Cheatography

# Physics Cheat Sheet by patrick

VCE Physics unit 1

### Atomic Notation ### Isotopes

 Isotope - An atom with the same number of protons but a different number of neutrons. Radioi­sotope - An isotope that is radioa­ctive and sometimes unstable. They decay.

 Ionising means that the radiation rips off electrons from nearby atoms that it passes. Three types of radiation: Alpha, Beta and Gamma Alpha passes through paper. Beta passes through thin metal. Gamma passes through thick metal. Alpha radiation is most ionising, so it loses it's energy very quickly.

 Alpha radiat­ion - Helium nucleus (2 protons and 2 neutrons), highly ionising, charge of 2+, heavy. Beta radiat­ion - fast moving electron, negatively charged electron, not as high ionising capabi­lities, charge of -1. Gamma radiat­ion - electr­oma­gnetic radiation, less ionising then alpha or beta. Dete­cting radiat­ion - Geiger­-Muller tube counts ions that are produced inside it, these create an electrical signal. Radi­ation in the body - Ionising radiation can damage or kill the cells, dna can replicate in the damaged form, cancers can form.

### Decay ### Neutron Bombar­dment ### Electric field

 The electric field in any region of space is defined as the electric force per unit charge: E = F/q the force on a charge of q in an electric field is given by F= qE

### Resistance Ohm’s law: ∆V=iR or V=Amps­*Ohms

A charge q moving through a potential difference ∆V will lose potential energy: ∆U=qV

### Electrical energy and power

 Electrical energy (Joules) = potential drop (volts) x current (amps) x times (seconds) E=VIt P=E/t were one watt = 1 joule per second E/t = VIT/t or P= VI Power (watts) = voltage (volts) x current (amps) P=VI How much energy does a 100W light bulb use in half an hour? P=100W and t=0.5h So E=100W x 0.5h = 50Wh or 0.05kWh To find power used Volts times Amps

### resistance 1/Rt = 1/R1 + 1/R2 + ...+1/Rn if in parallel.

### Formulas

 W1= q ∆V1 (J)..(.c)..(v) Power - rate of doing work P= w/t (joule­s/s­econds) Power w/t = q∆v/t P=i∆v Unit of energy w=pt (joules) = Watt*Sec New energy unit = kWh Electrical Energy 1kWh ≡ 1000*3600 1kWh ≡ 3.6*10^6J

### Half life

 Half­-life - the time taken for the radioa­ctivity of a specified isotope to fall to half its original value.

### Nuclear fissio­n/f­usion

 Fission- When a nucleus splits into two or more pieces usually after bombar­dment by neutrons. Fusion- A process taking place inside stars in which small nuclei are forced together to make larger nuclei. Energy is released in the process. Chain reaction - A series of nuclear fissions that may or may not be contro­lled. The neutrons that are released cause the reaction.

### Nuclear Fission Reactors

 Used to harness energy from Fission reactions. Neutrons released from Uraniu­m-235 when it undergoes fission are travelling at high speeds, this leads to a chain reaction which causes an explosion. The heat generated from the fission process is used to make steam which drives the turbine. Fuel rods- long, thin rods containing pellets of enriched uranium moderator- material that slows neutrons. control rods- rods made of a material that absorbs neutrons coolant- a liquid or gas to absorb the heat energy

### Nuclear Reactor ### Electric charge

 Conduc­tors: All metals, especially silver, gold, copper aluminium and any ionic solution. Moderate conduc­tors: Water and earth. Semi-c­ond­uctors: Silicon, Germanium and skin. Insula­tors: Plastics, polyst­yrene, dry air, glass, porcelain, cloth (dry) Moderate insula­tors: wood, paper, damp air, ice and snow.

### Electrical forces and fields For the forces between two charges q1 and q2 at a distance of r
k= 9.0 x 109 N m2 x C2

### Electric Current

 Electric current is the rate of transfer of charge: I=q/t where q is the charge transf­erred and t is the time taken. 1 ampere (A) = 1 coulomb per second (C*s-1) So 1 coulomb (C) = 1 ampere second (A*s) 1 volt = 1 joule per coulomb (1V = 1JC-1) 1 ohm = 1 volt per ampere (1ohm = 1VA-1)

### Resistance

 ∆VBatt­ery=i Rtotal ∆vBatt­ery­=∆V­1+∆V2 Therefore i*Rtotal = ∆V1+ ∆V2=iR­1+iR2 Rtotal = R1 + R2 R= V/I or V = IR

### Electric Circuits

 In any electric circuit the sum of all currents flowing into any point is equal to the sum flowing out of it. The total potential drop around a closed circuit must be equal to the total EMF (elect­rom­otive force, the energy provided by the cell)

### Symbols and devices ### Formulas

 Two loops Junction law Current in = current out at(a) itotal = i1 +i2 Parallel arrang­ement ∆V1= ∆V2 iTotal = ∆VBatt­ery­/Rtotal iTotal = i1 +i2 ∆VBatt­ery­/Rt­ota­l=∆­V1/­R1+­∆V2/R2 1/Rtotal = 1/R1 + 1/R2 Rtotal = (1/R1 +1/R2) R1=R2 =10Ohms 1/Rt = 1/10 + 1/10 = 2/10 = 1/5 RT= 5 ohms Voltage loop law One loop ∆Vbattery = ∆1+∆2 Voltage drop of battery must equal Sum of voltage drops around one loop.

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