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      <title>Jerry HL Concept &amp; Vocabulary Tracker by Joshua Hulks</title>
      <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9</link>
      <description></description>
      <language>en-us</language>
      <pubDate>2025-07-20 13:33:51 UTC</pubDate>
      <lastBuildDate>2025-10-20 15:26:00 UTC</lastBuildDate>
      <webMaster>hello@padlet.com</webMaster>
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      <item>
         <title>Week 2 Lesson 1</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525017207</link>
         <description><![CDATA[<p><strong>1) Neutron-to-proton ratio</strong></p><p>--&gt; Stable isotopes tend to lie within a narrow band on the graph of neutrons vs protons (left chart)</p><p><br/></p><p>--&gt; Lighter elements are more stable when the number of protons = number of neutrons</p><p><br/></p><p>--&gt; For heavier elements, more neutrons are needed to counteract the repulsive forces between many protons, so the N/Z ratio increases (N &gt; Z)</p><p><br/></p><p><strong>2) Nuclear Binding Energy Per Nucleon</strong></p><p>--&gt; A higher binding energy per nucleon means the nucleus is more tightly bound and stable</p><p><br/></p><p>--&gt; Heavier or lighter nuclei than iron-56 are less tightly bound, which is why light nuclei can fuse to become more stable and Heavy nuclei can split to become more stable</p>]]></description>
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         <pubDate>2025-07-21 11:39:01 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525017207</guid>
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      <item>
         <title>Explain which part of the Bohr model of the atom is an example of a binding energy and the significance of the minus sign.</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525969277</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-07-22 10:59:00 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525969277</guid>
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      <item>
         <title>Explain which part of Newton&#39;s theory of gravitation between two masses is an example of a binding energy and the significance of the minus sign.</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525969460</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-07-22 10:59:33 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525969460</guid>
      </item>
      <item>
         <title>Explain which part of the photoelectric effect is an example of a binding energy and the significance of the minus sign.</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525969908</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-07-22 11:00:40 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525969908</guid>
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      <item>
         <title>Explain which part of the nuclear fission process is an example of a binding energy and the significance of the minus sign.</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525970069</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-07-22 11:01:05 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525970069</guid>
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      <item>
         <title>Explain which part of the nuclear fusion process is an example of a binding energy and the significance of the minus sign.</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525970196</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-07-22 11:01:19 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525970196</guid>
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      <item>
         <title>Explain how the strong nuclear force contributes to the binding energy of a nucleus and the significance of the minus sign in this context.</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525970452</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-07-22 11:01:43 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3525970452</guid>
      </item>
      <item>
         <title>Week 2 Lesson 2</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3527030221</link>
         <description><![CDATA[<p>Explain why radioactive waste is dangerous to living organisms.</p><p><br/></p><p>Radioactive waste is dangerous to living organisms because it emits ionizing radiation, such as alpha, beta, and gamma rays. This ionizing radiation has enough energy to:</p><p><br/></p><p>1) break chemical bonds in DNA and other molecules inside cells, which can cause the cells to malfunction and die</p><p>2) alter DNA and lead to mutations which can result in cancer or be passed on if reproductive cells are affected</p><p>3) accumulate inside the body and irradiate tissues if inhaled, swallowed, or absorbed through the skin</p><p>4) remain active and hazardous for thousands of years</p><p><br/></p>]]></description>
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         <pubDate>2025-07-23 13:48:16 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3527030221</guid>
      </item>
      <item>
         <title>Negative Potential Energy</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3527033026</link>
         <description><![CDATA[<p>If potential energy is negative, then you need to add energy to move the object or system to a higher potential energy state (Potential Energy to Kinetic Energy).</p><p><br></p><p>Potential energy is stored energy due to an object`s position and a negative potential energy means the object is in a state where it is naturally attracted to a lower energy state. To move the object to a higher potential energy state, you need to counteract the attractive force and do work against it.</p>]]></description>
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         <pubDate>2025-07-23 13:52:01 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3527033026</guid>
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      <item>
         <title>Derive Newton’s 2nd Law from the definition of net force as the rate of change of a momentum of an object, when mass is constant and when the mass is constant.</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3528670642</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-07-25 10:13:00 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3528670642</guid>
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      <item>
         <title>Outline what is meant by &#39;impulse&#39;.</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3528670740</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-07-25 10:13:26 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3528670740</guid>
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      <item>
         <title>Impulse is the change in momentum of an object when a force is applied over time.</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3534591946</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-08-04 00:15:56 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3534591946</guid>
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      <item>
         <title>How does a wireless charger work?</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3534624536</link>
         <description><![CDATA[<p>The coils in the wireless charger induces a current to one direction because of the current flowing from the the positive charge to the negative charge, which creates a magnetic field around the coils of the wireless charger and the phone, and the direction of the magnetic field on the wireless charger would be OUT of the page and on the phone would be INTO the page according to the curl hand rule.</p><p><br/></p><p>Therefore, the induced current changes direction and the potential, which is directly proportional to the current would also change (Alternating Current), charging the phone wirelessly</p>]]></description>
         <enclosure url="" />
         <pubDate>2025-08-04 01:12:41 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3534624536</guid>
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      <item>
         <title></title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3535038864</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-04 12:04:31 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3535038864</guid>
      </item>
      <item>
         <title>Kinetic Energy</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3535384143</link>
         <description><![CDATA[<p>There are different formulae and applications for Kinetic Energy,</p><p><br/></p><p>E<sub>ktrans(lational)</sub> = 1/2mv<sup>2</sup></p><p>E<sub>krot(ational) </sub>= 1/2 Iw<sup>2</sup></p><p>E<sub>kvib(rational) </sub></p>]]></description>
         <enclosure url="" />
         <pubDate>2025-08-05 00:17:08 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3535384143</guid>
      </item>
      <item>
         <title>Potential</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3535391783</link>
         <description><![CDATA[<p>Potential at a point in the field is thhe energy needed to move 1 kg or 1 C from infinity to a point in the field.</p><p><br/></p><p>emf: electromotive force in volts (V) or J C<sup>-1</sup></p>]]></description>
         <enclosure url="" />
         <pubDate>2025-08-05 00:28:29 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3535391783</guid>
      </item>
      <item>
         <title>Connection Between SHM &amp; Lenz&#39;s Law</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3537299645</link>
         <description><![CDATA[<p>In SHM, the middle line in which an object oscillates is an equilibrium line, Lenz's Law does the same, the system wants to get back to equilibrium.</p><p><br/></p><p>F/m = a = -w<sup>2</sup>x</p>]]></description>
         <enclosure url="" />
         <pubDate>2025-08-07 00:26:15 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3537299645</guid>
      </item>
      <item>
         <title>Magnet oscillating on a spring</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540229611</link>
         <description><![CDATA[<p>N<strong>Φ is flux linkage</strong></p><p><br/></p><p>a. The induced emf is proportional to the rate of change of magnetic flux, and because the magnetic field is changing as the magnet passes through the solenoid, therefore the magnetic flux will change and the emf will be induced.</p><p><br/></p><p>b. Calculate the angular velocity of the magnet's oscillation.</p><p><br/></p><p><strong>F = ma and a = -ω<sup>2</sup>x</strong></p><p><strong>-kx = -mω<sup>2</sup>x</strong></p><p><strong>ω = sqrt(k/m)</strong></p><p><strong>ω = sqrt(20/0.2)</strong></p><p><strong>ω = 10</strong></p><p><br/></p><p>c. Determine an expressions for the induced emf in terms of B<sub>0</sub>, ω, and other relevant quantities.</p><p><br/></p><p><strong>emf<sub>induced </sub>= N x -Δ (BA cos (theta))/Δt</strong></p><p><strong>emf<sub>induced</sub> = NB<sub>0</sub>A (-si5n (ωt)) ω</strong></p><p><br/></p><p>d. Calculate the maximum induced current in the circuit</p><p><br/></p><p><strong>emf<sub>peak</sub> = ωNB<sub>0</sub>A</strong></p><p><strong>IR = ωNB<sub>0</sub>A</strong></p><p><strong>I = 0.005 A</strong></p><p><br/></p><p>e. Explain what happens to the induced EMF if the amplitude of the magnet's oscillation doubles.</p><p><br/></p><p><strong>The induced EMF will increase as the amplitude of the magnet's oscillation doubles because a larger amplitude of oscillation means the magnet moves through a greater distance within the coil, resulting in a larger change in magnetic flux and thus a higher induced EMF.</strong></p>]]></description>
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         <pubDate>2025-08-11 11:38:15 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540229611</guid>
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      <item>
         <title>The Photoelectric Effect</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540277118</link>
         <description><![CDATA[<p>The photoelectric effect refers to the release of electrons from the surface of metals as a result of photons hitting the surface.</p><p><br></p><p>The threshold frequency is the minimum frequency for photoelectric effect to take place. The threshold frequency can be found using the formula </p><p><br></p><p>The photoelectric effect is applied in the photovoltaic cells, which are found in solar panels.</p><p><br></p><p>Higher intensity, more electrons being emitted per second, meaning a larger current because current is the amount of electrons flowing through per second.</p>]]></description>
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         <pubDate>2025-08-11 12:01:06 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540277118</guid>
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      <item>
         <title>Binding Energy and minus sign in photoelectric effect</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540278649</link>
         <description><![CDATA[<p>K<sub>max </sub>​= hf − ϕ</p><p><br></p><p>The work function ϕ is the minimum energy required to free an electron from the metal's surface and this is the electron's <strong>binding energy</strong> to the material.</p><p><br></p><p>The equation subtracts ϕ because you must use up part of the photon's energy to overcome the binding energy, E = hf is photon energy. If hf &lt; ϕ, the minus sign makes K<sub>max</sub> negative, which means no electrons are being emitted.</p>]]></description>
         <enclosure url="" />
         <pubDate>2025-08-11 12:03:39 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540278649</guid>
      </item>
      <item>
         <title>Ionising Radiation</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540278754</link>
         <description><![CDATA[<p>In PET imaging, brighter colour indicates higher frequency.</p><p><br></p><p>PET imaging is when a radioactive substance is injected to a patient. So, the factor that primarily influences the choice for the substance would be the isotope's half-life, shorter half-life, less time to stay in the body.</p><p><br></p><p>Alpha particles --&gt; low penetration and high ionization ability (trapped in the skin)</p><p>Beta particles --&gt; moderate penetration and moderate ionization ability</p><p>Gamma particles --&gt; high penetration and low ionization ability</p>]]></description>
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         <pubDate>2025-08-11 12:03:53 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540278754</guid>
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      <item>
         <title>Stopping voltage </title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540279064</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-11 12:04:35 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540279064</guid>
      </item>
      <item>
         <title>Stefan Boltzmann Law </title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540279403</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-08-11 12:05:13 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540279403</guid>
      </item>
      <item>
         <title>Wien’s Displacement Law (wavelength)</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540279772</link>
         <description><![CDATA[]]></description>
         <enclosure url="" />
         <pubDate>2025-08-11 12:05:48 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540279772</guid>
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      <item>
         <title>Newton&#39;s theory of gravitation between two masses binding energy and minus sign</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540282895</link>
         <description><![CDATA[<p>E<sub>p</sub> = −(GMm)​/r</p><p><br/></p><p>The magnitude of the potential energy here is the amount of energy needed to be supplied in order to separate the two masses to an infinite distance apart where gravitational interaction becomes zero. This is the <strong>gravitational binding energy</strong>, the energy debt needed to be overcome to break the <strong>gravitational bond</strong>.</p><p><br/></p><p>The minus sign in the formula means the potential energy is negative when the two masses are bound together and by convention, <strong>zero potential energy</strong> is at r = infinity</p><p><br/></p><p>Since it takes a <strong>POSITIVE</strong> energy input to move from the current position to r = infinity, the potential energy at any finite r must be LESS THAN ZERO.</p><p><br/></p><p>A negative E<sub>p</sub> indicates the system is bound and objects cannot be separated without doing <strong>work</strong> against gravity. Minus sign = the gravitational system is bound and you must add energy to reach <strong>ZERO</strong> potential at infinity.</p>]]></description>
         <enclosure url="" />
         <pubDate>2025-08-11 12:10:17 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540282895</guid>
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         <title></title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540284755</link>
         <description><![CDATA[<p>The energy levels is given by the Bohr model, E = -13.6 eV/n<sup>2</sup> and the formula gives the binding energy of an electron in the nth orbit where the more negative the value, the stronger the binding to the nucleus.</p><p><br></p><p>The minus sign means the electron's energy is lower than zero, so it is not free. Therefore, you must add POSITIVE energy equal to the magnitude of that negative value to remove the electron from the atom or to ionize the atom. For example, to remove an electron from the first energy level, you need to add +13.6 eV of energy.</p>]]></description>
         <enclosure url="https://padlet-uploads-usc1.storage.googleapis.com/4142696871/a7702a73eecbcc9237ea6fe26d0ab057/image.png" />
         <pubDate>2025-08-11 12:12:40 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540284755</guid>
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         <title>SNF contributing to NBE of a nucleus and minus sign</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540344752</link>
         <description><![CDATA[<p>Protons repel each other via the <strong>electrostatic force</strong>. The SNF is a short-range, extremely strong attractive force between nucleons (protons and neutrons) and this force overcomes the proton-proton repulstion at VERY SHORT distances (&lt;1 fm) and holds the nucleons tightly together.</p><p><br/></p><p>Because of this attraction from the SNF, the total mass of a nucleus is less than the sum of  the individual masses of its separate nucleons and this <strong>mass defect Δm</strong> corresponds to the binding energy <strong>NBE = Δmc<sup>2</sup></strong>. This NBE is the energy you must supply to completely separate the nucleus into FREE protons and neutrons.</p><p><br/></p><p>The potential energy of a bound nucleus is NEGATIVE compared to the zero level for free, infinitely separated nucleons. the minus sign means that the nucleus is in a <strong>bound state</strong> and its energy is <em>lower</em> than the total energy of the separated nucleons. The <strong>binding energy</strong> is the POSITIVE amount you need to <strong>add</strong> to overcome the SNF and bring the nucleus to ZERO potential energy (unbound state).</p><p><br/></p><p>So, the minus sign clearly distinguish between BOUND systems with negative E<sub>p</sub> from UNBOUND systems with zero or positive E<sub>p</sub>.</p>]]></description>
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         <pubDate>2025-08-11 12:25:33 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540344752</guid>
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         <title>Nuclear fission process as BE and minus sign</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540358415</link>
         <description><![CDATA[<p>In a nucleus, <strong>binding energy</strong> is the energy that holds the nucleons (protons and neutrons) together, provided mainly by the <strong>strong nuclear force</strong>. </p><p><br></p><p>In fission:</p><ul><li><p>A heavy nucleus splits into two (or more) smaller nuclei.</p></li><li><p>The <strong>binding energy per nucleon</strong> of the fission products is <strong>greater</strong> than that of the original heavy nucleus.</p></li><li><p>This means the nucleons in the products are held more tightly together.</p></li></ul><p><br></p><p>The <strong>difference</strong> in total binding energy between the products and the original nucleus is released as kinetic energy of the split fragments, gamma rays, and energy by emitted neutrons.</p><p><br></p><p>A bound nucleus has <strong>negative potential energy</strong> compared to free nucleons at infinite separation (zero energy level). The minus sign means the system's energy is below zero, therefore the nucleus is in a <strong>bound state</strong>. To break it apart into free nucleons, you’d need to <strong>add positive energy</strong> equal to the magnitude of the binding energy.</p><p><br></p><p>In fission, when the products have <strong>more negative total potential energy</strong>, the change in energy is negative for the system, meaning energy is released to the surroundings.</p>]]></description>
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         <pubDate>2025-08-11 12:44:56 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540358415</guid>
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      <item>
         <title>Nuclear fusion BE and minus sign</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540368989</link>
         <description><![CDATA[<ul><li><p>In fusion (for example, hydrogen nuclei fusing into helium), the <strong>mass of the final nucleus</strong> is <strong>less than</strong> the total mass of the separate starting nuclei.</p></li><li><p>The <strong>mass defect</strong> has been converted into energy according to Einstein’s equation E = mc<sup>2</sup>.</p></li><li><p>The <strong>binding energy</strong> of a nucleus is the energy required to break it into its individual protons and neutrons.</p></li><li><p>In fusion, the product nucleus has <strong>greater binding energy per nucleon</strong> than the original nuclei, meaning its nucleons are more tightly bound.</p></li></ul><p><br/></p><p>The <strong>total energy of a bound system</strong> (like a nucleus) is <strong>negative</strong> relative to free nucleons at infinite separation.</p><ul><li><p>Energy of free nucleons (not bound) is set to 0.</p></li><li><p>When nucleons bind, the system’s potential energy decreases and the bound state has <strong>less energy</strong> than the free particles.</p></li><li><p>This lower energy is written as a <strong>negative value</strong>.</p></li></ul><p><br/></p><p>The minus sign means energy must be <em>added</em> (positive) to overcome the nuclear force and break the nucleus apart and a larger magnitude of negative binding energy means a more stable nucleus.</p>]]></description>
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         <pubDate>2025-08-11 13:00:30 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540368989</guid>
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      <item>
         <title>Slide 8-9 from D4 Lesson 3</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540802772</link>
         <description><![CDATA[<p>Q1) EMF = −(dΦ/dt​​)</p><p><br/></p><p>Four positions:</p><ol><li><p><strong>Above position</strong></p><ul><li><p>Magnetic flux is small and <em>constant</em> → No EMF.</p></li><li><p>Direction: Not applicable.</p></li></ul></li><li><p><strong>Below position</strong></p><ul><li><p>Magnetic flux is small and <em>constant</em> → No EMF.</p></li><li><p>Direction: Not applicable.</p></li></ul></li><li><p><strong>Left position</strong></p><ul><li><p>Magnetic flux increases when the switch is closed → EMF induced.</p></li><li><p>Direction: Field is going up, right-hand rule shows EMF going CCW.</p></li></ul></li><li><p><strong>Right position (same as left, symmetric)</strong></p><ul><li><p>Magnetic flux increases → EMF induced.</p></li><li><p>Direction: Symmetrical to the left position, so CCW.</p></li></ul></li></ol><p><br/></p><p>Q2) 7.5 V</p>]]></description>
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         <pubDate>2025-08-11 23:55:50 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540802772</guid>
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      <item>
         <title>Slide 10 D4 Lesson 3</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540805423</link>
         <description><![CDATA[<p>The position of the battery-less circuit which has the largest induced EMF would be either the left or right position, because according to Faraday's Law, EMF is directly proportional to the magnitude of sin <strong>θ</strong>, therefore when the angle of the magnetic flux is perpendicular to the circuit, sin <strong>θ </strong>will have the MAX value.</p>]]></description>
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         <pubDate>2025-08-11 23:59:09 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540805423</guid>
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      <item>
         <title>Self-inductance</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540841028</link>
         <description><![CDATA[<p>Self-inductance is the ability of a solenoid to oppose a current that is going through the solenoid. Because there is a magnetic field flowing through the coil, which induced a current, which induces a NEW/secondary magnetic field due to the primary current.</p><p><br/></p><p>I<sub>battery</sub> --&gt; induces B inside the solenoid --&gt; induces I --&gt; Induces secondary B</p><p><br/></p><p>EMF of the battery will create a change in flux, which induced a new or secondary EMF.</p><p><br/></p><p>EMF<sub>battery</sub> = -(d<strong>Φ</strong>/dt) = EMF<sub>induced</sub></p>]]></description>
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         <pubDate>2025-08-12 00:43:44 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3540841028</guid>
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      <item>
         <title>A particle starts from rest and moves with constant acceleration</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3542818538</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-14 00:18:09 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3542818538</guid>
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         <title>Question 6 of Question Booklet</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3542825342</link>
         <description><![CDATA[<p>The Resultant Force is the forces acting on box Q, which here is the push force from box P. But because it moves at constant speed, meaning the frictional force is strong enough to cancel the push force, therefore the resultant force must be 0 N. F<sub>resultant</sub> = F<sub>1</sub> - F<sub>2</sub></p>]]></description>
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         <pubDate>2025-08-14 00:27:20 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3542825342</guid>
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         <title>Question 17 of Question Booklet</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3542861297</link>
         <description><![CDATA[<p>A, because the weight W is going down</p><p><br/></p><p>AND the reaction force R when the lift is going upwards, the reaction force will change its direction and the initial direction of R going up will go down. Therefore, the 2 forces are acting on the same direction, so +.</p>]]></description>
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         <pubDate>2025-08-14 01:09:58 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3542861297</guid>
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      <item>
         <title>Complete this ASAP (then ask for the worked solution)</title>
         <author>joshuah51</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3544164302</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-15 10:35:58 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3544164302</guid>
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         <title>Question 26 of Question Booklet</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545359094</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-17 11:52:03 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545359094</guid>
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         <title>Question 6 of Question Booklet</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545359704</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-17 11:53:21 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545359704</guid>
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         <title>Question</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545412740</link>
         <description><![CDATA[<p>a. (i) Moment of inertia is a measure of a body's resistance to angular acceleration on a given axis.</p><p><br/></p><p>(ii) Net torque about any axis is zero for rotational equilibrium</p>]]></description>
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         <pubDate>2025-08-17 14:16:23 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545412740</guid>
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         <title>Diagram</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545582652</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-18 00:12:08 UTC</pubDate>
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         <title>Question 5 of Quiz</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545637244</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-18 01:14:09 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545637244</guid>
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         <title>Doppler Definition</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545640064</link>
         <description><![CDATA[<p>Doppler --&gt; the component of the relative velocity along the line of sight (v<sub>parallel</sub>)</p><p><br/></p><p>Δ<strong>λ = λ (v/c)</strong></p><p><br/></p><p><strong>larger wavelength shifts more</strong></p>]]></description>
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         <pubDate>2025-08-18 01:16:57 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3545640064</guid>
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         <title>Wave Interference Maximum and Minimum</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3546845428</link>
         <description><![CDATA[<p>Draw another wave in between the crests to represent the troughs, and you'll find out Q is in the middle of the meeting between two troughs, which gives a maximum at Q also due to constructive interference.</p><p><br/></p><p>Destructive interference or the minimum would be when the crest meets the trough.</p>]]></description>
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         <pubDate>2025-08-19 00:37:39 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3546845428</guid>
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         <title>Q5 HL Quiz</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3546870411</link>
         <description><![CDATA[<p>Light can travel by itself, therefore the wavelength of light is not affected by the medium, whereas for sound, if a source is stationary, the wavelength will be fixed due to the not changing medium.</p><p><br/></p><p>But frequency still changes because the closer the observer get to the source, the observer will hear more of the soundwave, which corresponds to the frequency.</p>]]></description>
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         <pubDate>2025-08-19 00:59:32 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3546870411</guid>
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         <title>Similar Question to Q5</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547508445</link>
         <description><![CDATA[<p>The difference between the movement of this train with the rotating device is the rotating device is constantly changing its angle every second, whereas for a moving train going towards and going past an observer, there will be a discrete jump of frequency heard.</p>]]></description>
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         <pubDate>2025-08-19 11:05:41 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547508445</guid>
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         <title>Intensity Distribution Question</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547514549</link>
         <description><![CDATA[<p>Key Effects of Increasing N:</p><ol><li><p><strong>Positions of maxima</strong> (where principal peaks occur):</p><ul><li><p>The <strong>angles</strong> of principal maxima depend only on d and λ, not on N.</p></li><li><p>nλ = d sin θ</p></li></ul></li></ol><ol start="2"><li><p><strong>Intensity of principal maxima</strong>:</p><ul><li><p>Maximum intensity scales as:</p><p>I<sub>max </sub>∝ N<sup>2</sup></p></li><li><p>If we double slits from 4 → 8, it should get brighter and not weaker</p></li></ul></li></ol><ol start="3"><li><p><strong>Intensity of secondary maxima</strong>:</p><ul><li><p>Secondary maxima appear between the main peaks.</p></li><li><p>Their relative intensity compared to principal maxima decreases as N increases.</p></li><li><p>So when N increases, <strong>secondary maxima become weaker</strong>.</p></li></ul></li></ol><ol start="4"><li><p><strong>Width of principal maxima</strong>:</p><ul><li><p>The central peaks become <strong>narrower (sharper)</strong> as N increases.</p></li><li><p>This is because constructive interference is “more selective” when there are more slits.</p></li><li><p>In fact, the angular width Δθ ∝ 1/N<sup>2</sup></p></li></ul></li></ol>]]></description>
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         <pubDate>2025-08-19 11:16:45 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547514549</guid>
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         <title>Selectivity in Interference</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547515052</link>
         <description><![CDATA[<p>When you have many slits, each slit sends out a wave. For constructive interference (a bright fringe) to occur, <strong>all the waves must line up in phase</strong> at that angle θ.</p><ul><li><p>With <strong>few slits</strong> (say 2 or 4), you don’t need the phase alignment to be <em>perfect</em> for the waves to still add up fairly strongly. This means that the intensity doesn’t drop very sharply when you move a little away from the central angle. → <strong>broader peaks</strong>.</p></li><li><p>With <strong>many slits</strong> (say 8, 10, 100), the requirement for perfect alignment becomes <em>stricter</em>. If you move even slightly away from the constructive angle, the phase difference builds up quickly between the many slits, and destructive interference cancels the light much more strongly. → <strong>narrower peaks</strong>.</p></li></ul><p>That’s why we say the system is <strong>“more selective”</strong>:</p><ul><li><p>It only “selects” angles that satisfy the constructive condition <em>very precisely</em>.</p></li><li><p>Any small deviation leads to destructive interference, cutting the intensity sharply.</p></li></ul>]]></description>
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         <pubDate>2025-08-19 11:17:30 UTC</pubDate>
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         <title>Memory jog</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547515591</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-19 11:18:36 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547515591</guid>
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      <item>
         <title>Incident light on rectangular block of glass in air</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547522048</link>
         <description><![CDATA[<ul><li><p>Light bends <strong>towards the normal</strong> when it enters a denser medium (air → glass).</p></li><li><p>Light bends <strong>away from the normal</strong> when it leaves a denser medium (glass → air).</p></li><li><p>In a rectangular block, since the sides are parallel, the emerging ray is <strong>parallel to the incident ray</strong> (but shifted sideways).</p></li></ul><p><br/></p><p>+++++</p><ul><li><p>When light enters a denser medium (glass), it slows down → bends <strong>towards the normal</strong>.</p></li><li><p>When it exits back to air, it speeds up again → bends <strong>away from the normal</strong>.</p></li><li><p>Since the two surfaces of the glass are parallel, the direction is restored, and the emergent ray is <strong>parallel to the incident ray</strong>, only shifted sideways.</p></li></ul><p><br/></p><p>So C is the answer</p>]]></description>
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         <pubDate>2025-08-19 11:29:22 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3547522048</guid>
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         <title>Deriving Newton&#39;s 2nd Law</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548833541</link>
         <description><![CDATA[]]></description>
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         <pubDate>2025-08-20 10:22:10 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548833541</guid>
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         <title>Q6 SL Quiz</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548880481</link>
         <description><![CDATA[<p>Earth is to the right. At the instant shown, the <strong>blue star (top)</strong> moves <strong>left</strong>, so relative to Earth it’s <strong>receding</strong> → <strong>redshift</strong> (λ increases → line shifts <strong>right</strong>). The <strong>red star (bottom)</strong> moves <strong>right</strong>, so it’s <strong>approaching</strong> → <strong>blueshift</strong> (λ decreases → line shifts <strong>left</strong>).</p><p><br/></p><p>Because they share the same circular orbit, their speeds and hence the magnitudes of their radial velocities at these positions (purely along the line of sight) are equal. So the shifts are equal in size but opposite in direction, bringing the two spectral lines <strong>toward each other</strong> on the wavelength scale. </p><p><br/></p><p>So D</p>]]></description>
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         <pubDate>2025-08-20 11:33:12 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548880481</guid>
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         <title>Wave Phenomena</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548885817</link>
         <description><![CDATA[<p>Sources S<sub>1</sub>​ and S<sub>2</sub> start <strong>π out of phase (the phase difference).</strong> At P we hear <strong>maximum intensity</strong> → <strong>constructive interference</strong>.</p><p><br/></p><p><strong>Condition for constructive interference with initial phase difference</strong></p><ul><li><p>Total phase difference at P must be an integer multiple of 2π</p></li></ul><p>Δd = 8.2 − 7.6 = 0.6&nbsp;m</p><p><br/></p><p>Δϕ<sub>total</sub> = π + 2π * Δd/λ = 2πn</p><p>Divide by π gives:</p><p>1+ 2Δd/λ = 2n</p><p>2Δd/λ = 2n - 1</p><p>λ = 2Δd / ( 2n-1 )</p><p>where 2n - 1 is <strong>always </strong>an odd integer.</p><p><br/></p><p>SO, λ = 1.2 / odd integer</p><p>smallest odd integer is 1, substituting 1 to the equation gives λ = 1.2 m</p><p><br/></p><p><br/></p><p><br/></p>]]></description>
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         <pubDate>2025-08-20 11:41:25 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548885817</guid>
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         <title></title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548910804</link>
         <description><![CDATA[<ul><li><p>From t=0 to t=10 s, the train approaches the station but is decelerating, so its speed decreases.</p></li><li><p>At t = 10 s, the train stops (speed = 0), so the frequency heard equals the emitted frequency (1000 Hz).</p></li><li><p>After t = 10 s, the train accelerates away from the station, so its speed increases in the opposite direction.</p></li></ul><p>Hence, the frequency heard will be lower than 1000 Hz and increasing as the train accelerates away.</p><p><br></p><p>SO:</p><ul><li><p>Before t = 10 s, frequency f is greater than 1000 Hz because the train is approaching.</p></li><li><p>At t = 10 s, frequency exactly equals 1000 Hz (train stops).</p></li><li><p>After t = 10 s, frequency f is less than 1000 Hz because the train is moving away.</p></li></ul><p><br></p><p>++++++</p><ul><li><p>The train starts at some high speed approaching the station at t = 0 s and then slows down until it stops at t = 10 s.</p></li><li><p>When the train is moving fast toward you, the frequency you hear is higher. </p></li><li><p>As the train slows down, its speed decreases, so the Doppler shift effect reduces.</p></li><li><p>This means the frequency you hear decreases over time as the train approaches because the train is getting slower even though it's still getting closer.</p></li><li><p>At t=10 s, the train stops, so the frequency equals the emitted frequency (1000 Hz).</p></li></ul><p>Only Graph D fits the equation</p>]]></description>
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         <pubDate>2025-08-20 12:14:46 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548910804</guid>
      </item>
      <item>
         <title>Question from Teacher Joshua: Changes during Refraction </title>
         <author></author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548918606</link>
         <description><![CDATA[<p>When light passes across a boundary into water or glass in undergoes refraction. Refraction doesn’t change the colour of light, but its wavelength does change during refraction. </p><p><br></p><p>Explain how is this possible.</p><p>Hint: 1. What stays constant during refraction and what changes?</p><ol start="2"><li><p>Think about topic E1 atomic physics and the cause of colour. </p></li></ol>]]></description>
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         <pubDate>2025-08-20 12:24:30 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3548918606</guid>
      </item>
      <item>
         <title>Energy Balance</title>
         <author></author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3553073753</link>
         <description><![CDATA[<p>The radiation coming from the black body cannot radiate out because it is a closed system and the black body is insulated, so the emitted radiation of the black body will completely go to the gray body due to its emissivity being 1. Therefore, there is some fraction of emissitivity going to be absorbed by the gray body, which makes the equation like so.</p>]]></description>
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         <pubDate>2025-08-25 01:06:23 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3553073753</guid>
      </item>
      <item>
         <title>Refractive Index</title>
         <author></author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3554511506</link>
         <description><![CDATA[<p>Colour in air: blue or green</p><p>Colour in glass: WOULD STAY THE SAME</p><p><br/></p><p>Even when the wavelength in glass changes, BECAUSE E = hf where if E changes, the colour would change as colour is the perception of the energy of light. THEREFORE, if f changes, then E changes and our perception of the energy would change, thus changing our perception of the colour.</p><p><br/></p><p>However, in this case, the frequency stays constant, so the energy would stay the same and therefore, the colour would stay the same.</p>]]></description>
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         <pubDate>2025-08-26 00:14:37 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3554511506</guid>
      </item>
      <item>
         <title></title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3554601166</link>
         <description><![CDATA[<p>Conducting sphere where the charges are only on the surface and they're free to move around the surface.</p><p><br/></p><p>Insulating sphere where the charges are spread throughout and they're not free to move.</p><p><br/></p><p>A sphere is HIGHLY symmetric, so all field strength would cancel out on all directions, which gives zero field strength until the radius is changed.</p>]]></description>
         <enclosure url="https://padlet-uploads-usc1.storage.googleapis.com/4142696871/4ae0a3ab9d629a84733e9a6488065d26/image.png" />
         <pubDate>2025-08-26 01:13:01 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3554601166</guid>
      </item>
      <item>
         <title>Terminal Voltage and EMF</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3557692236</link>
         <description><![CDATA[<ol><li><p>Kirchhoff's Law, real battery, EMF and Terminal voltage concept</p></li><li><p>Real battery has some internal resistance in it which adds on to its R<sub>total</sub> value. In contrast, ideal battery is assumed to have zero resistance to ensure its output voltage remains constant and is not affected by the current it supplies to the circuit.</p><p><br/></p><p>EMF or Electromotive Force is the maximum potential difference a power source can provide, which is measured when no current is flowing, representing the total energy per unit charge supplied by the source, often battery. In contrast, terminal voltage is the actual potential difference across the terminals of a source when current is flowing through it. Terminal voltage is always less than the EMF because some energy is "lost" as heat due to the internal resistance (r) of the source</p></li></ol>]]></description>
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         <pubDate>2025-08-28 00:11:22 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3557692236</guid>
      </item>
      <item>
         <title>Terminal Potential Difference</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3557752881</link>
         <description><![CDATA[<p>Current stays constant, so when using V = IR, the current will stay constant, hence V will increase proportionally to the R. Power would decrease as the resistance increase.</p>]]></description>
         <enclosure url="https://padlet-uploads-usc1.storage.googleapis.com/4142696871/458e6338f90f347fe7deb5e84cbb99ab/image.png" />
         <pubDate>2025-08-28 00:48:57 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3557752881</guid>
      </item>
      <item>
         <title>Spacetime positions in different reference frames</title>
         <author></author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3608170918</link>
         <description><![CDATA[<p>Difference between this Lorentz transformation and Galilean relativity:</p><p><br/></p><p>Galilean does not have Lorentz factor because it is not moving at 0.2c or above.</p><p><br/></p><p>Time is constant in Galilean, but here it is not constant, it is dilated.</p><p><br/></p><p>In Lorentz transformation, the t-coordinate is the instantaneous time seen on the clock.</p><p><br/></p><p>Two types of q: t-coordinate, time dilation</p>]]></description>
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         <pubDate>2025-09-29 01:01:41 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3608170918</guid>
      </item>
      <item>
         <title>Uncertainty for logarithmic function</title>
         <author></author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3620856766</link>
         <description><![CDATA[<p>Calculating uncertainty for logarithmic function is to be done using the formula of Unc. = Range/2</p>]]></description>
         <enclosure url="https://padlet-uploads-usc1.storage.googleapis.com/4512760337/f1cedf46b95a2cdc124b6c4cf6cbdf5d/image.png" />
         <pubDate>2025-10-07 00:27:07 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3620856766</guid>
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      <item>
         <title>NEED HELP!</title>
         <author></author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3635656623</link>
         <description><![CDATA[<p>The MS says it is C, but we thought it was D</p>]]></description>
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         <pubDate>2025-10-16 10:57:26 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3635656623</guid>
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      <item>
         <title>NEED HELP! (2)</title>
         <author></author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3635657504</link>
         <description><![CDATA[]]></description>
         <enclosure url="https://padlet-uploads-usc1.storage.googleapis.com/4571143074/04ff9ebdb27208039133c5752b7d0119/image.png" />
         <pubDate>2025-10-16 10:58:04 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3635657504</guid>
      </item>
      <item>
         <title>NEED HELP! </title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3639516983</link>
         <description><![CDATA[]]></description>
         <enclosure url="https://padlet-uploads-usc1.storage.googleapis.com/4142696871/4346113f9c526d6c753da15598b4c92e/image.png" />
         <pubDate>2025-10-19 16:05:04 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3639516983</guid>
      </item>
      <item>
         <title>NEED HELP💀☠️</title>
         <author>23085020</author>
         <link>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3641243211</link>
         <description><![CDATA[<p>The MS says the answer is D, but would the radius of the stars be the same because they are along the same diagonal that matches the main sequence star?</p>]]></description>
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         <pubDate>2025-10-20 15:24:37 UTC</pubDate>
         <guid>https://padlet.com/joshuah51/m33u8e37rc0zs7t9/wish/3641243211</guid>
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