Physics: Post your doubts here!

[Amirtha]

how do you do this?

 

[fatima_jawed]

Yes p32

How did you find the paper? Was it difficult?

 

[A*****]

Someone plz tell that what is wrong with statement A

 

[Hamnah Zahoor]

Someone plz tell that what is wrong with statement A

I guess the main focus point here is the Elastic potential energy lost is the area ' X ' for the graph here is force-extension thus, it show lost energy.
But there was another question in june 2013 12 where it asked what the area between the graphs meant and there the correct option was amount of thermal energy dissipated in the rubber.
http://physics-ref.blogspot.com/2018/02/a-rubber-cord-hangs-from-rigid-support.html
I don't know, I am getting confused myself.

 

[A*****]

I guess the main focus point here is the Elastic potential energy lost is the area ' X ' for the graph here is force-extension thus, it show lost energy.
But there was another question in june 2013 12 where it asked what the area between the graphs meant and there the correct option was amount of thermal energy dissipated in the rubber.
http://physics-ref.blogspot.com/2018/02/a-rubber-cord-hangs-from-rigid-support.html
I don't know, I am getting confused myself.

Maybe then it is the energy 'released as heat' to the surroundings...lol idk

 

[Hamnah Zahoor]

Maybe then it is the energy 'released as heat' to the surroundings...lol idk

Maybe .

 

[hellodjfos;s'ff]

Someone plz tell that what is wrong with statement A

This question is really easy and simple. You can easily answer it by understanding that the area under a graph is the work done for a F-x graph. So, the total area under the graph to reach e is the combo of X and Y, thus the answer is B. What I am about to say may confuse you, so you may choose to ignore it. A being incorrect would require a detailed explanation with a diagram(too lazy for that)... but the confusing part about it is that when the spring is going from e to 0, the area under the graph is not Y... it is the area above X(look at it clearly and understand it). So when you shade the area under the graphs for both occasions(1st occasion being 0 to e, 2nd occasion being e to 0). You will end up finding that both of them are equal... hence none of the elastic potential energy is converted to heat energy which tells you that A is wrong.

 

[hellodjfos;s'ff]

View attachment 63702
how do you do this?

Bro, if you don't know how to do this, may god bless you for the upcoming exam. Really easy question...
Anyways: firstly, efficiency is the (output/input)*100. In this case, the input is the KE while the output GPE is the input. You find the KE: 1/2*500*2^2. Then all these values in the efficiency formula, the KE as your input, m(9.81)(15) which is your GPE as your output and 10 as your efficiency. You will get the final answer 0.68 kg(B) which is most definitely correct(if it's wrong, I would opt to kill myself).

 

[Hamnah Zahoor]

This question is really easy and simple. You can easily answer it by understanding that the area under a graph is the work done for a F-x graph. So, the total area under the graph to reach e is the combo of X and Y, thus the answer is B. What I am about to say may confuse you, so you may choose to ignore it. A being incorrect would require a detailed explanation with a diagram(too lazy for that)... but the confusing part about it is that when the spring is going from e to 0, the area under the graph is not Y... it is the area above X(look at it clearly and understand it). So when you shade the area under the graphs for both occasions(1st occasion being 0 to e, 2nd occasion being e to 0). You will end up finding that both of them are equal... hence none of the elastic potential energy is converted to heat energy which tells you that A is wrong.

Can you please draw the diagram? I am not getting the idea clearly.

 

[hellodjfos;s'ff]

There

 

[A*****]

There

So how do you explain the area bw the two graphs?

 

[Aparichit Bekti]

View attachment 63702
how do you do this?

Q. no. 18 Answer:

Kinetic Energy = (1/2)mv^2
= (1/2) x (500) x (2.0)^2
= 1000 J

Since the pump is 10% efficient so,
output energy = 10% of 1000 J
= 100 J
As this energy is used to lift the water, it means that the kinetic energy is converted to potential energy. So,

mgh=100
or m = 100 / (gh)
= 100 / (9.81 x 15)
= 100 / 147.15
= 0.67957866123003737682636765205573
~ 0.68

Therefore, option B is the correct answer.

 

[Krypton36]

Anything about Paper 34???

 

[ba-lock-ey]

Anything about Paper 34???

Focus on electricity. Mechanics is pretty straightforward but with electricity, you have to know all about connections.

 

[Sijanshrestha]

Focus on electricity. Mechanics is pretty straightforward but with electricity, you have to know all about connections.

Is electricity part hard?? What can be the limitations for mechanics experiment??

 

[Krypton36]

How do we show if a constant is within the limits of experimental accuracy or not? What method would be best for such questions??

 

[hellodjfos;s'ff]

So how do you explain the area bw the two graphs?

You can see both the areas are equal... So the elastic potential energy was same when the spring went from 0 to e and e to 0. So... none of the elastic energy is converted into heat... so A can't be correct since there isn't any energy converted into heat. You can explain this ans more than this.

 

[hellodjfos;s'ff]

How do we show if a constant is within the limits of experimental accuracy or not? What method would be best for such questions??

You use a specified criteria like: if the constants have a percentage difference of 20%, they are within limits of experimental accuracy. The criteria you specify is upto you.

 

[Krypton36]

You use a specified criteria like: if the constants have a percentage difference of 20%, they are within limits of experimental accuracy. The criteria you specify is upto you.

Okay! So if I get the percentage uncertainty of a measured value as 14% and my constants have %uncertainty of 25% then my value is proportional within the limits right??

I’ve been doing this! Just got confused...
Just out of curiousity, I wanted to know what you do at your side of the globe! Lol!

 

[anyone888888]

Guys any thing new about physics p34?