to Mars only has to reach a speed of 45,000 km/hr to get away from the earth&#8217;s gravity. So the earth is moving very fast around the sun !</p>
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		<title>Neutrinos faster than light?</title>
		<link>http://www.physicsandcalculus.com/2012/neutrinos-faster-than-light</link>
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		<pubDate>Tue, 10 Jan 2012 07:33:30 +0000</pubDate>
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				<category><![CDATA[Physics]]></category>
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		<description><![CDATA[Whats going on with Neutrinos? A few students have asked me recently about the news concerning neutrinos. Here’s some info and some speculation……. Here is what has happened: Physicists at CERN in Geneva, Switzerland produced neutrinos that were shot through the earth and detected 732 km away in the Gran Sasso Laboratory in Italy. The [...]]]></description>
			<content:encoded><![CDATA[<p>Whats going on with Neutrinos?</p>
<p>A few students have asked me recently about the news concerning neutrinos. Here’s some info and some speculation…….</p>
<p>Here is what has happened: Physicists at CERN in Geneva, Switzerland produced neutrinos that were shot through the earth and detected 732 km away in the Gran Sasso Laboratory in Italy. The diagram below shows the path of the neutrinos.  What is a neutrino? It is a very, very, very small particle with (we think) almost no mass. It passes through most material, which is why it can go through the earth and still be detected. What happened in 2011 is that the neutrino’s speed was calculated by the Europeans at CERN and found to be faster than the speed of light. According to our understanding of modern physics no particle can accelerate to a speed faster than light. So the scientists who made the observation are asking for other scientists with similar facilities in the USA and Japan to confirm the results.</p>
<p>Here are three possible explanations (and there are more):<br />
1.	The calculations are somehow in error and further experiments will correct the calculations so that the neutrino is not going faster than light.<br />
2.	The calculations are correct and the neutrino is travelling faster than light. If this turns out to be the case, the famous physicist Jim Al-Khalili has said he would eat his boxer shorts on live TV as posted in his blog <a href="http://www.jimal-khalili.com/blog/faster-than-the-speed-of-light.html">here</a>  . So you can tell from this post that many scientists feel that the speed of light is in fact a hard speed limit as shown by calculations of Einstein’s theory of Relativity. I agree with this point about the speed limit as well.<br />
3.	But there is a third possibility. The results are correct BUT …the neutrino is still not traveling faster than light. How could this be?  Answer: The neutrino has access to other dimensions. I believe there is strong evidence that very small things like electrons and neutrinos have access to more dimensions vs larger objects of the size of say, golf balls. When we look at a golf ball we see the whole thing. But when we look at an electron we don’t see the whole electron. We see part of it. The rest of it exists in another dimension or dimensions that only very small things have access to. So the neutrinos moving from Switzerland to Italy are able to take a shortcut through these extra dimensions. This interpretation would say that the data is correct, but the neutrino is not travelling faster than light, instead it is accessing an additional dimension or dimensions and only seems to us to be moving faster than light.<br />
If this third possibility turns out to be the correct interpretation, then the next step is to devise an experiment that differentiates between accessing additional dimensions and faster than light travel.  There is one caveat to the third explanation though. We detected neutrinos from a supernova in 1987. If they really do access other dimensions with the same time difference we see at CERN, then they should have arrived years earlier then they did. So perhaps the fact that these neutrinos pass through the earth is what changes things. Mass being around makes the difference perhaps.<br />
Can you think of a good experiment to do next?<br />
When we get to A-Level Quantum Mechanics this year I’ll touch on a few of the concepts related to very small objects. But stay tuned to the neutrino story – it may turn out to be very interesting. On the other hand it might just be a mistake, but we shall see…….<br />
<a href="http://www.physicsandcalculus.com/wp-content/uploads/2012/01/neutrinos.png"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2012/01/neutrinos-300x164.png" alt="" title="neutrinos" width="300" height="164" class="aligncenter size-medium wp-image-1619" /></a></p>
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		<title>Lower Secondary Questions</title>
		<link>http://www.physicsandcalculus.com/2011/lower-secondary-questions</link>
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		<pubDate>Sun, 27 Nov 2011 03:28:55 +0000</pubDate>
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		<description><![CDATA[Beyond PSLE &#8211; Lower Secondary Science Here’s the kind of question in Lower Secondary Physics that gives students trouble….. Question: A helium neon laser produces red light with a wavelength of 633 nanometers. Express the wavelength in micrometers. This question requires knowledge of the metric system and how to handle powers of ten. These are [...]]]></description>
			<content:encoded><![CDATA[<p><a href="http://www.physicsandcalculus.com/wp-content/uploads/2011/11/HeNe.gif"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2011/11/HeNe.gif" alt="" title="HeNe" width="107" height="222" class="alignright size-full wp-image-1607" /></a><br />
<strong>Beyond PSLE &#8211; Lower Secondary Science </strong><br />
<br />
Here’s the kind of question in Lower Secondary Physics that gives students trouble…..<br />
<br />
<strong>Question: A helium neon laser produces red light with a wavelength of 633 nanometers. Express the wavelength in micrometers.</strong><br />
<br />
This question requires knowledge of the metric system and how to handle powers of ten. These are both a strong focus of our workshops that we run over the holidays.<br />
<br />
For Lower Secondary Science, students need to memorize the Metric System Prefixes. The sooner they master this, the better. Here they are…<br />
<br />
<a href="http://www.physicsandcalculus.com/wp-content/uploads/2011/11/metric-system.gif"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2011/11/metric-system-300x120.gif" alt="" title="metric system" width="450" height="180" class="aligncenter size-medium wp-image-1608" /></a><br />
<br />
A nanometer is a power of ten to -9. That means nine zeros in front of the decimal place. A micrometer is ten to -6. That means six zeros in front of the decimal place. To go from nanometers to micrometers, move the decimal 3 places to the left.</p>
<p><strong>Answer: 633 nanometers = 0.633 micrometers</strong><br />
<br />
For information on our Holiday workshops and our 2012 Lower Secondary Science Classes, contact us at the link below&#8230;..<br />
<b><br />
<a href="http://www.physicsandcalculus.com/singapore/contact">Contact Us</a></p>
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		<title>O-Level Questions</title>
		<link>http://www.physicsandcalculus.com/2011/o-level-questions</link>
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		<pubDate>Tue, 06 Sep 2011 16:13:11 +0000</pubDate>
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		<description><![CDATA[O-Level Practice Question 2: A helium balloon has a volume of 621 cm3 at ground level where the atmospheric pressure is 101,000 Pa. The balloon rises and expands to 636 cm3. At this point, what is the height of the balloon above ground ? Take the density of air to be 1.25 kg/m3 and constant [...]]]></description>
			<content:encoded><![CDATA[<p><strong>O-Level Practice Question 2:</strong>  A helium balloon has a volume of 621 cm3 at ground level where the atmospheric pressure is 101,000 Pa. The balloon rises and expands to 636 cm3. At this point, what is the height of the balloon above ground ? Take the density of air to be 1.25 kg/m3 and constant over the height h.<br />
<a href="http://www.physicsandcalculus.com/wp-content/uploads/2011/09/OLev2.gif"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2011/09/OLev2-179x300.gif" alt="" title="OLev2" width="90" height="150" class="alignright size-medium wp-image-1601" /></a><br />
<FONT COLOR="#2554C7"> In doing this problem, realize that the pressure inside the balloon must be equal to the pressure outside (atmospheric pressure at a specific height).<br />
<br />
The first step is to find the pressure at height h. Although not explicitly stated, we have to assume that temperature is a constant, which means we can use the following equation,<br />
<br />
(P1)(V1) = (P2)(V2)<br />
101,000 Pa (621 cm3) = P2 (636 cm3)<br />
P2 = 98,618 Pa<br />
<br />
The <strong>difference</strong> in pressure between the top and the bottom will be equal to (density)gh. This is the same concept behind saying the pressure under a column of liquid is (density)gh. Go deeper under a lake and pressure increases due to the height of the column of liquid above. Go deeper in the atmosphere and you have the same effect. </p>
<p>101,000 Pa &#8211; 98,618 Pa = 2382 Pa = (1.25 kg/m3) (10 m/s2) h</p>
<p><strong>h = 191 m</strong><br />
</FONT></p>
<p><strong>O-Level Practice Question 1: </strong> A rocket with a mass of 1,000 kg takes off from the surface of another planet using a thrust of 30,000 N. An acceleration of 1.5 m/s2 is achieved.<br />
<br />
A) What is the resultant force on the rocket?<br />
<a href="http://www.physicsandcalculus.com/wp-content/uploads/2011/09/OLev1.gif"><img src="http://www.physicsandcalculus.com/wp-content/uploads/2011/09/OLev1.gif" alt="" title="OLev1" width="317" height="167" class="alignright size-full wp-image-1597" /></a><br />
<FONT COLOR="#2554C7">(1000 kg) (1.5 m/s2)= 1500 N</FONT><br />
<br />
B) What is the weight of the rocket?<br />
<br />
<FONT COLOR="#2554C7">Resultant Force = Thrust – Weight<br />
1500 N = 30,000 N – W<br />
W = 28,500 N</FONT><br />
<br />
C) What is the acceleration of gravity on the other planet?<br />
<br />
<FONT COLOR="#2554C7">Weight = ma<br />
28,500 N = (1000 kg) a<br />
<strong>a = 28.5 m/s2</strong></FONT><br /></p>
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		<title>Old Subscription Post</title>
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		<pubDate>Thu, 12 May 2011 03:23:45 +0000</pubDate>
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		<pubDate>Thu, 12 May 2011 03:06:20 +0000</pubDate>
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		<description><![CDATA[Our Physics Tutorials cover the entire Introductory Sequence at the University Level. The emphasis is Problem Solving.]]></description>
			<content:encoded><![CDATA[<p><span style="font-size: medium;">Our Physics Tutorials cover the entire Introductory Sequence at the University Level. The emphasis is Problem Solving.</span><br />
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		<description><![CDATA[We take you through the basics first and then move to the harder problems.]]></description>
			<content:encoded><![CDATA[<p>We take you through the basics first and then move to the harder problems.<br />
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		<description><![CDATA[Tutorials for both the Calculus and the Non-Calculus based classes.]]></description>
			<content:encoded><![CDATA[<p>Tutorials for both the Calculus and the Non-Calculus based classes.<br />
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