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An electron has a kinetic energy E and a de Broglie wavelength λ. The kinetic
energy is increased to 4E. Why is the new de Broglie wavelength λ/2??
energy is increased to 4E. Why is the new de Broglie wavelength λ/2??
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#2
(Original post by HK4)
An electron has a kinetic energy E and a de Broglie wavelength λ. The kinetic
energy is increased to 4E. Why is the new de Broglie wavelength λ/2??
An electron has a kinetic energy E and a de Broglie wavelength λ. The kinetic
energy is increased to 4E. Why is the new de Broglie wavelength λ/2??
KE = 0.5mv^2 => 4KE => 2v
λ = h/mv
As v has doubled, λ will half.
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(Original post by britishtf2)
Kinetic energy increased implies increased speed (as mass stays constant, assuming no relativistic effects).
KE = 0.5mv^2 => 4KE => 2v
λ = h/mv
As v has doubled, λ will half.
Kinetic energy increased implies increased speed (as mass stays constant, assuming no relativistic effects).
KE = 0.5mv^2 => 4KE => 2v
λ = h/mv
As v has doubled, λ will half.
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#4
(Original post by HK4)
Thanks for the quick reply, I'm not too good at Physics.
Thanks for the quick reply, I'm not too good at Physics.
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(Original post by britishtf2)
Don't worry, and feel free to PM me any questions.
Don't worry, and feel free to PM me any questions.
An electric motor of input power 100 W raises a mass of 10 kg vertically at a steady speed of 0.5 m s −1 . What is the efficiency of the system? Answer is 50%
A string passes through a smooth thin tube. Masses m and M are attached to the ends of the string. The tube is moved so that the mass m travels in a horizontal circle of constant radius r and at constant speed v. What is equal to M ?
Answer is 𝑚𝑣^2/𝑟𝑔
Here are the other question are didn't quite get in the paper, if you or anyone else could have a go at explaining that would be great
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#6
(Original post by HK4)
An object falls freely from rest. After falling a distance d its velocity is v. What is its velocity after it has fallen a distance 2d? Answer is √2 v
An electric motor of input power 100 W raises a mass of 10 kg vertically at a steady speed of 0.5 m s −1 . What is the efficiency of the system? Answer is 50%
A string passes through a smooth thin tube. Masses m and M are attached to the ends of the string. The tube is moved so that the mass m travels in a horizontal circle of constant radius r and at constant speed v. What is equal to M ?
Answer is 𝑚𝑣^2/𝑟𝑔
Here are the other question are didn't quite get in the paper, if you or anyone else could have a go at explaining that would be great
An object falls freely from rest. After falling a distance d its velocity is v. What is its velocity after it has fallen a distance 2d? Answer is √2 v
An electric motor of input power 100 W raises a mass of 10 kg vertically at a steady speed of 0.5 m s −1 . What is the efficiency of the system? Answer is 50%
A string passes through a smooth thin tube. Masses m and M are attached to the ends of the string. The tube is moved so that the mass m travels in a horizontal circle of constant radius r and at constant speed v. What is equal to M ?
Answer is 𝑚𝑣^2/𝑟𝑔
Here are the other question are didn't quite get in the paper, if you or anyone else could have a go at explaining that would be great
change in mgh = change in 0.5mv^2 (ignoring pos/neg)
-Try it from here. I will post the full solution at the bottom for you (goes for all of them)
2) Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
(See below for full)
3) See posts 7 and 8.
1, ans)
change in mgh = change in 0.5mv^2
v^2 = 2gh
v = root(2gh)
Change in height doubles and 2g stays the same, so v increases by a factor of root(2)
2, ans)
Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
Useful energy out = Ke+Change in mgh per second
Ke = 0.5*10*0.5^2 = 1.25
Change in mgh in one second = mg*delta(h) = mg*vt = 10*9.8*0.5*1 = 49
49+1.25 = 50.25
Efficiency = 100%* 50.25/100 = 50.25% = 50% 2sf
3, ans) See post 8.
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(Original post by britishtf2)
1) Equate energies, and let d = h
change in mgh = change in 0.5mv^2 (ignoring pos/neg)
-Try it from here. I will post the full solution at the bottom for you (goes for all of them)
2) Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
(See below for full)
3) Have you got a diagram for this? i.e. Is the tube vertical/horizontal?
1, ans)
change in mgh = change in 0.5mv^2
v^2 = 2gh
v = root(2gh)
Change in height doubles and 2g stays the same, so v increases by a factor of root(2)
2, ans)
Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
Useful energy out = Ke+Change in mgh per second
Ke = 0.5*10*0.5^2 = 1.25
Change in mgh in one second = mg*delta(h) = mg*vt = 10*9.8*0.5*1 = 49
49+1.25 = 50.25
Efficiency = 100%* 50.25/100 = 50.25% = 50% 2sf
3, ans) Will do once I know what the set up is.
1) Equate energies, and let d = h
change in mgh = change in 0.5mv^2 (ignoring pos/neg)
-Try it from here. I will post the full solution at the bottom for you (goes for all of them)
2) Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
(See below for full)
3) Have you got a diagram for this? i.e. Is the tube vertical/horizontal?
1, ans)
change in mgh = change in 0.5mv^2
v^2 = 2gh
v = root(2gh)
Change in height doubles and 2g stays the same, so v increases by a factor of root(2)
2, ans)
Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
Useful energy out = Ke+Change in mgh per second
Ke = 0.5*10*0.5^2 = 1.25
Change in mgh in one second = mg*delta(h) = mg*vt = 10*9.8*0.5*1 = 49
49+1.25 = 50.25
Efficiency = 100%* 50.25/100 = 50.25% = 50% 2sf
3, ans) Will do once I know what the set up is.
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#8
(Original post by HK4)
Thanks a lot, http://filestore.aqa.org.uk/resource...-74081-SQP.PDF its question 29, I thought I'd attached an image sorry.
Thanks a lot, http://filestore.aqa.org.uk/resource...-74081-SQP.PDF its question 29, I thought I'd attached an image sorry.
The weight of mass M is providing the centripetal force of the mass m. It is the only force providing it, so it is equal to it.
Try to work from that.
Answer:
Mg = (mv^2)/r
M = (mv^2)/rg
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#9
(Original post by britishtf2)
1) Equate energies, and let d = h
change in mgh = change in 0.5mv^2 (ignoring pos/neg)
-Try it from here. I will post the full solution at the bottom for you (goes for all of them)
2) Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
(See below for full)
3) See posts 7 and 8.
1, ans)
change in mgh = change in 0.5mv^2
v^2 = 2gh
v = root(2gh)
Change in height doubles and 2g stays the same, so v increases by a factor of root(2)
2, ans)
Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
Useful energy out = Ke+Change in mgh per second
Ke = 0.5*10*0.5^2 = 1.25
Change in mgh in one second = mg*delta(h) = mg*vt = 10*9.8*0.5*1 = 49
49+1.25 = 50.25
Efficiency = 100%* 50.25/100 = 50.25% = 50% 2sf
3, ans) See post 8.
1) Equate energies, and let d = h
change in mgh = change in 0.5mv^2 (ignoring pos/neg)
-Try it from here. I will post the full solution at the bottom for you (goes for all of them)
2) Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
(See below for full)
3) See posts 7 and 8.
1, ans)
change in mgh = change in 0.5mv^2
v^2 = 2gh
v = root(2gh)
Change in height doubles and 2g stays the same, so v increases by a factor of root(2)
2, ans)
Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
Useful energy out = Ke+Change in mgh per second
Ke = 0.5*10*0.5^2 = 1.25
Change in mgh in one second = mg*delta(h) = mg*vt = 10*9.8*0.5*1 = 49
49+1.25 = 50.25
Efficiency = 100%* 50.25/100 = 50.25% = 50% 2sf
3, ans) See post 8.
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#10
(Original post by flyingthin)
How did you learn first question, book doesnt tell you and my physics teacher is trash
How did you learn first question, book doesnt tell you and my physics teacher is trash
GPE = mgh = KE = 0.5mv^2
Hope this was of help, or you managed to figure it out yourself.
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