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Physics Prelim Q's (1 Viewer)

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1. Water has a refractive index of 1.33

a) Calculate the speed of light in water. (2.26?)

b) A ray of light in air is incident on water at an angle of 25 degrees. Calculate the refracted angle of light in the water. (The answer is 18.5 degrees?)

Need working for all questions.

Will be posting up more in this thread
 

Aerath

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a) v1n1 = v2n2
v1 = 3x108 m/s (speed of light in air)
n1 = 1 (refractive index of air
n2 = 1.33 (supplied)

3x108 x 1 = v2 x 1.33
v2 = 3x108x 1 / 1.33 = 2.26 x 108 m/s

b) Look here
 
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Shoom

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7.3 x 10^19 electrons passed through a certain point in 4 mins calculate the charge,

Can someone do that please.

If a positive coulomb of 1 charge (or something) was placed next to a positive coulomb of 2 charges (or something)
(it was +1 and +2 though)
Draw the magnetic field around them and the null point

Please do these for me.
 
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A popular Sydney radio station transmits at a frequency of 104.9 MHz. What is the wavelength of this radio station? (I know you apply the formula of speed = frequency x wavelength, but how would you know the speed?)
 

ratcher0071

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DiligentStudent said:
A popular Sydney radio station transmits at a frequency of 104.9 MHz. What is the wavelength of this radio station? (I know you apply the formula of speed = frequency x wavelength, but how would you know the speed?)
All waves operate at the speed of light (i think)

velocity(m/s)=frequency(Hz) x wavelength(m)
wavelength(m)=velocity(m/s)/frequency (Hz)
wavelength (m)=(3.6 x 10^8) / (104.9 MHz x 1000 kHz x 1000 Hz)
wavelength(m)= 3.43 metres

somebody please verify!
 

ratcher0071

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Shoom said:
7.3 x 10^19 electrons passed through a certain point in 4 mins calculate the charge
One coulomb (C) is the charge equal to the total charge on 6.28 x 10^18 electrons.

If the charge on 1 electron is -1.6 x 10^-19 Coulombs
then the charge on 7.3 x 10^19 electrons is -11.68 Coulombs
:cold: This might be wrong. :cold:
 

bubblesss

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ratcher0071 said:
All waves operate at the speed of light (i think)

velocity(m/s)=frequency(Hz) x wavelength(m)
wavelength(m)=velocity(m/s)/frequency (Hz)
wavelength (m)=(3.6 x 10^8) / (104.9 MHz x 1000 kHz x 1000 Hz)
wavelength(m)= 3.43 metres

somebody please verify!
when do we use the speed of light as 330m/s?????:uhoh: yeh but i think ur right.
 

Shoom

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Ratcher how did you get that?

Show working please thanks for help.

Could you try the other question?

Ok I get the 11.68 C but what do we do about the four minutes?

Do I multiply this answer by 240 seconds?
 
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ratcher0071

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Shoom said:
Ratcher how did you get that?

Show working please thanks for help.

Could you try the other question?

Ok I get the 11.68 C but what do we do about the four minutes?

Do I multiply this answer by 240 seconds?

I'm not really sure what you do about the 4 minutes, you'd have to give me the full question.

The 2nd question, i don't know.
 

H4rdc0r3

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DiligentStudent said:
A popular Sydney radio station transmits at a frequency of 104.9 MHz. What is the wavelength of this radio station? (I know you apply the formula of speed = frequency x wavelength, but how would you know the speed?)
3x10^8 / 104.9 x 10^6
 

electrolysis

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bubblesss said:
when do we use the speed of light as 330m/s?????:uhoh: yeh but i think ur right.
speed of sound is 330m/s
speed of light is 3x10^8
 

x.Exhaust.x

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May as well post my questions up here as well :)

1. Explain why parabolic reflectors are used to collect signals from outer space.
2. Explain how electrical potential difference and electrical charge are related.

And what are all the necessary formulae needed to be remembered for the exam?
 

12o9

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x.Exhaust.x said:
May as well post my questions up here as well :)

1. Explain why parabolic reflectors are used to collect signals from outer space.
2. Explain how electrical potential difference and electrical charge are related.

And what are all the necessary formulae needed to be remembered for the exam?
1. To bring the signals to a focus in order to obtain a clearer signal (by the inverse square law, the signals will be scattered).
 

ratcher0071

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x.Exhaust.x said:
May as well post my questions up here as well :)

1. Explain why parabolic reflectors are used to collect signals from outer space.
2. Explain how electrical potential difference and electrical charge are related.

And what are all the necessary formulae needed to be remembered for the exam?
2) E=qV
Electric Field = electrical charge x electrical potential difference
 

darusan92

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x.Exhaust.x said:
May as well post my questions up here as well :)

2. Explain how electrical potential difference and electrical charge are related.

And what are all the necessary formulae needed to be remembered for the exam?
Electric potential difference is the force needed to move per unit of charge. You get that from the formulae E=F/q
 

ratcher0071

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darusan92 said:
Electric potential difference is the force needed to move per unit of charge. You get that from the formulae E=F/q
I don't think so.
E=F/q
is
Electric Field Strength=Force/charge
 

12o9

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lol. I had that question in my half yearly.

Transverse waves cannot be transmitted through fluid media as the particles in a transverse wave propagate perpendicular to the direction of occilation. In water this isn't possible.

Yeah sounds dodge, but I got the marks for it =/.
 

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