pranish1
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Hello there, I am really struggling to get the right answers with the following multiple choice questions!!
If anyone could help me by explaining their thought process on getting the right answer, that would be really helpful

the answers are:
D
D
A
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britishtf2
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(Original post by pranish1)
Hello there, I am really struggling to get the right answers with the following multiple choice questions!!
If anyone could help me by explaining their thought process on getting the right answer, that would be really helpful

the answers are:
D
D
A
1) (10/10+r) = 1.26/1.45
Rearrange for r.

2) Force = Mass*Acceleration. You can find acceleration from v^2 = u^2 + 2as

3) It will take half the time of flight to reach its maximum height.
Use v = u + at for vertical motion. v = 0 at max height. Then use v^2 = u^2 + 2as to find height.
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Rexar
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When you connect a voltmeter over a cell this gives you the emf. So the emf is 1.45V.
When the external circuit is connected the lost volts is the emf-PD over the supply, 1.45-1.26 = 0.19V.
Internal resistance is the lost volts. So Ir = 0.19V
The current in the circuit is (1.45)/(10+r)
So (1.45)/(10+r) times by r = 0.19V. Rearrange and you get 1.45r=0.19(10+r). 1.26r=1.9. R=1.508 and thus is D.
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Rexar
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That should really be a 5 mark question and not 1 mark, lol
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britishtf2
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(Original post by Rexar)
That should really be a 5 mark question and not 1 mark, lol
Nah. 2 at most. It's just (10*1.45)/(1.26) - 10 = r
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Rexar
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For the second question use newtons second law. F=deltaMV/deltaT. Use Surat to find the time taken for the car to crumple and thus the impact time. This works out at 3/50 seconds
Then use F=(500*(10-0)) / (3/50) and this equals 83333 so this the answer is D
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Rexar
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(Original post by britishtf2)
Nah. 2 at most. It's just (10*1.45)/(1.26) - 10 = r
Wow my method was allot more complicated then yours. I am intrigued as to how you got that equation, I really dislike electricity
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britishtf2
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(Original post by Rexar)
Wow my method was allot more complicated then yours. I am intrigued as to how you got that equation, I really dislike electricity
The emf does not change. Voltage is "used up" by a component due to a linear relationship dependant on the percentage of resistance that the component uses. In the original scenario, the switch is open so the only component is the resistor. Therefore, all the emf is used in the resistor and none in the battery due to its internal resistance.

When the switch is closed, current flows. The voltage across the resistor drops, telling you there is an internal resistance as stated. The ratio between the voltage passing through the resistor and total, and between resistor and total are the same. This is to say the following:

Vr/Vt = Rr/Rt

We are given Vr as 1.26.
Vt is equal to the emf, so equal to the voltage across the resistor with no flowing current.
Rr is 10.
Rt is Rr+Ir = 10+Ir

1.26/1.45 = 10/(10+Ir)
10+Ir = (10*1.45)/(1.26)
Ir = (10*1.45)/(1.26) - 10 as given.
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Seth17
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(Original post by britishtf2)
1) (10/10+r) = 1.26/1.45
Rearrange for r.

2) Force = Mass*Acceleration. You can find acceleration from v^2 = u^2 + 2as

3) It will take half the time of flight to reach its maximum height.
Use v = u + at for vertical motion. v = 0 at max height. Then use v^2 = u^2 + 2as to find height.
For question 3 can't you just use s=vt-1/2at^2?
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britishtf2
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(Original post by Seth17)
For question 3 can't you just use s=vt-1/2at^2?
You don't have the speed at the beginning. Once you have worked out the original speed, you can also use s = ut + 0.5at^2 [it's + not -, just the acceleration is minus].
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Seth17
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(Original post by britishtf2)
You don't have the speed at the beginning. Once you have worked out the original speed, you can also use s = ut + 0.5at^2 [it's + not -, just the acceleration is minus].
As you said before at maximum height v=0 leaving you with s=-1/2at^2. So plugging a=-1.6 it then becomes s=-1/2(-1.6)t^2?
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britishtf2
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(Original post by Seth17)
As you said before at maximum height v=0 leaving you with s=-1/2at^2. So plugging a=-1.6 it then becomes s=-1/2(-1.6)t^2?
Ohh. Misread the v as a u - kinda forgot that version of the equation existed to be honest. Yes, you can do that.
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Eimmanuel
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(Original post by britishtf2)
The emf does not change. Voltage is "used up" by a component due to a linear relationship dependant on the percentage of resistance that the component uses. In the original scenario, the switch is open so the only component is the resistor. Therefore, all the emf is used in the resistor and none in the battery due to its internal resistance.

When the switch is opened, current flows. The voltage across the resistor drops, telling you there is an internal resistance as stated. The ratio between the voltage passing through the resistor and total, and between resistor and total are the same. This is to say the following:

Vr/Vt = Rr/Rt

We are given Vr as 1.26.
Vt is equal to the emf, so equal to the voltage across the resistor with no flowing current.
Rr is 10.
Rt is Rr+Ir = 10+Ir

1.26/1.45 = 10/(10+Ir)
10+Ir = (10*1.45)/(1.26)
Ir = (10*1.45)/(1.26) - 10 as given.
I am sorry that I may be a pain in the ....

Your conclusion of
Vr/Vt = Rr/Rt
is a just simple potential divider formula in the disguise.
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britishtf2
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(Original post by Eimmanuel)
I am sorry that I may be a pain in the ....

Your conclusion of
Vr/Vt = Rr/Rt
is a just simple potential divider formula in the disguise.
Yes, but I thought it would be worthwhile explaining why, rather than just saying "Use this cos it's right." oe
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Eimmanuel
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(Original post by britishtf2)
Yes, but I thought it would be worthwhile explaining why, rather than just saying "Use this cos it's right." oe
I believe that potential divider formula should be well explained in most of the A level physics texts.

If you had referred the following deduction
"Voltage is "used up" by a component due to a linear relationship dependant on the percentage of resistance that the component uses."
from the potential divider formula, I believe your argument is more easy to follow. Anyway that is me.

By the way, how does current flows in an open circuit?
"... When the switch is opened, current flows...."
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britishtf2
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(Original post by Eimmanuel)
I believe that potential divider formula should be well explained in most of the A level physics texts.

If you had referred the following deduction
"Voltage is "used up" by a component due to a linear relationship dependant on the percentage of resistance that the component uses."
from the potential divider formula, I believe your argument is more easy to follow. Anyway that is me.

By the way, how does current flows in an open circuit?
"... When the switch is opened, current flows...."
Hmm, thanks for the criticism. Always good to get other people's points of view on how to word things.

As for the quote, that is a typing mistake by me: meant to read "... switch is closed...".
Thanks for pointing that out; changed now.
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