OCR A level Physics help required!!
Hello, I need help with understand the below question.....
I dont understand how to do the following question..
I Understand the energy supplied is w = vq ...... but im unsure how to calculate the energy dissipated in the internal resistance of the charger and car battery..... could someone please explain? also, how would i calculate energy gained (opposite of this question, for future reference).
here's the entire paper and its a part of Q 2...
http://www.ocr.org.uk/Images/58045-question-paper-unit-g482-electrons-waves-and-photons.pdf
I dont understand how to do the following question..
I Understand the energy supplied is w = vq ...... but im unsure how to calculate the energy dissipated in the internal resistance of the charger and car battery..... could someone please explain? also, how would i calculate energy gained (opposite of this question, for future reference).
here's the entire paper and its a part of Q 2...
http://www.ocr.org.uk/Images/58045-question-paper-unit-g482-electrons-waves-and-photons.pdf
(edited 11 years ago)
Scroll to see replies
Original post by SexyNerd
I Understand the energy supplied is q = vq ...... ???????
I Understand the energy supplied is q = vq ...... ???????
Total energy supplied in joules = volts x current x time (i.e. Watt seconds).
i.e. 14 x 2.5 x (6 x 60 x 60) = 0.756 x106 joules.
To calculate the energy dissipated in the resistances use I2R and the energy formula above:
Vres = Icharge x Rtot = Icharge(Rcharger + Rbattery)
Pres = Vres x Icharge
Energy dissipated = Pres x tseconds
(edited 11 years ago)
Original post by uberteknik
Total energy supplied in joules = volts x current x time (i.e. Watt seconds).
i.e. 14 x 2.5 x (6 x 60 x 60) = 0.756 x106 joules.
To calculate the energy dissipated in the resistances use I2R and the energy formula above:
Vres = Icharge x Rtot = Icharge(Rcharger + Rbattery)
Pres = Vres x Icharge
Energy dissipated = Pres x tseconds
i.e. 14 x 2.5 x (6 x 60 x 60) = 0.756 x106 joules.
To calculate the energy dissipated in the resistances use I2R and the energy formula above:
Vres = Icharge x Rtot = Icharge(Rcharger + Rbattery)
Pres = Vres x Icharge
Energy dissipated = Pres x tseconds
thanks... how do i calculate the minimum power required to lift something.... the total gravitational potential energy is 60037200..... mass 1 = 200 kg, which carries 80 individual mass of 60 kg.. vertical height is 900 m, 5 minutes in time..........
Original post by uberteknik
Total energy supplied in joules = volts x current x time (i.e. Watt seconds).
i.e. 14 x 2.5 x (6 x 60 x 60) = 0.756 x106 joules.
To calculate the energy dissipated in the resistances use I2R and the energy formula above:
Vres = Icharge x Rtot = Icharge(Rcharger + Rbattery)
Pres = Vres x Icharge
Energy dissipated = Pres x tseconds
i.e. 14 x 2.5 x (6 x 60 x 60) = 0.756 x106 joules.
To calculate the energy dissipated in the resistances use I2R and the energy formula above:
Vres = Icharge x Rtot = Icharge(Rcharger + Rbattery)
Pres = Vres x Icharge
Energy dissipated = Pres x tseconds
so I^2 R..... would be (2.5)^2 x (.8) = 5
Original post by joostan
Sure 
Original post by SexyNerd
so I^2 R..... would be (2.5)^2 x (.8) = 5
That gives the power dissipated.
Now multiply by t seconds to get the total energy used in dissipation.
Original post by SexyNerd
thanks... how do i calculate the minimum power required to lift something.... the total gravitational potential energy is 60037200..... mass 1 = 200 kg, which carries 80 individual mass of 60 kg.. vertical height is 900 m, 5 minutes in time..........
Are you serious? This is GCSE stuff.
GPE = Mtot x gravity x height. i.e. this is the total energy needed to lift that mass 900m against gravity.
Also
Energy used = Power x Time
So Power = Energy/Time (1Watt = 1 Joule/second)
This gives the raw power needed in a straight line vertical lift executed in 5 minutes. i.e. using no other mechanical advantage which would reduce the power at the expense of a longer time.
(edited 11 years ago)
have any of you got a way of remembering the wavelengths of EM spectra? i'm just find it really hard to memorize them 
Original post by sophiekutie
have any of you got a way of remembering the wavelengths of EM spectra? i'm just find it really hard to memorize them
Try:
Reading
Music
Is
Virtually
Useless for
Xylophones and
Glockenspiels
i.e.:
Radio waves
Micro waves
Infra red
Visible
Ultra violet
X-rays
Gamma rays
Original post by sophiekutie
have any of you got a way of remembering the wavelengths of EM spectra? i'm just find it really hard to memorize them
This is my way:
Reinvest
Money
Into
Vaults
Until
eXtras
Grow
Original post by SexyNerd
the first question...
Seems to have been answered
Original post by sophiekutie
have any of you got a way of remembering the wavelengths of EM spectra? i'm just find it really hard to memorize them
Run
Miles
In
Very
X-Treme
Games
Original post by uberteknik
x
Original post by krisshP
x
Original post by joostan
x
ahh thanks, i get the order of them, it's just the wavelengths (e.g Ultraviolet between 10^-5 to 10^-7)
apparently for my AS exam i need to know a typical wavelength for each, and i need to be able to tell what type of wave it is based on just wavelength.
thanks for your help
Original post by sophiekutie
ahh thanks, i get the order of them, it's just the wavelengths (e.g Ultraviolet between 10^-5 to 10^-7)
apparently for my AS exam i need to know a typical wavelength for each, and i need to be able to tell what type of wave it is based on just wavelength.
thanks for your help
apparently for my AS exam i need to know a typical wavelength for each, and i need to be able to tell what type of wave it is based on just wavelength.
thanks for your help
No real shortcut other than to learn them
Original post by uberteknik
Are you serious? This is GCSE stuff.
GPE = Mtot x gravity x height. i.e. this is the total energy needed to lift that mass 900m against gravity.
Also
Energy used = Power x Time
So Power = Energy/Time (1Watt = 1 Joule/second)
This gives the raw power needed in a straight line vertical lift executed in 5 minutes. i.e. using no other mechanical advantage which would reduce the power at the expense of a longer time.
GPE = Mtot x gravity x height. i.e. this is the total energy needed to lift that mass 900m against gravity.
Also
Energy used = Power x Time
So Power = Energy/Time (1Watt = 1 Joule/second)
This gives the raw power needed in a straight line vertical lift executed in 5 minutes. i.e. using no other mechanical advantage which would reduce the power at the expense of a longer time.
well i didnt do my gcse's.....how about moments.... plank is 2m.. mass a (50kg) is .2 from pivot... how far must mass b (5kg) be from pivot to be in equilibrium?
Original post by joostan
.....
(edited 11 years ago)
Original post by SexyNerd
well i didnt do my gcse's.....how about moments.... plank is 2m.. mass a (50kg) is .2 from pivot... how far must mass b (5kg) be from pivot to be in equilibrium?
http://www.gcse.com/fm/moments.htm
Original post by uberteknik
yes thank you, I've learnt how to do moments, but I'm stuck on this particular problem.... can you help me please? and thanks for the help in regards to the last question.
(edited 11 years ago)
Original post by SexyNerd
yes thank you, I've learnt how to do moments, but I'm stuck on this particular problem.... can you help me?
What are you stuck on?
Original post by uberteknik
What are you stuck on?
i dont get the correct answer.... use clock wise as positive from pivot .. b is on right side of pivot, a is on the left....
so
5g(x) - 50g(.20) = 0 (equal to 0 because its in equilibrium)
5g(x) = 10g
x = 10/5
x = 2
which is not correct, can you see where my problem is?
Original post by SexyNerd
i dont get the correct answer.... use clock wise as positive from pivot .. b is on right side of pivot, a is on the left....
so
5g(x) - 50g(.20) = 0 (equal to 0 because its in equilibrium)
5g(x) = 10g
x = 10/5
x = 2
which is not correct, can you see where my problem is?
so
5g(x) - 50g(.20) = 0 (equal to 0 because its in equilibrium)
5g(x) = 10g
x = 10/5
x = 2
which is not correct, can you see where my problem is?
You have the correct answer so you have to say equilibrium cannot be achieved with the stated variables. The only way of balancing is to either increase the length of the plank, or increase the mass (b), or decrease the distance of mass(a) from the pivot, or decrease mass (a).
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