# physics multiple choice help needed plz Watch

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questions are in the attachments and so are the answers.

please guide me with these problems

thanks in advance

and sorry if it seems long winded

please guide me with these problems

thanks in advance

and sorry if it seems long winded

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#2

Hey ho! Glad to be of service. Well, I hope I'll be of service... xD

I'll take you through the working, so, let's see.

15)

There's balanced forces vertically, 300 kN forward horizontally.

It wants the rate of change of kinetic energy.

Logically, we only have one or two equations to use, the key one being Power = Force * Velocity, with both in equal direction

We have a force of 300 kN, and a velocity currently of 40 ms^-1

Hence multiply for the rate of change, which should be the answer you gave, C.

13)

Torque, as far as I know, is just the moment. Different exam board presumably.

This is force * distance.

It's a couple, hence force * perpendicular distance between them will be the moment on an object.

Y is the orthogonal distance, hence, F*Y

B is the correct answer

14)

The ball is thrown over a flat field. It wants us to describe its motion.

Let's make several assumptions: acceleration, due to g, is entirely vertical and constant. No air resistance. The horizontal velocity of the ball will remain constant.

A Velocity is a vector. Direction has changed. Screw this!

B Technically, yes. There is no horizontal acceleration, and hence it is constant.

C g is constant, as we assumed, thus this is balderdash.

D Horizontal isn't going to be zero, is it? After all, we assumed no air resistance, and g is orthogonal to the horizontal component of v, as we assumed.

10)

We need to analyse this one well, as these sorts of questions get funner if you have to deal with numbers, and if you're having trouble with it, it might be worth it.

The sum of the momenta before equals the sum after. We assume no net force.

Hence, m*u + M(0) = mu = sum before

Hence afterwards, the total is mu.

Vector diagrams can show this.

The horizontal component

A Angle alpha for mu/2 seems incorrect, doesn't it? We need to translate the direction of mu/2 shown, and the angle alpha is incorrect for this.

B Seems fine. Alpha is in correct location, as is beta, and the vectors all connect up in the end forming a total momenta of mu. I like this one.

C Sweet mother of mercy what is going on here. The angles

D Angles fine, I like this so far. Wait, WHAT?! The directions! The sum is zero, which is incorrect.

11)

Velocity will decrease linearly, hence a constant negative gradient. Seems to want air resistance considered. A,D out of the question. C or B are plausible, but C shows a more reasonable tapering of the line with regard to how fast acceleration may fall off.

7)

A Uniform acceleration, hence a constant negative gradient for velocity against time. This can't be it.

B Seems plausible.

C Uniform acceleration, nope.

D I can't even.

5)

*Googles Galvanometer* Ah, it measures current.

A With an infinite p.d, there is no way in hell there's no current.

B If it's zero, there can't be a current... V = IR, hence either R is zero, in which case infinite current, or I is zero.

C Infinite resistance, zero current, sounds good.

D No way; if zero resistance, I can bet charges are going to be dancing all over that.

25)

We're talking about harmonic frequencies here. We're maintaining the stationary wave. Hence, an integer multiple of 92 Hz, our first harmonic frequency.

I'd go for B personally. Check to see if they're all multiples of 92 in the questions. Regardless of how we put it, he's going to have a wave oscillating in there,and I can see why you chose three (1/4 of the wave in the bugle, etc) but the wave is just going to continue (I'd think) slightly outside of the bugle. To be honest, I can't perfectly explain it, but B just seems logical because I can just visualise the wave oscillating at its first harmonic, tapering off outside the bugle.

19)

E = (FL)/(AdL) = doubled.

The length of copper is equal to that of the steel.

The area of copper is four times that of the steel.

The Youngs Modulus of steel is twice that of copper.

There is a constant force between them.

This means that for a standard parameter of copper, identical to that of the steel, the steel would extend by half as much. The area of the copper is four times that of the steel however, hence, we're dealing with the copper extending one quarter as far in comparison. The steel will hence extend twice that of copper.

Hence, B is the answer.

38)

If someone chose C, I would find the closest bridge and just, jump from it. Heck, I don't know what's worse, D or C. They'd have forgotten the anti!

Hope I helped. If you get an answer for that Bugle question, hook me up! :P

EDIT : That vector question too.

BACK TO POPCORN AND SUSHI GINGER AND TEA AND COFFEE AND A NIGHT OF CRAMMING

I'll take you through the working, so, let's see.

15)

There's balanced forces vertically, 300 kN forward horizontally.

It wants the rate of change of kinetic energy.

Logically, we only have one or two equations to use, the key one being Power = Force * Velocity, with both in equal direction

We have a force of 300 kN, and a velocity currently of 40 ms^-1

Hence multiply for the rate of change, which should be the answer you gave, C.

13)

Torque, as far as I know, is just the moment. Different exam board presumably.

This is force * distance.

It's a couple, hence force * perpendicular distance between them will be the moment on an object.

Y is the orthogonal distance, hence, F*Y

B is the correct answer

14)

The ball is thrown over a flat field. It wants us to describe its motion.

Let's make several assumptions: acceleration, due to g, is entirely vertical and constant. No air resistance. The horizontal velocity of the ball will remain constant.

A Velocity is a vector. Direction has changed. Screw this!

B Technically, yes. There is no horizontal acceleration, and hence it is constant.

**WE HAVE A WINNER!**C g is constant, as we assumed, thus this is balderdash.

D Horizontal isn't going to be zero, is it? After all, we assumed no air resistance, and g is orthogonal to the horizontal component of v, as we assumed.

10)

We need to analyse this one well, as these sorts of questions get funner if you have to deal with numbers, and if you're having trouble with it, it might be worth it.

The sum of the momenta before equals the sum after. We assume no net force.

Hence, m*u + M(0) = mu = sum before

Hence afterwards, the total is mu.

Vector diagrams can show this.

The horizontal component

*must*equal mu, and the two other vectors must be aligned directionally and connect up to the end vertex of mu.A Angle alpha for mu/2 seems incorrect, doesn't it? We need to translate the direction of mu/2 shown, and the angle alpha is incorrect for this.

B Seems fine. Alpha is in correct location, as is beta, and the vectors all connect up in the end forming a total momenta of mu. I like this one.

**:P**C Sweet mother of mercy what is going on here. The angles

*and*directions are up the whazoo! Mother mary of jane *shoots self*D Angles fine, I like this so far. Wait, WHAT?! The directions! The sum is zero, which is incorrect.

**Okay you have me...**Why isn't it B?! The vector mu should be equal to the vectors shown... There must be a markscheme error.**We're gonna need some help... :-;**11)

Velocity will decrease linearly, hence a constant negative gradient. Seems to want air resistance considered. A,D out of the question. C or B are plausible, but C shows a more reasonable tapering of the line with regard to how fast acceleration may fall off.

7)

A Uniform acceleration, hence a constant negative gradient for velocity against time. This can't be it.

B Seems plausible.

C Uniform acceleration, nope.

D I can't even.

5)

*Googles Galvanometer* Ah, it measures current.

A With an infinite p.d, there is no way in hell there's no current.

B If it's zero, there can't be a current... V = IR, hence either R is zero, in which case infinite current, or I is zero.

C Infinite resistance, zero current, sounds good.

D No way; if zero resistance, I can bet charges are going to be dancing all over that.

25)

We're talking about harmonic frequencies here. We're maintaining the stationary wave. Hence, an integer multiple of 92 Hz, our first harmonic frequency.

I'd go for B personally. Check to see if they're all multiples of 92 in the questions. Regardless of how we put it, he's going to have a wave oscillating in there,and I can see why you chose three (1/4 of the wave in the bugle, etc) but the wave is just going to continue (I'd think) slightly outside of the bugle. To be honest, I can't perfectly explain it, but B just seems logical because I can just visualise the wave oscillating at its first harmonic, tapering off outside the bugle.

19)

E = (FL)/(AdL) = doubled.

The length of copper is equal to that of the steel.

The area of copper is four times that of the steel.

The Youngs Modulus of steel is twice that of copper.

There is a constant force between them.

This means that for a standard parameter of copper, identical to that of the steel, the steel would extend by half as much. The area of the copper is four times that of the steel however, hence, we're dealing with the copper extending one quarter as far in comparison. The steel will hence extend twice that of copper.

Hence, B is the answer.

38)

If someone chose C, I would find the closest bridge and just, jump from it. Heck, I don't know what's worse, D or C. They'd have forgotten the anti!

Hope I helped. If you get an answer for that Bugle question, hook me up! :P

EDIT : That vector question too.

BACK TO POPCORN AND SUSHI GINGER AND TEA AND COFFEE AND A NIGHT OF CRAMMING

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#3

(Original post by

questions are in the attachments and so are the answers.

please guide me with these problems

thanks in advance

and sorry if it seems long winded

**assassinbunny123**)questions are in the attachments and so are the answers.

please guide me with these problems

thanks in advance

and sorry if it seems long winded

(Original post by

Hey ho! Glad to be of service. Well, I hope I'll be of service... xD

I'll take you through the working, so, let's see.

15)

There's balanced forces vertically, 300 kN forward horizontally.

It wants the rate of change of kinetic energy.

Logically, we only have one or two equations to use, the key one being Power = Force * Velocity, with both in equal direction

We have a force of 300 kN, and a velocity currently of 40 ms^-1

Hence multiply for the rate of change, which should be the answer you gave, C.

13)

Torque, as far as I know, is just the moment. Different exam board presumably.

This is force * distance.

It's a couple, hence force * perpendicular distance between them will be the moment on an object.

Y is the orthogonal distance, hence, F*Y

B is the correct answer

14)

The ball is thrown over a flat field. It wants us to describe its motion.

Let's make several assumptions: acceleration, due to g, is entirely vertical and constant. No air resistance. The horizontal velocity of the ball will remain constant.

A Velocity is a vector. Direction has changed. Screw this!

B Technically, yes. There is no horizontal acceleration, and hence it is constant.

C g is constant, as we assumed, thus this is balderdash.

D Horizontal isn't going to be zero, is it? After all, we assumed no air resistance, and g is orthogonal to the horizontal component of v, as we assumed.

10)

We need to analyse this one well, as these sorts of questions get funner if you have to deal with numbers, and if you're having trouble with it, it might be worth it.

The sum of the momenta before equals the sum after. We assume no net force.

Hence, m*u + M(0) = mu = sum before

Hence afterwards, the total is mu.

Vector diagrams can show this.

The horizontal component

A Angle alpha for mu/2 seems incorrect, doesn't it? We need to translate the direction of mu/2 shown, and the angle alpha is incorrect for this.

B Seems fine. Alpha is in correct location, as is beta, and the vectors all connect up in the end forming a total momenta of mu. I like this one.

C Sweet mother of mercy what is going on here. The angles

D Angles fine, I like this so far. Wait, WHAT?! The directions! The sum is zero, which is incorrect.

11)

Velocity will decrease linearly, hence a constant negative gradient. Seems to want air resistance considered. A,D out of the question. C or B are plausible, but C shows a more reasonable tapering of the line with regard to how fast acceleration may fall off.

7)

A Uniform acceleration, hence a constant negative gradient for velocity against time. This can't be it.

B Seems plausible.

C Uniform acceleration, nope.

D I can't even.

5)

*Googles Galvanometer* Ah, it measures current.

A With an infinite p.d, there is no way in hell there's no current.

B If it's zero, there can't be a current... V = IR, hence either R is zero, in which case infinite current, or I is zero.

C Infinite resistance, zero current, sounds good.

D No way; if zero resistance, I can bet charges are going to be dancing all over that.

25)

We're talking about harmonic frequencies here. We're maintaining the stationary wave. Hence, an integer multiple of 92 Hz, our first harmonic frequency.

I'd go for B personally. Check to see if they're all multiples of 92 in the questions. Regardless of how we put it, he's going to have a wave oscillating in there,and I can see why you chose three (1/4 of the wave in the bugle, etc) but the wave is just going to continue (I'd think) slightly outside of the bugle. To be honest, I can't perfectly explain it, but B just seems logical because I can just visualise the wave oscillating at its first harmonic, tapering off outside the bugle.

19)

E = (FL)/(AdL) = doubled.

The length of copper is equal to that of the steel.

The area of copper is four times that of the steel.

The Youngs Modulus of steel is twice that of copper.

There is a constant force between them.

This means that for a standard parameter of copper, identical to that of the steel, the steel would extend by half as much. The area of the copper is four times that of the steel however, hence, we're dealing with the copper extending one quarter as far in comparison. The steel will hence extend twice that of copper.

Hence, B is the answer.

38)

If someone chose C, I would find the closest bridge and just, jump from it. Heck, I don't know what's worse, D or C. They'd have forgotten the anti!

Hope I helped. If you get an answer for that Bugle question, hook me up! :P

EDIT : That vector question too.

BACK TO POPCORN AND SUSHI GINGER AND TEA AND COFFEE AND A NIGHT OF CRAMMING

**Callicious**)Hey ho! Glad to be of service. Well, I hope I'll be of service... xD

I'll take you through the working, so, let's see.

15)

There's balanced forces vertically, 300 kN forward horizontally.

It wants the rate of change of kinetic energy.

Logically, we only have one or two equations to use, the key one being Power = Force * Velocity, with both in equal direction

We have a force of 300 kN, and a velocity currently of 40 ms^-1

Hence multiply for the rate of change, which should be the answer you gave, C.

13)

Torque, as far as I know, is just the moment. Different exam board presumably.

This is force * distance.

It's a couple, hence force * perpendicular distance between them will be the moment on an object.

Y is the orthogonal distance, hence, F*Y

B is the correct answer

14)

The ball is thrown over a flat field. It wants us to describe its motion.

Let's make several assumptions: acceleration, due to g, is entirely vertical and constant. No air resistance. The horizontal velocity of the ball will remain constant.

A Velocity is a vector. Direction has changed. Screw this!

B Technically, yes. There is no horizontal acceleration, and hence it is constant.

**WE HAVE A WINNER!**C g is constant, as we assumed, thus this is balderdash.

D Horizontal isn't going to be zero, is it? After all, we assumed no air resistance, and g is orthogonal to the horizontal component of v, as we assumed.

10)

We need to analyse this one well, as these sorts of questions get funner if you have to deal with numbers, and if you're having trouble with it, it might be worth it.

The sum of the momenta before equals the sum after. We assume no net force.

Hence, m*u + M(0) = mu = sum before

Hence afterwards, the total is mu.

Vector diagrams can show this.

The horizontal component

*must*equal mu, and the two other vectors must be aligned directionally and connect up to the end vertex of mu.A Angle alpha for mu/2 seems incorrect, doesn't it? We need to translate the direction of mu/2 shown, and the angle alpha is incorrect for this.

B Seems fine. Alpha is in correct location, as is beta, and the vectors all connect up in the end forming a total momenta of mu. I like this one.

**:P**C Sweet mother of mercy what is going on here. The angles

*and*directions are up the whazoo! Mother mary of jane *shoots self*D Angles fine, I like this so far. Wait, WHAT?! The directions! The sum is zero, which is incorrect.

**Okay you have me...**Why isn't it B?! The vector mu should be equal to the vectors shown... There must be a markscheme error.**We're gonna need some help... :-;**11)

Velocity will decrease linearly, hence a constant negative gradient. Seems to want air resistance considered. A,D out of the question. C or B are plausible, but C shows a more reasonable tapering of the line with regard to how fast acceleration may fall off.

7)

A Uniform acceleration, hence a constant negative gradient for velocity against time. This can't be it.

B Seems plausible.

C Uniform acceleration, nope.

D I can't even.

5)

*Googles Galvanometer* Ah, it measures current.

A With an infinite p.d, there is no way in hell there's no current.

B If it's zero, there can't be a current... V = IR, hence either R is zero, in which case infinite current, or I is zero.

C Infinite resistance, zero current, sounds good.

D No way; if zero resistance, I can bet charges are going to be dancing all over that.

25)

We're talking about harmonic frequencies here. We're maintaining the stationary wave. Hence, an integer multiple of 92 Hz, our first harmonic frequency.

I'd go for B personally. Check to see if they're all multiples of 92 in the questions. Regardless of how we put it, he's going to have a wave oscillating in there,and I can see why you chose three (1/4 of the wave in the bugle, etc) but the wave is just going to continue (I'd think) slightly outside of the bugle. To be honest, I can't perfectly explain it, but B just seems logical because I can just visualise the wave oscillating at its first harmonic, tapering off outside the bugle.

19)

E = (FL)/(AdL) = doubled.

The length of copper is equal to that of the steel.

The area of copper is four times that of the steel.

The Youngs Modulus of steel is twice that of copper.

There is a constant force between them.

This means that for a standard parameter of copper, identical to that of the steel, the steel would extend by half as much. The area of the copper is four times that of the steel however, hence, we're dealing with the copper extending one quarter as far in comparison. The steel will hence extend twice that of copper.

Hence, B is the answer.

38)

If someone chose C, I would find the closest bridge and just, jump from it. Heck, I don't know what's worse, D or C. They'd have forgotten the anti!

Hope I helped. If you get an answer for that Bugle question, hook me up! :P

EDIT : That vector question too.

BACK TO POPCORN AND SUSHI GINGER AND TEA AND COFFEE AND A NIGHT OF CRAMMING

I would provide a slight different explanation for Q7. You have a kinematics equation for constant acceleration which links final speed with distance travelled.

Use this relationship you should be able to work out the correct choice.

For question 10, the question should be B. Have a look at this website for elastic collision.

http://www.physics.brocku.ca/PPLATO/h-flap/phys2_5.html

What is the answer in marksheme?

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#4

(Original post by

.....

25)

We're talking about harmonic frequencies here. We're maintaining the stationary wave. Hence, an integer multiple of 92 Hz, our first harmonic frequency.

I'd go for B personally. Check to see if they're all multiples of 92 in the questions. Regardless of how we put it, he's going to have a wave oscillating in there,and I can see why you chose three (1/4 of the wave in the bugle, etc) but the wave is just going to continue (I'd think) slightly outside of the bugle. To be honest, I can't perfectly explain it, but B just seems logical because I can just visualise the wave oscillating at its first harmonic, tapering off outside the bugle.

…

**Callicious**).....

25)

We're talking about harmonic frequencies here. We're maintaining the stationary wave. Hence, an integer multiple of 92 Hz, our first harmonic frequency.

I'd go for B personally. Check to see if they're all multiples of 92 in the questions. Regardless of how we put it, he's going to have a wave oscillating in there,and I can see why you chose three (1/4 of the wave in the bugle, etc) but the wave is just going to continue (I'd think) slightly outside of the bugle. To be honest, I can't perfectly explain it, but B just seems logical because I can just visualise the wave oscillating at its first harmonic, tapering off outside the bugle.

…

For a closed and open pipe instrument, the fundamental frequency is

Only the odd harmonics frequency will occur in such instrument.

where

*n*is odd integer.

So the answer has to be C for Q25.

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#5

**assassinbunny123**)

questions are in the attachments and so are the answers.

please guide me with these problems

thanks in advance

and sorry if it seems long winded

For Q14 which is the pressure question and I believe @Callicious had answered Q8 instead of Q14 in the above.

You just need to apply the defining equation for pressure difference

is 20.0 m

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#6

(Original post by

For a closed and open pipe instrument, the fundamental frequency is

Only the odd harmonics frequency will occur in such instrument.

where

So the answer has to be C for Q25.

**Eimmanuel**)For a closed and open pipe instrument, the fundamental frequency is

Only the odd harmonics frequency will occur in such instrument.

where

*n*is odd integer.So the answer has to be C for Q25.

The MS said B though, so I'm even more confused than ever... this stuff isn't on my syllabus (the open air instrument formula) so my working was probably incorrect for that answer.

Turns out the answer was B for Q10, the me-in-the-middle-of-the-night probably just derped up reading it and got paranoid ;-;

Thanks for the help! WAIT! WAIT! One last thing... doesn't matter, just more thanks for the better explanation for Q7, I never really thought about using formulae to justify the question.

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#7

(Original post by

My brains ;-;

The MS said B though, so I'm even more confused than ever... this stuff isn't on my syllabus (the open air instrument formula) so my working was probably incorrect for that answer.

....

**Callicious**)My brains ;-;

The MS said B though, so I'm even more confused than ever... this stuff isn't on my syllabus (the open air instrument formula) so my working was probably incorrect for that answer.

....

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#8

(Original post by

Do you mean Q25 that the MS said B? But I saw the OP answer to Q25 is C.

**Eimmanuel**)Do you mean Q25 that the MS said B? But I saw the OP answer to Q25 is C.

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Thx.

could you please explain why the answer is 20J and not 16J.

since force and distance is proportional to workdone. so wouldn't doubling force and distance quadruple the work done.. or have i missed something.

could you please explain why the answer is 20J and not 16J.

since force and distance is proportional to workdone. so wouldn't doubling force and distance quadruple the work done.. or have i missed something.

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**Callicious**)

Hey ho! Glad to be of service. Well, I hope I'll be of service... xD

I'll take you through the working, so, let's see.

15)

There's balanced forces vertically, 300 kN forward horizontally.

It wants the rate of change of kinetic energy.

Logically, we only have one or two equations to use, the key one being Power = Force * Velocity, with both in equal direction

We have a force of 300 kN, and a velocity currently of 40 ms^-1

Hence multiply for the rate of change, which should be the answer you gave, C.

13)

Torque, as far as I know, is just the moment. Different exam board presumably.

This is force * distance.

It's a couple, hence force * perpendicular distance between them will be the moment on an object.

Y is the orthogonal distance, hence, F*Y

B is the correct answer

14)

The ball is thrown over a flat field. It wants us to describe its motion.

Let's make several assumptions: acceleration, due to g, is entirely vertical and constant. No air resistance. The horizontal velocity of the ball will remain constant.

A Velocity is a vector. Direction has changed. Screw this!

B Technically, yes. There is no horizontal acceleration, and hence it is constant.

**WE HAVE A WINNER!**

C g is constant, as we assumed, thus this is balderdash.

D Horizontal isn't going to be zero, is it? After all, we assumed no air resistance, and g is orthogonal to the horizontal component of v, as we assumed.

10)

We need to analyse this one well, as these sorts of questions get funner if you have to deal with numbers, and if you're having trouble with it, it might be worth it.

The sum of the momenta before equals the sum after. We assume no net force.

Hence, m*u + M(0) = mu = sum before

Hence afterwards, the total is mu.

Vector diagrams can show this.

The horizontal component

*must*equal mu, and the two other vectors must be aligned directionally and connect up to the end vertex of mu.

A Angle alpha for mu/2 seems incorrect, doesn't it? We need to translate the direction of mu/2 shown, and the angle alpha is incorrect for this.

B Seems fine. Alpha is in correct location, as is beta, and the vectors all connect up in the end forming a total momenta of mu. I like this one.

**:P**

C Sweet mother of mercy what is going on here. The angles

*and*directions are up the whazoo! Mother mary of jane *shoots self*

D Angles fine, I like this so far. Wait, WHAT?! The directions! The sum is zero, which is incorrect.

**Okay you have me...**Why isn't it B?! The vector mu should be equal to the vectors shown... There must be a markscheme error.

**We're gonna need some help... :-;**

11)

Velocity will decrease linearly, hence a constant negative gradient. Seems to want air resistance considered. A,D out of the question. C or B are plausible, but C shows a more reasonable tapering of the line with regard to how fast acceleration may fall off.

7)

A Uniform acceleration, hence a constant negative gradient for velocity against time. This can't be it.

B Seems plausible.

C Uniform acceleration, nope.

D I can't even.

5)

*Googles Galvanometer* Ah, it measures current.

A With an infinite p.d, there is no way in hell there's no current.

B If it's zero, there can't be a current... V = IR, hence either R is zero, in which case infinite current, or I is zero.

C Infinite resistance, zero current, sounds good.

D No way; if zero resistance, I can bet charges are going to be dancing all over that.

25)

We're talking about harmonic frequencies here. We're maintaining the stationary wave. Hence, an integer multiple of 92 Hz, our first harmonic frequency.

I'd go for B personally. Check to see if they're all multiples of 92 in the questions. Regardless of how we put it, he's going to have a wave oscillating in there,and I can see why you chose three (1/4 of the wave in the bugle, etc) but the wave is just going to continue (I'd think) slightly outside of the bugle. To be honest, I can't perfectly explain it, but B just seems logical because I can just visualise the wave oscillating at its first harmonic, tapering off outside the bugle.

19)

E = (FL)/(AdL) = doubled.

The length of copper is equal to that of the steel.

The area of copper is four times that of the steel.

The Youngs Modulus of steel is twice that of copper.

There is a constant force between them.

This means that for a standard parameter of copper, identical to that of the steel, the steel would extend by half as much. The area of the copper is four times that of the steel however, hence, we're dealing with the copper extending one quarter as far in comparison. The steel will hence extend twice that of copper.

Hence, B is the answer.

38)

If someone chose C, I would find the closest bridge and just, jump from it. Heck, I don't know what's worse, D or C. They'd have forgotten the anti!

Hope I helped. If you get an answer for that Bugle question, hook me up! :P

EDIT : That vector question too.

BACK TO POPCORN AND SUSHI GINGER AND TEA AND COFFEE AND A NIGHT OF CRAMMING

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#11

(Original post by

you were right about question 10. i think you misread my terrible scribbling.

**assassinbunny123**)you were right about question 10. i think you misread my terrible scribbling.

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(Original post by

Pff, no one scribbles worse than me!

**Callicious**)Pff, no one scribbles worse than me!

Do u know what a null indicator means???

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#13

(Original post by

still not sure about question 5.... the markscheme says D

Do u know what a null indicator means???

**assassinbunny123**)still not sure about question 5.... the markscheme says D

Do u know what a null indicator means???

"an instrument used to measure an electrical quantity by adjusting known quantities in the circuit until a reading of zero is obtained."

Never heard the term before, so I just went from the idea that the Galvanometer was an ammeter. You'll need to get someone else to help on that one, no clue! ;-;

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#15

(Original post by

Thx.

could you please explain why the answer is 20J and not 16J.

since force and distance is proportional to workdone. so wouldn't doubling force and distance quadruple the work done.. or have i missed something.

**assassinbunny123**)Thx.

could you please explain why the answer is 20J and not 16J.

since force and distance is proportional to workdone. so wouldn't doubling force and distance quadruple the work done.. or have i missed something.

KE_final - KE_initial = work done

KE_final - 4.0 = 4Fs

where 4Fs = 16 J (which is what you have calculated. Well done!)

KE_final = 16 J + 4.0 J = 20 J

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Are these questions from AQA or other spec ?

**Lawlerfhfbdj**)Are these questions from AQA or other spec ?

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#17

**assassinbunny123**)

still not sure about question 5.... the markscheme says D

Do u know what a null indicator means???

It just mean to indicate whether there is current between the two points. Do you know why galvanometer is used instead of "normal ammeter"?

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I think they are from CIE or I may be wrong.

**Eimmanuel**)I think they are from CIE or I may be wrong.

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http://encyclopedia2.thefreedictiona...Null+Indicator

It just mean to indicate whether there is current between the two points. Do you know why galvanometer is used instead of "normal ammeter"?

**Eimmanuel**)http://encyclopedia2.thefreedictiona...Null+Indicator

It just mean to indicate whether there is current between the two points. Do you know why galvanometer is used instead of "normal ammeter"?

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i thought the answer is A but its apparently B.

my thinking was that in A the contact force is correct on both the wall and the AC

my thinking was that in A the contact force is correct on both the wall and the AC

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