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Drake Glau

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Thank you very much people!

More questions? :)

[HL, Gravitation and Orbital Motion] A satellite is orbiting the earth in a circular orbit. Which one of the following properties of the satellite does not remain constant?

A. Kinetic energy

B. Gravitational potential energy

C. Angular momentum

D. Velocity

I think A and B should be both constant since the R is not changing. So it's either C or D. What is angular momentum, though? How is it different from momentum? I think velocity should be constant but may be not due to acceleration and air resistance... (even though I think air resistance is negligible)

I'm not quite sure what angular momentum is either... But I think velocity shouldn't be constant since velocity is a vector quantity and a satellite changes direction all the time... Thus even if it's going at the same speed it's velocity is not constant due to constant change of direction, perhaps?

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I'm not quite sure what angular momentum is either... But I think velocity shouldn't be constant since velocity is a vector quantity and a satellite changes direction all the time... Thus even if it's going at the same speed it's velocity is not constant due to constant change of direction, perhaps?

OHHH haha yeah I forgot :) thank you! So velocity changes with orbital position.. but if we talk about speed, it remains constant right? I get that now, thank you! :)

Btw I finished my MCQ holiday homework so you guys should be happy as I won't ask more questions soon (I may need help with lab report later but that's another business) so thank you guys very much for helping me! :)

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  • 3 weeks later...

Any physics students out there please help!!!

A lead bullet is fired into an iron plate where is deforms and stops. As a result, the temperature of the lead increases by an amount ΔT. For a lead bullet having twice the mass but the same spee d of impact, what would be the best estimate of its temperature increase?

A. 1/2 ΔT

B. ΔT

C. √2 ΔT

D. 2 ΔT

Thanks in advance

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  • 2 weeks later...

I have a question which I'm struggling to get to grips with. This question is in four parts (I've done the first part, so I'll show you the other three parts):

1) Monochromatic light of wavelength 5.4×10^-7 m is incident on a metal surface which has a work function of 1.40 × 10^-19 J.

a) Calculate the maximum kinetic energy of an electron emitted from the surface.

I was thinking to use KE= 0.5mv^2, but I don't know the velocity of the electron, although I know that the mass of one electron is 9.1 × 10^-31 kg.

b) Calculate the maximum speed of the emitted electron.

I have no idea how to do this.....

c) Calculate the de Broglie wavelength of the fastest electrons.

I know that the general equation for the de Broglie wavelength for a particle that has mass is h/√2πmKT, where m is the mass, K is the Boltzmann's constant, T is the temperature, and h is Planck's constant. I also know that I could use the mass, 9.1 × 10^-31 kg for this equation, but I don't know the temperature. Can the temperature be worked out using the work function and the wavelength? Or am I just getting everything completely wrong here? :D

Thanks for your help!!

D.

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  • 2 weeks later...

I added a pdf version so that its easier to read, so you might look at that rather then this if you would like

Information provided:

λ = 5.4 x 〖10〗^(-7)m

ɸ (work function) = 1.40 x 〖10〗^(-19) J

Einstein’s equation to find the energy carried by a photon:

E=hf

Electrons absorb different amounts of photons. If the photon is large enough it will give electrons enough energy to leave the surface. This energy is known as the work function. If there is any extra energy it will be kept by the electron as kinetic energy.

Hence the incoming energy of the photon will equal the energy required for the electron to leave the surface plus the kinetic energy

hf= ɸ+KE

Now frequency can be written as:

f= c/λ = (3.0 x 〖10〗^8)/(5.4 x 〖10〗^(-7) ) = …….

(Sorry don’t have a calculator at the moment :). so you will have to do the calculating)

Now your question says to find the KE, so let’s re-arrange the equation:

KE= hf- ɸ

h = plank constant = 6.63 x 〖10〗^(-34)

Therefore KE = ((6.63 x 〖10〗^(-34)) x f) - (1.40 x 〖10〗^(-19))

b) Now the next question is asking about the speed

KE = 1/2 mv^2

Make v the subject

V = √ 2KE/m

m = mass of an electron

c) De Broglie’s equation is λ= h/p

p being momentum

Now KE = 1/2 mv^2 and p=mv

v = p/m

substitute in the KE equation and we get

KE = p^2/2m

Make p the subject

p = √(KE x 2m)

After getting p we can find the De Broglie wavelength λ= h/p

question.pdf

Edited by ∑fan
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From the May 08 SL paper 2:

The initial volume of the gas is 0.20 m3 at a temperature of 27C and a pressure of 100,000 Pa. After compression the volume of the gas is 0.070 m3 and the temperature is 57 C. Determine the final pressure of the gas.

I know it has something to do with PV=nRT but I'm not sure what to do..

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well you have

v= 0.20 m3

T= 27C

P1 =100,000 Pa

after compression you get

v= 0.070 m3 T=57 C

so what we do first is find a value that will be the same in both of them. This will be the number of moles.

PV=nRT

PV/RT = n

R is the gas constant

(100,000 x 0.20)/ (8.31 x 27) = 89.14 moles

so now we have

v= 0.070 m3

T=57 C

R=8.31

n =89.14 moles

substitute the values into

P2=nRT/v and we get the the final pressure :D

final pressure is 603,174 Pa

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well you have

v= 0.20 m3

T= 27C

P1 =100,000 Pa

after compression you get

v= 0.070 m3 T=57 C

so what we do first is find a value that will be the same in both of them. This will be the number of moles.

PV=nRT

PV/RT = n

R is the gas constant

(100,000 x 0.20)/ (8.31 x 27) = 89.14 moles

so now we have

v= 0.070 m3

T=57 C

R=8.31

n =89.14 moles

substitute the values into

P2=nRT/v and we get the the final pressure :)

final pressure is 603,174 Pa

Thanks, that's roughly what I got :blum:

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Alex aren't you meant to find the percentage uncertainty of each value and then add them?

e.g

Time (s) ± 0.05

1.00

2.00

(0.05/1.00) x 100 = 5%

(0.05/2.00) x 100 = 2.5%

total percentage uncertainty = 7.5%

average time = (1.00 + 2.00)/2 = 1.5

then you find 7.5% of 1.5 and then you get your uncertainty.

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  • 3 weeks later...

So, my IA is currently for option G (electromagnetic waves) and I decided to "prove" snell's law. I found a glass rectangular prism and a laser for this and all that good stuff but was wondering if anyone could verify that my thought process behind the math isn't out of whack...

K, so snell's law is nsin(theta1)=nsin(theta2)

For my diagram theta 2 is being used so I'm going to call it theta3 for snell's law so

nsin(theta1)=nsin(theta3)

theta3=theta1+theta2

The way I planned on getting theta2 was law of sines (sin a)/A=(sin b)/B but using a drawn in triangle (as you can see in the diagram) so (sin90)/h=(sin theta2)/x

So theta2=sin-1[x(sin90)]/h

All in all I got snell's law down to nsin(theta1)=nsin[sin-1(xsin(90)/h)+theta1]

This all make sense or did I screw it all up? :o

post-31095-0-92709600-1300845618_thumb.j

Edited by Drake Glau
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  • 3 weeks later...

I need help on solving this:

This question is about Kinematics.

Someone is walking in the street, when suddenly he sees that a woman is falling down of the last floor of a building, so he uses his superpowers and starts flying with a constant acceleration of 15m/s2. The woman is falling down from a distance of 300m. Then,

a) What would be the length that the woman fell down up to when she was rescued by the man?

b) How much will the man take to save the woman?

c) What would be the speed of the man and the woman when the man rescues the woman?

Thank you very much for your help! I tried solving it, but i can't get the ammount of metres the woman fell down...

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Assuming he was standing directly underneath her when he started flying vertically upwards, that he started flying at the exact moment she started falling, and that acceleration by gravity is 10ms-2

Man's distance from the last floor = 300 - Distance he flies

= 300 - at2/2

= 300 - 15t2/2

Woman's distance from the last floor = 10t2/2

At the point he rescues her, both distances are equal:

300 - 15t2/2 = 10t2/2

300 = 25t2/2

24 = t2

t = 4.90s

It seems I had to solve b) before solving a) :blink:

a) Woman's distance from last floor = 10t2/2

= 10t2/2

= 10*24/2

= 120m

b) Time taken to reach woman = 4.90s

c) Speed of man = at

= 15*4.9

= 3010ms-1 to 3sf.

This final speed is too high and would kill instead of saving the girl. Should I take into consideration the effect of gravitational pull on the man? Which would cause he's (effective) acceleration to be 5ms-2 not 15ms-2 and his speed upon rescue to be much less (31.6ms-1). Maybe this had been taken into consideration when his acceleration was given, another assumption I made to answer the problem. This is not a good problem as I had to make many assumptions...

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