Physics: Post your doubts here!

[Mr."S"]

Was the graph posiive or negative

 

[hancy pandey]

Was the graph posiive or negative

Negative

 

[Mr."S"]

My graph was positive ,the value of v was decreasing with increasing number of capacitors so v^-1 was increasing

 

[Mts.99]

yeah i got positive graph too

 

[Mr."S"]

yeah i got positive graph too

U just made my day man

 

[Mts.99]

U just made my day man

glad to!

 

[ba-lock-ey]

Positive graph. so did everyone else.

 

[HKHOHOHOHOHL]

IT was easy . what about you

How did you find the paper? Was it difficult?

 

[dwightedknight]

I am taking AS exams this summer. My AS batch felt it was really easy and A2 batch found it slightly difficult(this could be because they had a P5 clash so they studied more for P5 I guess).

I need some genius help for paper 22.

For the the graph area question (was it question 2). The question was to find the displacement of the object at t = 24s from A. Let the object be at C when t = 24s. Area above the graph was the distance AB. Area below the graph was the distance BC. AC = AB - BC. So I got something like 45 but not 72.

Regarding the upthrust question, as the rod is lifted from water into the air, does the power output of the motor increase or decrease? I wrote that it increases because pressure at top of the rod remains roughly the same (air pressure is roughly the same for small changes in height) while pressure at the bottom decreases rapidly. So pressure difference is lower and hence upthrust is lower. Which means tension is higher and power output is higher.

Energy dissipated in a battery is the energy lost due to internal resistance right?

 

[ba-lock-ey]

I need some genius help for paper 22.

For the the graph area question (was it question 2). The question was to find the displacement of the object at t = 24s from A. Let the object be at C when t = 24s. Area above the graph was the distance AB. Area below the graph was the distance BC. AC = AB - BC. So I got something like 45 but not 72.

Regarding the upthrust question, as the rod is lifted from water into the air, does the power output of the motor increase or decrease? I wrote that it increases because pressure at top of the rod remains roughly the same (air pressure is roughly the same for small changes in height) while pressure at the bottom decreases rapidly. So pressure difference is lower and hence upthrust is lower. Which means tension is higher and power output is higher.

Energy dissipated in a battery is the energy lost due to internal resistance right?

I like to remove everything from my head once the exam is done but I'd like to say that overall displacement is Larger area minus smaller area from what I know. Can't believe people are still discussing this paper.

 

[hellodjfos;s'ff]

I need some genius help for paper 22.

For the the graph area question (was it question 2). The question was to find the displacement of the object at t = 24s from A. Let the object be at C when t = 24s. Area above the graph was the distance AB. Area below the graph was the distance BC. AC = AB - BC. So I got something like 45 but not 72.

Regarding the upthrust question, as the rod is lifted from water into the air, does the power output of the motor increase or decrease? I wrote that it increases because pressure at top of the rod remains roughly the same (air pressure is roughly the same for small changes in height) while pressure at the bottom decreases rapidly. So pressure difference is lower and hence upthrust is lower. Which means tension is higher and power output is higher.

Energy dissipated in a battery is the energy lost due to internal resistance right?

Don't remember my answers from the paper. I think the output power was higher. And yes energy dissipated is work done against internal resistance.

 

[angryrider_56]

I need some genius help for paper 22.

For the the graph area question (was it question 2). The question was to find the displacement of the object at t = 24s from A. Let the object be at C when t = 24s. Area above the graph was the distance AB. Area below the graph was the distance BC. AC = AB - BC. So I got something like 45 but not 72.

Regarding the upthrust question, as the rod is lifted from water into the air, does the power output of the motor increase or decrease? I wrote that it increases because pressure at top of the rod remains roughly the same (air pressure is roughly the same for small changes in height) while pressure at the bottom decreases rapidly. So pressure difference is lower and hence upthrust is lower. Which means tension is higher and power output is higher.

Energy dissipated in a battery is the energy lost due to internal resistance right?

Q1) yaar ye toh theek hi hoga
2) 61°
b 0.75
Q3) 12 s
72m
0.5 acceleration
45 frictional force
6.7 angle
870 frequency
Frequency increases as speed decreases
Q4) difference between 2 points in phase
Points with max displacement
1888.88888
12 seconds
Q5) density nhi yaad
200 N
360 N i guess was the tension.
Power 7.2
Power increases as upthrust decreases and more force required. Speed constant.
Work is done by upthrust
Q6) charge is conserved
3750
3 V
1.3 resistance
Higher area
Lower resistivity
Power dissipated would be more as total resistance decreases and current increases.
Q7) circle electro and neutrino
Anti electron neutrino
Not isotopes as difference in proton number
Udd to uud

 

[angryrider_56]

Q1) yaar ye toh theek hi hoga
2) 61°
b 0.75
Q3) 12 s
72m
0.5 acceleration
45 frictional force
6.7 angle
870 frequency
Frequency increases as speed decreases
Q4) difference between 2 points in phase
Points with max displacement
1888.88888
12 seconds
Q5) density nhi yaad
200 N
360 N i guess was the tension.
Power 7.2
Power increases as upthrust decreases and more force required. Speed constant.
Work is done by upthrust
Q6) charge is conserved
3750
3 V
1.3 resistance
Higher area
Lower resistivity
Power dissipated would be more as total resistance decreases and current increases.
Q7) circle electro and neutrino
Anti electron neutrino
Not isotopes as difference in proton number
Udd to uud

these were the answers of one of my friends which he sent me via whatsapp......However there is an arguement over two questions, the power question in the circuit whether it was 6000W or 3750W and the time between two loudest points question which was either 6, 9, or 12....

 

[dwightedknight]

these were the answers of one of my friends which he sent me via whatsapp......However there is an arguement over two questions, the power question in the circuit whether it was 6000W or 3750W and the time between two loudest points question which was either 6, 9, or 12....

Time is 12s pretty sure. For Q3, I didn't get 72m. I got something like 45m.

 

[angryrider_56]

Time is 12s pretty sure. For Q3, I didn't get 72m. I got something like 45m.

look man for this question I also got 12seconds....but then the question said the time between two loudest areas, meaning the time between the two antinodes....the tube was close end meaning it had 2 antinodes and 2 nodes.....the entire thing was 12 seconds so I halved it and got 6seconds hopefully the paper releases soon so that teachers would solve it. and for the displacement, it was definitely 72 no doubt in that.

 

[Kanekii]

Need help with this.

 

[hellodjfos;s'ff]

look man for this question I also got 12seconds....but then the question said the time between two loudest areas, meaning the time between the two antinodes....the tube was close end meaning it had 2 antinodes and 2 nodes.....the entire thing was 12 seconds so I halved it and got 6seconds hopefully the paper releases soon so that teachers would solve it. and for the displacement, it was definitely 72 no doubt in that.

I don't know why you would divide 12 by 2... the answer was surely 12. If you divide the answer by 2, it gives you the time taken for the wave to go from an antinode to a node. I am certain it's 12s, because I contacted my teacher about that question.

 

[hellodjfos;s'ff]

I'll do the march 18 paper 1 in a couple of days, so I will reply to those questions in a couple of days bro.

 

[dwightedknight]

Need help with this.View attachment 63755 View attachment 63756 View attachment 63757 View attachment 63758

For 14, downwards force is mg/5 + mg*sin(10 degree). Upwards force is also mg/5 + mg*sin(10 degree). Power is upward force * speed.

For 21, answer is D. If the waves moves left to right, the rightmost region of the wave will be first experienced by the particle. Then the leftmost region. So answer is the mirror image of the given wave.

For 26, remember that distance from slit to screen and slit separation remains constant. If you know frequency of light, you can find the wavelength.

For 27, the path difference is 1.5*lambda.

 

[dwightedknight]

I'll do the march 18 paper 1 in a couple of days, so I will reply to those questions in a couple of days bro.

Are you applying to Cambridge or Imperial college? You can get in easily.