H.1.1
Describe what is meant by a frame of reference.
The frame of reference illustrates the idea that the motion of an object is dependent on the P.O.V of the observer. For example, sat from my pov is not moving, however from the pov of space sat is moving incredibly fast as he is moving at the speed at which the planet he is on is rotating.
The point of view of an observer is called the frame of reference.
H. 1.2
Galilean Transformation is a method of mathematically quantifying the relative velocities of two objects in motion relative to each other. For example if you take an object travelling at 0.6c to the east and object travelling towards the other objects to the west then the Galilean transformation would be 1.2c. However this is wrong as the velocity exceeds the speed of light which is not possible.
Johsan hehe giggle Physics
Sunday, March 20, 2011
Tuesday, February 22, 2011
Astrophysics E.2 - E.4
E.2.1
Fusion is the main energy source of stars. For fusion to take place high pressures and high temperatures are needed.As temperature is proportional to energy, this means that high temperatures are needed to gain high energies to overcome the electrostatic repulsion between two atoms. High pressures are needed to bring the atoms close enough together for them to bond.
E.2.2
A stable star is essentially a ball of hot gases. As temperature increases gases expand, which leads us to believe that a star should be forever expanding however all particles mass have their own gravitational attraction. This means that once a star gets to a certain size the attraction of gravity between particles becomes equal and opposite to that of the radiation pressure. This is when the star ceases to continue growing in size.
E.2.3
Define luminosity
The total power radiated by a star is called its luminosity
E 2.4
The brightness of a star when brought to a distance of 10 parsecs away from the observer is called the apparent brightness.
E 2.5
The Stefan Boltzmann law defines that luminosity is proportional to the Temperature^4
to compare the luminosities of different stars you must use the following equation:
E 3.4
d=1/p
Fusion is the main energy source of stars. For fusion to take place high pressures and high temperatures are needed.As temperature is proportional to energy, this means that high temperatures are needed to gain high energies to overcome the electrostatic repulsion between two atoms. High pressures are needed to bring the atoms close enough together for them to bond.
E.2.2
A stable star is essentially a ball of hot gases. As temperature increases gases expand, which leads us to believe that a star should be forever expanding however all particles mass have their own gravitational attraction. This means that once a star gets to a certain size the attraction of gravity between particles becomes equal and opposite to that of the radiation pressure. This is when the star ceases to continue growing in size.
E.2.3
Define luminosity
The total power radiated by a star is called its luminosity
E 2.4
The brightness of a star when brought to a distance of 10 parsecs away from the observer is called the apparent brightness.
E 2.5
The Stefan Boltzmann law defines that luminosity is proportional to the Temperature^4
to compare the luminosities of different stars you must use the following equation:
E 2.6
Wein's law states that lambder max (the most commonly emitted wavelength - the most intense wavelength) is inversely proportional to temperature.
This means that surface temperature of a star can be determined by the maximum wavelength of the light emitted.
The graph above illustrates wavelength against intensity, lambder max is the most intense wavelength emitted (all wavelengths are emitted however we are looking for the most commonly emitted wavelength). The most intense wavelength defines the temperature of the star.
The colours on the graph illustrate what colour the star will be according to its lambder max.
This is Wein's Law.
E 3.1 A parsec is defined as the distance between the earth and a distance stellar object when the angle subtended on us by the distant object is equal to two arc seconds over a time period of a year.
E 3.2
Dont understand, but basically d=1/p where d is the distance between the star and the earth and p is the parallax angle in arc seconds.
E 3.3
The parallax method only works for stellar objects that are roughly less than 300 parsecs away from the observer. This is because past this distance the angle tends to closely to zero for us to measure accurately.
d=1/p
E 3.5
The apparent magnitude scale is a comparative scale of apparent brightness of different stellar object of which the lower the value the brighter the object. If the magnitude is over the value of 6 - the naked eye limit - the object is not visible without a telescopic lens.
The apparent magnitude is dependent on the luminosity and the distance of the star.
E 3.6
The absolute magnitude is the brightness of a star at a distance of 10 parsecs.
E 3.7 & 3.8
Solve questions, when it's not quarter to 12 and when im not about to collapse from physical and mental stress.
Sunday, February 20, 2011
Astrophysics
E.1.1
Planets orbit the sun, moons orbit planets.
Planets in our solar system
(remember this from the video, where the camera goes from the sun past all the planets and that Neptune is last)
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto is not a planet, it is a proto planet (dwarf planet)
A stellar cluster is made from a group of stars brought together by gravity.
Constellations are patterns are stars brought identified using imagination
Planets orbit the sun, moons orbit planets.
Planets in our solar system
(remember this from the video, where the camera goes from the sun past all the planets and that Neptune is last)
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto is not a planet, it is a proto planet (dwarf planet)
E.1.2
Constellations are patterns are stars brought identified using imagination
E.1.3
E.1.4
E 1.5
Stars appear to move across the night sky, this is the relative motion between the earth and stellar objects.
Sunday, January 30, 2011
13.2.3 & 13.2.4
13.2.3 - Describe one piece of evidence for the existance of nuclear energy levels.
If you look at the graph below: We can see that the kinetic energy (of the emitted alpha particle) is only ever certain values, this means that the data is discreet. This means that the Mass defect (the energy given off by the nucleus) is also discreet. If these two values are discreet then it suggests that there are discreet energy levels in both the atom and the electron orbitals.
Another way of proving that there are discreet energy levels, is when gamma radiation is emitted from the daughter nucleus. When an atom undergoes alpha decay, it emits alpha particles with discreet energy levels. The atom can either either emit these alpha particles with maximum kinetic energy which leaves the daughter nucleus in a ground state, or it can emit the alpha particles with minimum kinetic energy which leaves the daughter nucleus in an excited state. When this happens gamma radiation is emitted, and this radiation is of a discreet frequency which is further evidence for the existence of nuclear energy levels.
13.2.4 - Describe Beta decay, including the existence of the neutrino
When Beta decay occurs a neutron changes into a proton and an electron is released from the atom. The atomic number increases as there is now an extra proton, however the mass number stays the same.
When Beta-plus decay occurs a proton changes into a neutron and a positron is released from the atom. The atomic number decreases as there is now one less proton, but the mass number stays the same.
When scientist were looking at this, they realised that with every experiment the energy from mass defect was not equal to the energy from the electrons and the positrons. Which meant suggested that there was energy being lost in a form that wasn't originally detected. For this reason scientists proposed there was being another particle emitted during the beta and beta-plus decay. This particle has no charge and extremely low mass, which it is so hard to detect.
For Beta decay an electron is released with an anti neutrino
For Beta-plus decay a positron is released with a neutrino
Saturday, January 29, 2011
Tuesday, November 30, 2010
Subscribe to:
Posts (Atom)