Physics HL/SL Help

[IB Smacher Glau]

thnxs DeBrogliez I did exactly what you did on my practice paper but I wondered that it's supposed to by h/(sqrt(2*m*q*V)) and not wavelength divided by the the sqrt

[DeBrogliez]

I see that DeBrogliez himself has answered your question!

Hahaha you are right, that's why I chose this username

I have two questions please

first one is about topic 10

how do we differentiate between adiabatic and isothermal processes from a P-V graph?

I know that adiabatic process should have a steeper graph, but why?

Also, what if I don't have both graphs to compare, how can I know? (In other words, only one graph is drawn)

Should I use the fact the the pressure and volume should be inversely proportional in isothermal process? (The product of the pressure and volume on any point on the graph should be the same?)

My second question is about topic 9

Is is true that the orbital speed equals to the square root of the magnitude of the gravitational potential at a given distance from a planet?

While the escape speed of the planet is equal to the square root of twice the magnitude of the gravitational potential at the surface of that planet?

I'm a little bit confused

Edited May 8, 2012 by DeBrogliez

[IB Smacher Glau]

DeBrogliez kef 7lak

I studied thermodymics 2day open page 188 in the Tskos book and you will know why !

[IB Smacher Glau]

In the multiple choice question I see that some answers need to be guessed ! ( I will wear my lucky shirt after 2moro )

Another thing I want to ask about what order of mangitudes we need to memorize I know we should memorize order of magnitude of EMS I memorize it as GAXUVIMR ! (Gamma, X-Ray etc )

[DeBrogliez]

DeBrogliez kef 7lak

I studied thermodymics 2day open page 188 in the Tskos book and you will know why !

Unfortunately, I don't have that book!

I rarely use the book actually

Almost dependent on the teacher notes

In the multiple choice question I see that some answers need to be guessed ! ( I will wear my lucky shirt after 2moro )

Another thing I want to ask about what order of mangitudes we need to memorize I know we should memorize order of magnitude of EMS I memorize it as GAXUVIMR ! (Gamma, X-Ray etc )

I think we also have to know the order of the diameter of the nucleus/atom

Which I never memorized, and probably will never

It is usually the first question in paper 1

And I am ready to lose that mark lol

[HiggsHunter]

Is is true that the orbital speed equals to the square root of the magnitude of the gravitational potential at a given distance from a planet?

While the escape speed of the planet is equal to the square root of twice the magnitude of the gravitational potential at the surface of that planet?

I'm a little bit confused

For a mass m orbiting at distance r from the centre of a planet of mass M at speed v we can equate the force due to gravity (GMm)/rˆ2 to the centripetal force (mvˆ2)/r, so v = sqrt(GM/r). (Note that it is independent of m)

At the escape speed, the sum of the KE ((1/2)mvˆ2) and the PE (-GMm/r) of the mass is zero, so v = sqrt(2GM/r). (Again, it is independent of m)

Hence the escape speed is sqrt(2) times the orbital speed at the same height from the surface of the planet.

Are you surprised that it is so much higher? Remember that the escape speed is high enough for a vehicle to escape from the gravity of the planet ballistically (like Jules Verne's moon ship!). A vehicle can distance itself from a planet at lower speed as long as its rocket motor or thrusters keep providing a propulsive force.

The Voyager probes have exceeded the solar escape velocity at their present distance from the sun, so even though they are still being slowed by the sun's gravitational field, they will never return to the solar system.

[SB26]

i 'm still puzzled with the meaning of

- gravitational potential (v=potential energy / m)

- gravitational potential energy ( -GMm/r)

- electric field strength

- electric potential

- electric potential energy

Don't quite really understand the definition in textbooks, could someone please help me clarify it or give me some simple analogy to clear things up

Thx a lot!!

[Award Winning Boss]

For the conclusion in the physics IA... what am I meant to say?

That rhymed!

[IB Smacher Glau]

For the conclusion in the physics IA... what am I meant to say?

That rhymed!

Just end it like what anyone of the IBS family would end a lab end it #ByTheCrazyWay

haa just compare your results with the actual results and don't forget to check the criteria everything is mentioned there !!

[eelnedross]

For the conclusion in the physics IA... what am I meant to say?

That rhymed!

Report any interesting patterns you find. Also, state the relationship found in the graph that you made. Besides that, you should obviously compare it to the known value if you've determined the value of a constant or the nature of a physical relationship. If none of this works, just check the IB criteria for IAs.

[fan]

6. The surface temperature of the sun is 5800K and the radius of the sun is 6.96x10⁸m.

(i)Assuming the sun is an ideal black body radiator what is the power radiated ?

so Area of a sphere is 4pr²

power radiated = Power radiated = eσAT⁴

for an ideal black body e=1

so I get my area as 6.09 x10^18 and my power as 2.0x10^15 W

but answer is 3.91x10²⁶W :/

[HiggsHunter]

6. The surface temperature of the sun is 5800K and the radius of the sun is 6.96x10⁸m.

(i)Assuming the sun is an ideal black body radiator what is the power radiated ?

so Area of a sphere is 4pr²

power radiated = Power radiated = eσAT⁴

for an ideal black body e=1

so I get my area as 6.09 x10^18 and my power as 2.0x10^15 W

but answer is 3.91x10²⁶W :/

I think that you have just made a slip in your arithmetic, as your reasoning is correct.

Here is the calculation for you to check:

P = 5.67 x (10ˆ-8) x 6.09 x (10ˆ18) x 1.132 x (10ˆ15)

= 3.91 x 10ˆ26 W

[fan]

ahh ok thanks!! Could you help with another question.

A point charge Q₁ = -4.nC is at the origin, and a second point charge Q₂ =+6.00 nC is on the x-axis at x=8.0m. Find the electric field(magnitude and direction) at each of the following points on the x-axis

a) x=0.2m

b) x=-0.2m

[HiggsHunter]

ahh ok thanks!! Could you help with another question.

A point charge Q₁ = -4.nC is at the origin, and a second point charge Q₂ =+6.00 nC is on the x-axis at x=8.0m. Find the electric field(magnitude and direction) at each of the following points on the x-axis

a) x=0.2m

b) x=-0.2m

To find the electric field at any location due to several point charges you just have to take the vector sum of the fields due to each charge, which you can derive from Coulomb's law as E = kQ/rˆ2

The direction of the component fields is outwards from the positive charge and inwards to the negative one.

For the figures in your question, the points in (a) and (b) are so near charge Q1 that the contribution to the field strength from the remote charge Q2 is very small indeed.

[Rigel]

Need help with this question:

A cell is connected to two identical lamps connected in parallel. The lamps are rated at 1,25V, 300mW. The cell has an emf of 1,5V and an internal resistance of 1.2ohms. Determine whether the lamp will light normally.

Also, on a whole different note, could someone please explain to me for what do you use a potential divider? And how should it be connected in electrical circuits?

Thank you very much.

[HiggsHunter]

Need help with this question:

A cell is connected to two identical lamps connected in parallel. The lamps are rated at 1,25V, 300mW. The cell has an emf of 1,5V and an internal resistance of 1.2ohms. Determine whether the lamp will light normally.

Also, on a whole different note, could someone please explain to me for what do you use a potential divider? And how should it be connected in electrical circuits?

Thank you very much.

At the rated voltage the pair of lamps in parallel would draw a total current of 2(0.3/1.25) A = 0.48 A. But at this current the output voltage from the battery would be only 1.5 - (0.48 x 1.2) v = 0.924 v, insufficient for the lamps to light normally.

Potential dividers have many uses in electrical circuits. Can you give more details of the question? Or are you just asked to give an example?

[aahmedov]

Can someone help with physics nuclear reaction problem?

1 H + 7 Li = 4 He + 4 He

1 3 2 2

Find the energy released. Atomic mass of lithium is 7.016u

[hedwig]

A typical family home uses a total of about 15 kWh of electrical energy per day and needs to heart about 45 kg (litres) of water from 12^oC to 80^oC for showers etc. If a house is in an area where it can receive, on average, about 400 W m^-2 of solar radiation for about 8 hours per day calculate:

a) The collecting area of the solar water heater if it has an efficiency of 40%;

b) The collecting area of the photovoltaic cells if their efficiency is 18%. (1 kWh = 3.6 x 10⁶J)

(Note: these calculations do not include energy used to heat the house during the winter.)

I'd appreciate it if you could explain step by step how to solve this and mention the formulas you used as well. We didn't have time to cover topic 8 in the school year so we have to do it on our own in the summer. I'm rather lost and the textbook is no help... Thank you!

[SerUmbras]

A typical family home uses a total of about 15 kWh of electrical energy per day and needs to heart about 45 kg (litres) of water from 12^oC to 80^oC for showers etc. If a house is in an area where it can receive, on average, about 400 W m^-2 of solar radiation for about 8 hours per day calculate:

a) The collecting area of the solar water heater if it has an efficiency of 40%;

b) The collecting area of the photovoltaic cells if their efficiency is 18%. (1 kWh = 3.6 x 10⁶J)

(Note: these calculations do not include energy used to heat the house during the winter.)

I'd appreciate it if you could explain step by step how to solve this and mention the formulas you used as well. We didn't have time to cover topic 8 in the school year so we have to do it on our own in the summer. I'm rather lost and the textbook is no help... Thank you!

Okay, I'll see what I can do.

a) The energy needed to heat 45 litres of water (80-12) = 68^o C is found using Q=mc(delta)T

Q = energy = ?

m = mass = 45

c = specific heat capacity = 4.2

(delta)T = change in temperature = 68^o.

Q = 45 x 4.2 x 68 = 12852 kJ

This is the energy needed. Now, if the solar heater has efficiency of 40%, only 40% of the energy incident is gained. So;

Energy needed (to heat water) by 1/efficiency = energy that needs to be collected.

12852(1/0.4) = 32130 kJ

Now that we've got that, we find how much energy the heater can get over the day.

At 400 W/m^-2 for 8 hours, the total energy per metre² is the 400 by the number of seconds in 8 hours

400(8 x 60 x 60) = 11520000 J = 11520 kJ

Area required = energy needed/enegry per unit area

A = 32130/11520 = 21.1m²

b) The energy used (and thus required) is 15kWh (15kW per hour) which equals 15(3.6 x 10⁶J) = 54000kJ per hour. Over 1 day (24 hours) that becomes 1296000kJ (1296MJ) of enegry needed per day.

Using the value for the energy we need and the efficiency of the photovoltaic cells (18%), we can find the actual energy needed .

1296 x 10⁶(1/0.18) = 7200 MJ

Using the energy gained per square metre per day from part a) (11520 kJ = 11.52MJ), we can find the area needed.

Area required = energy needed/enegry per unit area

A = 7200000000/11520 = 625000m²

(which, to be completely honest, is ridiculous and impossible and just what the IB would ask...)

So, hope that helped in your understanding!

GL, HF, DD

Edited August 25, 2012 by The Not-Zetta Slow One

[HiggsHunter]

A typical family home uses a total of about 15 kWh of electrical energy per day and needs to heart about 45 kg (litres) of water from 12^oC to 80^oC for showers etc. If a house is in an area where it can receive, on average, about 400 W m^-2 of solar radiation for about 8 hours per day calculate:

a) The collecting area of the solar water heater if it has an efficiency of 40%;

b) The collecting area of the photovoltaic cells if their efficiency is 18%. (1 kWh = 3.6 x 10⁶J)

(Note: these calculations do not include energy used to heat the house during the winter.)

I'd appreciate it if you could explain step by step how to solve this and mention the formulas you used as well. We didn't have time to cover topic 8 in the school year so we have to do it on our own in the summer. I'm rather lost and the textbook is no help... Thank you!

b) The energy used (and thus required) is 15kWh (15kW per hour) which equals 15(3.6 x 10⁶J) = 54000kJ per hour. Over 1 day (24 hours) that becomes 1296000kJ (1296MJ) of enegry needed per day.

Using the value for the energy we need and the efficiency of the photovoltaic cells (18%), we can find the actual energy needed .

1296 x 10⁶(1/0.18) = 7200 MJ

Using the energy gained per square metre per day from part a) (11520 kJ = 11.52MJ), we can find the area needed.

Area required = energy needed/enegry per unit area

A = 7200000000/11520 = 625000m²

(which, to be completely honest, is ridiculous and impossible and just what the IB would ask...)

So, hope that helped in your understanding!

GL, HF, DD

The photovoltaic cells of efficiency 18% can supply (0.4 x 0.18 x 8) = 0.576 kWh of electrical energy per square metre per day.

Hence the area of cells required to provide 15kWh per day is 15/0.576 = 26 m2.