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OCR Physics A G484 - The Newtonian World - 11th June 2015

Reply 280

imfrassin

2

why does change of state, change internal energy but no change temperature?

Can somebody explain to me please

Reply 281

Elcor

14

Original post by whyalwaysme??

Vector x vector= scalar, think of it as + and - where + is scalar and - vector

I do FP3 and a vector x a vector is another vector which is perpendicular to the two vectors. This isn't important for G484, though.

Reply 282

Tiwa

7

Original post by imfrassin

why does change of state, change internal energy but no change temperature?

Can somebody explain to me please

When you heat an object, the kinetic energy of the molecules increases since KE \propto T . When the object changes state, the potential energy of the molecules increases, thus increasing it's internal energy. When an object changes state, you only alter the potential energy, not the kinetic energy hence temperature doesn't change. Hope that helps.

Edit: please excuse the typos. I meant to say "thus increasing it's internal energy", not the potential energy. :tongue:

(edited 8 years ago)

Reply 283

imfrassin

2

Original post by Tiwa

When you heat an object, the kinetic energy of the molecules increases Since KE \propto T

. When the object changes state, the potential energy of the molecules increases, thus increasing it's potential energy. When an object changes state, you only alter the potential energy, not the kinetic energy hence temperature doesn't change. Hope that helps.

Exactly the explanation I was looking for!

Reply 284

Tiwa

7

Original post by imfrassin

Exactly the explanation I was looking for!

You're very welcome!

Reply 285

sagar448

4

Original post by BrokenS0ulz

Is the following correct?
scalar x scalar = scalar
vector x vector = vector
scalar x vector = vector

Yes for most cases but for things such as force x displacement it gives you energy which is not a vector.

Reply 286

BrokenS0ulz

3

Original post by Elcor

I do FP3 and a vector x a vector is another vector which is perpendicular to the two vectors. This isn't important for G484, though.

Well there was a question asking about why momentum was a vector and it was 2 marks. So you had to say that it has magnitude and direction + the product of a vector and a scalar is a vector.

Reply 287

BrokenS0ulz

3

Could somebody help? I know you've got to do pv(before) = pv(after) but how do you get the volume afterwards? The correct answer is 83.5kPa

Reply 288

sagar448

4

Original post by BrokenS0ulz

Could somebody help? I know you've got to do pv(before) = pv(after) but how do you get the volume afterwards? The correct answer is 83.5kPa

Can I know what paper this is please?

Reply 289

L'Evil Fish

OP

18

Original post by BrokenS0ulz

Could somebody help? I know you've got to do pv(before) = pv(after) but how do you get the volume afterwards? The correct answer is 83.5kPa

It gives you the change in volume for one section, so that difference is the other part

Reply 290

randlemcmurphy

11

Original post by BrokenS0ulz

Could somebody help? I know you've got to do pv(before) = pv(after) but how do you get the volume afterwards? The correct answer is 83.5kPa

Surely the volume in the pump chamber cannot be greater than the air space....

EDIT: If the answer is 83.5 kPa, this means the volume (V2) has to be 4.1497...x10^-4 m^3, which I can't get with the data given....

(edited 8 years ago)

Reply 291

sagar448

4

Original post by BrokenS0ulz

Could somebody help? I know you've got to do pv(before) = pv(after) but how do you get the volume afterwards? The correct answer is 83.5kPa

6.5x10^-4 - 3.5x10^-4 = result

Then result x 99kPa and divide it by 3.5x10^-4

You get the answer. :smile:

Reply 292

randlemcmurphy

11

Original post by sagar448

6.5x10^-4 - 3.5x10^-4 = result

Then result x 99kPa and divide it by 3.5x10^-4

You get the answer. :smile:

You get 84.8 kPa, and intially the pressure is 99kPa, this is when the pump is just being pulled up, so surely the volume of the air space will still be 3.5x10^-4 m^3

(edited 8 years ago)

Reply 293

sagar448

4

Original post by randlemcmurphy

You get 84.8 kPa, and intially the pressure is 99kPa, this is when the pump is just being pulled up, so surely the volume of the air space will still be 3.5x10^-4 m^3

Alright let's go step by step. I took them away from each other because the 6.5 is the volume in the pump which means it has taken some from the bottle so in order to find what it's taken FROM the bottle you take them away. You multiply the result with 99kPa because that is the pressure IN the bottle only and then divide by the air space of the bottle because then that'll give you the pressure after one upstroke.

:biggrin:

Reply 294

BrokenS0ulz

3

Original post by sagar448

Can I know what paper this is please?

From some questions I found here http://www.physicsandmathstutor.com/physics-revision/ocr-unit-4/

Original post by randlemcmurphy

Surely the volume in the pump chamber cannot be greater than the air space....

EDIT: If the answer is 83.5 kPa, this means the volume (V2) has to be 4.1497...x10^-4 m^3, which I can't get with the data given....

Yeah the mark scheme said you had to get 4.15x10^-4 m^3, which I also don't understand. I think I'll just forget this question, doesn't make sense to me, no methods seem to give 83.5kPa, could be an error in the mark scheme.

Reply 295

sagar448

4

Ok so this is kind of bugging me for the past hour. A force acting parallel on an object can cause it to speed up and a force acting anti-parallel can cause it to slow down. A force acting perpendicular causes the object to change direction. What I don't understand is when a force is acting parallel on an object but it's DECREASING or is CONSTANT does this mean the object is slowing down? or speeding up?

A question where the force was acting parallel on the the object but it was constant yet the answer said the object was speeding up, why is that? :C

Reply 296

Makashima

16

hm, in SHM so frequency and time period is independent from the amplitude right? so if the time period doubles, then velocity falls by the factor of 4??? in order to remain same amplitude

im a bit confuse with the impacts by changes in the frequency, amplitude and time period on oscillation and the total energy and velocity

if frequency was to doubled or halves...velocity decrease???

XD idk wat im sayin, anyone care to explain? thanks

Reply 297

randlemcmurphy

11

Original post by Makashima

hm, in SHM so frequency and time period is independent from the amplitude right? so if the time period doubles, then velocity falls by the factor of 4??? in order to remain same amplitude

im a bit confuse with the impacts by changes in the frequency, amplitude and time period on oscillation and the total energy and velocity

if frequency was to doubled or halves...velocity decrease???

XD idk wat im sayin, anyone care to explain? thanks

If I was you I would just refer to the equations in the formula booklet and have a look at the relationship, and then work from there.

So for your first question, if the time period doubles ( and assuming every other variable stays constant), then we have v_max=(2pif)A and hence we can see that v_max is directly proportional to frequency, as f=1/T, it follows that if the time period is doubled, the velocity is halved.

Reply 298

sagar448

4

Original post by Makashima

hm, in SHM so frequency and time period is independent from the amplitude right? so if the time period doubles, then velocity falls by the factor of 4??? in order to remain same amplitude

im a bit confuse with the impacts by changes in the frequency, amplitude and time period on oscillation and the total energy and velocity

if frequency was to doubled or halves...velocity decrease???

XD idk wat im sayin, anyone care to explain? thanks

Ok so just think of it in terms of equations. Since you know Vmax = (2pi/T)A so if the period was to double then that would mean the velocity will be halved if amplitude remains constant. Or if period is doubled then the amplitude is doubled if Vmax remains constant, but period is not affected by amplitude, amplitude is affected by period. The only thing period is affected by is if it's a pendulum oscillation it is affected by length and the weight of the pendulum. If it's a spring SHM then it's affected by the mass and the spring constant.

KE energy is max when velocity is max and this is when the displacement is 0. But KE is zero when displacement is max because velocity is 0 and acceleration is max.

GPE is max when acceleration is max and displacement is max but KE is zero because velocity is zero.

Ask if you don't understand. :biggrin:

Reply 299

sagar448

4

Re-Posting because I really need this answered....

Ok so this is kind of bugging me for the past hour. A force acting parallel on an object can cause it to speed up and a force acting anti-parallel can cause it to slow down. A force acting perpendicular causes the object to change direction. What I don't understand is when a force is acting parallel on an object but it's DECREASING or is CONSTANT does this mean the object is slowing down? or speeding up?

A question where the force was acting parallel on the the object but it was constant yet the answer said the object was speeding up, why is that? :C

OCR Physics A G484 - The Newtonian World - 11th June 2015

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