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Physics: Post your doubts here!

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For this you need to know what the R in the formula -GM\R implies. The R is the distance of the object whose potential you are calculating (i.e the object projected) from the centre of the object exerting the gravitational force (i.e Earth)
Initially the object was at the surface of the earth so the value of R initially was taken to be the radius of the earth(i.e 6. 4*10^6 m).After its motion it reached a height of 1.3 * 10^7 m so its distance from the centre of the Earth is the radius of the earth + the height it attains (i.e 6.4*10^6 + 1.3*10^7) m= 1.94 *10^7 m.
DONT FORGET TO CONVERT KM TO M otherwise YOU WONT GET THE CORRECT ANSWER
Hope this clears it up!!!!:)
 
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AOA
Can someone answer my query.... Light is made of photons (packets of energy) right and is Wave also a part of electromagnetic spectrum ... So are all types of waves within the electromagnetic spectrum made of photons ? Or what is exactly an electromagnetic wave . How might electrons , energy or photons be made into a wave .
O'L student.


In short, yes, the energy of all waves in the electromagnetic spectrum is carried by the photon; it is the quanta of electromagnetic radiation, and as such it carries the energy possessed in every wave, and it interacts with other particles by transferring this energy, etc.

In fact, electrons can be "made into" waves, since they exist in the form of waves; de Broglie, in his thesis, considered the possibility of matter having wave like properties, and he derived the following equation, that relates the momentum of a material object to its wavelength:

Wavelength = h/p
Where h = Planck's Constant
p = momentum of matter - based object concerned.

Usually, heavier objects have such small wavelengths that they can be ignored and assumed to not even exist; however, for matter, on the atomic level (e.g. electrons, etc), the values of momentum for the concerned particles are so small that the de Broglie wavelength is actually noticeable (by experiment) - for example, fast moving electrons can be diffracted; since only waves can be diffracted, the electrons can be proved to be possessing some sort of wave - like property.

Hope this helped!
Good Luck for all your exams!
 
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a)
State the relation between electric field strength E and potential V.
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this is a question in m/j 2007 paper4 .....i dont understand the ms ...te answers says field stregnth = potential gradient .....and the graph given above is of field stregnth againt distance?!
 
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Guys i need a "HELP" pls
I have practical exam at 3:00 i have a Q do we have to wear a coat for that as chemistry and biology !!!!
 
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Hi everyone I gave the AS practical paper 33 this thursday and I had a confusion about the electricity question. There was an ammeter connected to a battery and a rheostat in series and a voltmeter in parallel across the rheostat. They gave us wires but I accidentally used the ammeter wire as a slider for the rheostat instead of using an extra wire. Since it is in series with the rheostat will it be different then if a wire was used as a slider?View attachment 26206
 
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a)
State the relation between electric field strength E and potential V.
..........................................................................................................................................
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this is a question in m/j 2007 paper4 .....i dont understand the ms ...te answers says field stregnth = potential gradient .....and the graph given above is of field stregnth againt distance?!
electric field is the force per unit postive charge and the potential is the energy given up by a unit charge when it moves from one point to another in the circuit :)
 
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hey someone please explain this graph
 

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to get the percentage uncertainity of an average, the ER says "Very few candidates choose to use half the range of repeated readings" Explainn???????? :(
 
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and for Q2 a and b for http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w09_qp_33.pdf the ER says

a) Many candidates (over 50%) failed to repeat their readings ofd.Also many failed to read thediameterto the nearest 0.1mm (or 0.01mm) using a set of vernier callipers.
b)Many candidates used 0.05mm (half a division) for the absolute uncertainty indinstead of 0.1mm.Correct ratio ideas were often used. Very few candidates choose to use half the range of repeated readings.

For percentage uncertainity you use the least count or 1/2 the least count?
 
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Need help with A2 Physics

Guyz i need help in the following question frm year 2003 Physics Paper 4 Q1 (c) :%)
C(i): Change in gravitational Potential = Potential at Earth surface(GM/R) - Potential at Altitude(GM/r) where (r=R + Altitude)
C(ii) Kinetic Energy = Change in Potential Energy
0.5m(v)2 = GMm/R
0.5m(v)2 = GM/R x m
o.5m(v)2 = Change in gravitational potential ( Ci) x m
"m" in Kinetic energy and the "m" in Potential energy will cancel out.
Therefore:
0.5(v)2 = Change in gravitational Potential.

Hope this was useful!!!!
Good luck for the exams.:)
 
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