HSC Physics Paper Options and Explanations

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Utilizing the appropriate hand rule (or the left hand rule for an electron), the magnetic area should be into the web page to offer a balancing drive downwards.

The magnitude might be calculated by equating

$$qE = qvB$$

$$B = frac{E}{v} = 0.5 T$$

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This drive is the centripetal drive of round movement; by equation ( qvB = frac{mv^{2}}{r}), we will conclude that v is proportional to r.

Therefore the radius will improve.

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As g is bigger on Earth, a given interval might be achieved with an extended size in comparison with Mars, therefore the graph for Earth might be above that of Mars.

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Weight on Earth is w = mg = 70 x 9.8 = 686 N

Weight on the Moon is w = mg = 70 x 1.6 = 112 N

This can end in an induced EMF (Faraday’s Regulation) within the disc and eddy currents.

The eddy currents will produce a magnetic area to oppose the change within the flux (Lenz’s Regulation) which can end in a drive and a torque on the disc.

The disc will rotate in the identical route because the magnet

The picture is shaped by raster scanning, i.e. by scanning the beam throughout the display from pixel to pixel and altering the ability of the beam (e.g. the no. of electrons) because the beam hits every pixel to alter the depth of that pixel. As soon as a picture is shaped, the scanning repeats to refresh the picture

To ensure that there to be zero internet drive on Y, it should be drawn to X by the identical energy as it’s drawn to Z, therefore the present in X is to the appropriate.

$$F = frac{kI_{1}I_{2}L}{d}$$

$$frac{kI_{X}I_{Y}L}{d_{XY}} =frac{kI_{Y}I_{Z}L}{d_{YZ}} $$

$$frac{I_{X}}{0.75} =frac{20}{320} $$

$$I_{X} = 50 A$$

As electrons transfer by the lattice they collide with constructive ions. These collisions are inelastic on common and the electrons lose power to the lattice (transformed to warmth).

Via their interplay with the lattice one electron loses one phonon of power and the second electron absorbs one phonon. Therefore there isn’t any internet power loss for the Cooper pair.

$$g = frac{F_{G}}{m}$$

Path is the route of drive on a mass

$$E = frac{F_{E}}{q}$$

Path is the route of drive on a constructive cost

Pressure and acceleration are fixed in magnitude and route.

Part of velocity perpendicular to the sphere stays fixed. Particle accelerates in route of drive.

$$v_{x} = -1.6 m/s $$

The vertical acceleration is fixed so the vertical velocity adjustments at a relentless fee.

The preliminary velocity is 4.2 m/s, the ultimate velocity is -4.8 m/s.

The gradient of the graph offers the acceleration: ( a_{y} =  -9.5 m/s^{2} )

$$= frac{-GMm}{r_{f}} – frac{-GMM}{r_{i}}$$

$$= -GMm [frac {1}{r_{f}} – frac{1}{r_{i}}]$$

$$= -6.67 occasions 10^{-11} occasions  7.35 occasions 10^{22} occasions 20 [frac{1}{1740 000+500 000} – frac{1}{1740000}]$$

$$= 1.257… occasions 10^{7} J$$

$$= 1.26 occasions 10^{7} J$$ 

( KE_{Preliminary} = frac{1}{2} mv^{2} = frac{1}{2} occasions 20 occasions 1200^{2} = 1.44 occasions 10^{7}  J )

(KE_{Closing} = 1.44 occasions 10^{7} – 1.26 occasions 10^{7} = 1.8 occasions 10^{6} J )

(frac{1}{2} mv^{2} = 1.8 occasions 10^{6} J )

(v = 424 m/s)

• Electrons within the photocathode (A) will every take in one photon.
• Assuming the photon power is larger than the work perform (frequency is above the brink frequency for that steel), the power absorbed from the photons will permit the electron to beat the work perform and be ejected from the steel.
• The electrical area between A and B will speed up the electrons by the vacuum within the photocell in the direction of B, leading to a present by the circuit.

Photo voltaic Cells:

• Combining the n and p-type supplies will kind a p-n junction.
• Electrons from the n-type will diffuse into the p-type, inflicting a lack of cost carriers, and inflicting the n-type to grow to be negatively charged and the p-type to grow to be positively cost within the junction, leading to an electrical area being shaped. This area is named the depletion layer.
• When mild strikes the depletion layer (assuming the photon power is bigger than the bandgap) it may be absorbed by an electron within the valence band, which turns into excited to the conduction band, forsaking a gap.
• The electrical area will speed up the outlet into the p layer and the electron into the n layer. This cost separation creates a possible distinction between the p and n kind which leads to present flowing from p to n by the circuit

Our understanding of the valence-conduction band interplay, and significantly the band hole has enabled the event of semiconductor know-how. Particularly, the connection between the power of the band hole and the wavelength of emitted or absorbed photons have prompted using semiconductors in mild technology (LEDs) and power technology from mild (Photo voltaic Panels).

For these innovations to be applied, nevertheless an advance in know-how was required: purification of semiconductors and doping to regulate the band hole of semiconductors. With the power to engineer a exact, managed band-gap mild emitters and absorbers working at particular wavelengths might be generated, resulting in the wide range of colored LEDs and the excessive effectivity of photo voltaic panels, leading to a considerably elevated utilization of semiconductors.

(ii) The appliance of a powerful magnetic area adjustments the spin orientation of hydrogen nuclei within the physique. From an preliminary random state, the nuclei align both parallel or anti-parallel. A slight majority of nuclei are parallel – this constitutes the decrease power configuration.

As a result of differing impedances of the 2 supplies, each waves B and A will expertise some reflection on the fat-kidney interface. The reflection coefficient is a product completely of the variations and sums of the impedances, no matter which is the incident and transmissive medium – there ought to be no distinction between the transmission of wave A and wave B. 

(ii) Impedance of kidney tissue, ( Z = rho occasions v = 1050 occasions 1560 = 1.638 occasions 10^{6}rayls)

$$R=frac{(Z_{kidney} – Z_{fats})^{2}}{(Z_{kidney} + Z_{fats})^{2}}$$

Substitute R = 0.01

$$0.01=frac{(Z_{kidney} – Z_{fats})^{2}}{(Z_{kidney} + Z_{fats})^{2}}$$

$$0.1=frac{Z_{kidney} – Z_{fats}}{Z_{kidney} + Z_{fats}}$$

$$0.1(Z_{kidney} + Z_{fats})=Z_{kidney} – Z_{fats}$$

$$0.9Z_{kidney}=1.1Z_{fats}$$

Substitute  ( _{kidney} = 1.638 occasions 10^{6} rayls)

( Z_{fats} = 1.340 occasions 10^{6} rayls )

• Emission Sort: I-124 is a positron emitter – these particles will quickly self-annihilate and generate a pair of photons, normally gamma rays. These can be utilized in positron emission tomography for high-quality medical imaging. As well as, positrons have a low penetration and acceptable ionisation energy.  Therefore I-124’s emission is suitable.
• Half life: from the information, we will see that iodine-124 has a half life of roughly 100 hours. This can be a appropriate half-life for a medical imaging isotope; not so lengthy that it stays lively within the physique and causes extended hurt, however not so brief as to grow to be a logistic impediment.
• Chemical compatibility – Iodine is a hint component discovered within the human physique; it’s unlikely to have any detrimental chemical / toxicity results.

The Doppler impact is used  in Doppler sonography and ultrasonography. Acoustic waves reflecting off a shifting goal will expertise a change in frequency proportional to the speed and route of the goal. Therefore this permits the stream of fluids within the goal to be measured.

Particular functions embrace:

• Detection of blood clots by figuring out stream slowdowns / stoppages
• Detection of irregular aortic flows
• Monitoring foetal coronary heart fee
• Measuring a ‘stream velocity waveform’ – the cycle of quick and sluggish blood stream as the guts pumps.

(ii) Trigonometric parallax measurements are restricted by the decision of the telescope:

(Delta theta = frac{1.22 lambda}{D})

The place D is the aperture diameter, (lambda) is the noticed wavelength and (Delta theta) is the dimensions of the Ethereal disc.

A parallax might be noticed if the shift in obvious place might be resolved in keeping with Rayleigh’s criterion.

Atmospheric transparency is sweet for less than radio and optical wavelengths. Shorter wavelengths (UV) are blocked by the environment so the smaller parallaxes measurable can solely be noticed from area.

Floor-based optical telescopes at any altitude additionally cope with atmospheric turbulence that distorts incoming wavefronts and blurs the picture. This decreases efficient decision therefore limiting parallax measurements making decision worse than the above equation for (Delta theta.)

(ii) From graph B:

(T = 7 – 1 = 6) days
From the T-L graph, absolute magnitude is (M = -3.2).

$$M = m – 5 log (frac{d}{10})$$
$$d = 10^{(frac{(m-M+5)}{5})}$$
$$frac{(m-M+5)}{5} = frac{(16.3 + 3.2 +5)}{5} = 4.9$$
$$d = 10^{4.9}$$
(d = 79 400 textual content{computer}) (3 sig. fig.)

Elevated radiation stress expands the outer layers inflicting cooling and the star is now a crimson supergiant.

Carbon from the helium fusion can fuse within the core whereas hydrogen and helium fuse in surrounding shells (outer shells with lighter parts). The core area is like shells of an onion.

Oxygen fusion ought to happen yielding a bigger, redder, crimson supergiant.

The star ends life as a supernova explosion (creating many different parts) whereas the core of oxygen, neon and different parts collapses right into a neutron star.

Be aware: A flowchart may additionally be ample so long as it didn’t compromise on element.

Greater spatial decision over photographic plates yields higher parallax and distance measurements.

Greater accuracy and reliability in photometry yields improved understanding of how color pertains to different properties of stars (luminosity and so on) with wider frequency protection.

Spectroscopy sees the best advantages. Photographic spectra are 2D photographs the place relative spectral line intensities can’t be decided quantitatively precisely. Photoelectric detectors usually are not simply extra delicate to a wider vary of wavelengths (therefore yielding extra chemical info from different spectral traces) however permit 1D spectra to be produced simply the place spectral line intensities might be quantified exactly. This improves calculations of redshift (or blueshift) for radial velocities and line broadening because of Doppler broadening (thermal and rotational) and stress broadening. Throughout the board, moreover, CCDs yield sooner knowledge assortment, decreased knowledge loss and improved knowledge comparability.

Stellar emission spectra are noticed by trying on the environment of the star in opposition to the backdrop of area. Meaning observing sizzling fuel in opposition to a colder background. The ensuing spectrum exhibits emission traces of assorted species (chemical composition) and which traces of which species are seen signifies the temperature of the stellar environment. In a faculty lab, one can observe a fuel lamp (e.g hydrogen or mercury) or fluorescent mild tubes by a spectroscope in a darkened room.

Blackbody Spectra:

These are produced by blackbody radiators similar to stellar cores. Since such objects are by no means with no surrounding environment, they aren’t seen straight however because the underlying envelope of stellar absorption spectra.

Blackbody spectra alone yield no chemical info (is a easy spectrum by definition) they point out precisely the floor temperature of the star from the distinctive peak wavelength of the spectrum, the place ( lambda_{peak}) is proportional to (frac{1}{temperature}). This implies larger temperature yields shorter peak wavelength (hotter, bluer stars).

In a faculty lab, one might observe the glowing tungsten filament of an incandescent mild bulb by a spectroscope in a darkened room.

Absorption Spectra:

These are produced by observing comparatively cooler fuel in opposition to a brighter and warmer background. Therefore stellar absorption spectra are obtained by observing a star straight.

The result’s a blackbody spectrum (from the inside stellar core) superimposed with darkish traces or dips. The height of the blackbody yields temperature whereas the absorption traces yield chemical composition since all atoms are ions have distinctive attribute wavelengths of absorption (and emission). Which line of which species additionally signifies temperature.

In a faculty lab, one can observe a white mild by a clear field of fuel (e.g. hydrogen, oxygen, air) or by dispersing a pattern in a flame and observing the flame with a vibrant backlight utilizing a spectroscope in a darkened room (e.g. AAS in Chemistry).

(ii) Bohr’s mannequin proposes that electrons exist in particular, secure orbitals across the nucleus. These orbitals have a particular power. Electrons at a excessive power orbital could decay to a decrease power orbital by emitting a photon of sunshine with an power equal to the distinction of the 2 orbitals; as a result of all of those orbitals are discrete, particular ranges, there are just a few photon energies (therefore wavelengths) that might be emitted.

(ii) Mass of Reactants: 4.0012+9.0122 = 13.0134 amu.

Mass of Merchandise: 12 + 1.0087 = 13.0087 amu.

Mass defect = Mass of Reactants  –  Mass of Merchandise = 0.0047 amu.

Convert to kg by multiplying by (1.661 occasions 10^{-27} = 7.8067 occasions 10^{-30} kg. )

Convert to joules through (E = mc^{2} = 7.0260 times10^{-13} J )

Pauli’s proposal of the neutrino was an answer to the power distribution of beta particles in beta decay – for a given decay the overall power launched was fixed, and it was anticipated that the kinetic power of the launched beta particle could be fixed – as it’s by far the lighter of the 2 merchandise, it ought to obtain the entire launched power. This was not discovered to be the case – the power of beta particles different considerably – suggesting a violation of conservation of power. Pauli instructed an unobserved particle could also be accountable, carrying some power away from the response with out being detected – the neutrino. Because it was undetected till this stage, it should have no cost, low mass and minimal interactions with different particles

• supply X are doubtless alpha particles – their cloud path is brief and large
• supply Y are doubtless beta particles – their cloud trails are lengthy and skinny

Every nucleus of supply X present process alpha decay would lose 4 nucleons whole; 2 neutrons and a pair of protons.

Supply Y is present process beta decay, most probably beta-minus (as positron decay would produce brief tracks because of the positron annihilating itself with an electron).

In every nucleus of Y present process this decay, one neutron is transformed right into a proton, with an electron (beta-minus particle) being emitted together with an anti-neutrino.

There are six quarks (u, d, c, s, t, b) and 6 leptons (e, mu, tau, and three neutrinos), in addition to their antimatter counterparts. The elemental forces are the sturdy drive (mediated by gluons) that acts on quarks and gluons, the weak drive (mediated by W and Z bosons) that acts on quarks and leptons, and the electromagnetic drive that acts on charged particles (mediated by photons). (The usual mannequin doesn’t clarify gravity.)

Out present understanding of the atom is that it’s composed of a nucleus consisting of positively charged protons and impartial neutrons, surrounded by negatively charged electrons in discrete power ranges.  The electrons work together with the nucleus by the electromagnetic drive mediated by photons. The electrons are drawn to the nucleus by this drive which binds them to it. 

Within the nucleus, the protons and neutrons (nucleons) are examples of hadrons – composite particles manufactured from three quarks (uud for protons, and udd for neutrons). The quarks in every hadron are held collectively by the sturdy drive, mediated by gluons. The nucleons are held collectively by the (residual) sturdy drive which is engaging on the typical separation between nucleons. In a secure nucleus, the (residual) sturdy drive overcomes the electrostatic repulsion between the protons (because of their constructive cost) and is ready to bind each protons and neutrons to the nucleus. The neutrons are required as they improve the attraction because of the sturdy drive with out rising repulsion because of the electrical drive, and therefore they contribute to the steadiness of the nucleus.