Physics: Post your doubts here!

[Xylferion]

If we have to only estimate then we dont include least count?

Alright here's an example. I'll be using the ranges specified in the ms to prove that your uncertainty must be between 0.2<_<0.5

Let's say our times were: 6.72, 7.41, 8.23, 9.16, 7.34

First find the mean value.

That is 6.72, 7.41, 8.23, 9.16, 7.34 / 5 ======> 7.772

Now find the absolute uncertainty using ( Max - Min / No. of Readings ):

(9.16-6.72) / 5 = 0.488 s

0.488 s = well within the range of 0.2 <---> 0.5

Percentage uncertainty ---> 0.488/7.772 * 100 = 6.3 %

Hope that helped

 

[Midnight dream]

Alright here's an example. I'll be using the ranges specified in the ms to prove that your uncertainty must be between 0.2<_<0.5

Let's say our times were: 6.72, 6.75, 6.69, 6.74, 6.81

First find the mean value.

That is 6.72 + 6.75 + 6.69 + 6.74 + 6.81 / 5 ======> 33.71/5 = 6.742

Now find the absolute uncertainty using ( Max - Min / No. of Readings ):

(6.81 - 6.69) / 5 =0.024

0.024 / 6.742 * 100 = 0.356 ---> Well within the range of 0.2 <>0.5

When you're measuring something, you have to have several readings, so as to average them out and approach an accurate value.
You need to have times within 0.01 seconds of each other, of which you can then use to find an accurate timing for 10 oscillations.

Anyways, hope that helped

Thanks ,it did because i am very bad at it.

 

[Xylferion]

Thanks ,it did because i am very bad at it.

Sorry they wanted the absolute uncertainty not the percentage uncertainty to be between 0.2 and 0.5, so your times would have to be within 1 s of each of other. I changed the values, have a look at them.

 

[Midnight dream]

Sorry they wanted the absolute uncertainty not the percentage uncertainty to be between 0.2 and 0.5, so your times would have to be within 1 s of each of other. I changed the values, have a look at them.

But one value is 6.72 and the other is 9.16, isn't it a bit extreme.?

 

[Xylferion]

But one value is 6.72 and the other is 9.16, isn't it a bit extreme.?

All your releases on the swings won't be accurate, the whole point of releasing them at different times is to get an accurate value. Don't make it extreme as in 16 seconds for 10 swings and then 42 seconds for the other, between 6 and 9 is not extreme.

 

[Midnight dream]

All your releases on the swings won't be accurate, the whole point of releasing them at different times is to get an accurate value. Don't make it extreme as in 16 seconds for 10 swings and then 42 seconds for the other, between 6 and 9 is not extreme.

Okay. i will you my real readings when I performed the experiment.

I got 7.91, 7.94, 7.77. And when used the formula you suggested my absolut uncertainity was around o.o85.This means that my absolute uncertainity is wrong .

 

[The Sarcastic Retard]

Okay. i will you my real readings when I performed the experiment.

I got 7.91, 7.94, 7.77. And when used the formula you suggested my absolut uncertainity was around o.o85.This means that my absolute uncertainity is wrong .

I got 0.02

 

[The Sarcastic Retard]

Okay. i will you my real readings when I performed the experiment.

I got 7.91, 7.94, 7.77. And when used the formula you suggested my absolut uncertainity was around o.o85.This means that my absolute uncertainity is wrong .

It depends on your reading that u consider. BTW how did u end up with 0.085?

 

[Xylferion]

Okay. i will you my real readings when I performed the experiment.

I got 7.91, 7.94, 7.77. And when used the formula you suggested my absolut uncertainity was around o.o85.This means that my absolute uncertainity is wrong .

I can for one confirm that, your values need not belong to a single value like 7. The ms even says your times should be between 6 and 20. This has also been the case for all the stopwatch based experiments I've done. They give you a large range because you're not expected to release the object in question from the same exact position, so there'll be a difference in the times for each oscillation.

6 seconds refers to the people who pull the object back slightly. 20 being the people who pull it all the way. I'd recommend making sure your values +- 1 from each other.

 

[eliyeap]

Than

You cannot find the time period from a displacement/distance graph, but you can find the wavelength.

Firstly though, form any relations you can between the values that have changed. They tell you that waves move "forward" by 0.25T.

T = 1/ƒ
0.25T = 1/4ƒ

The above indicates that the frequency will quadruple, If the waves move forward by 0.25T.

v = ƒλ
λ = v/ƒ

If ƒ is quadrupled, you will take a quarter of lambda.

1/4λ = v/4ƒ

From the graph itself, you can see that the wavelength of 1 wave is 80 cm. If the waves move forward by 0.25T, they move forward by 0.25λ according to the above relations.

0.25λ = 80/4 = 20 cm

The waves will move forward by 20 cm. So your peaks move from 20 and 60 ---> 40 and 80, and so do all the other points on the wave.

View attachment 52260

Hope that made sense!

Thanks a bunch!!

 

[Midnight dream]

Keyword is "estimate" not "calculate", you need to have several readings.

Max - Min / Number of readings = absolute uncertainty

Find the mean value for t as well.

Absolute uncertainty / Mean value * 100 = Percentage uncertainty.

And isn't absolute uncertainity with
Max - Min / 2 =Absolute uncertainity ?

 

[Midnight dream]

I got 0.02

i did
7.91 - 7.77 / 2
isn't this way how we do it?

 

[The Sarcastic Retard]

i did
7.91 - 7.77 / 2
isn't this way how we do it?

I face lot of troubles in practical papers.
I generally do it like 0.2/(reading) * 100 for %UC :/ though idk if its correct or not acc to above post but i get marks in my exams.. Recently I scored 33/40

 

[TheJDOG]

9702 ON 2011 43 question no:- 3 c. ii and ii
Please answer with a clear explanation.
View attachment 52259 View attachment 52258

Hi, for 3c)i) the cube loses contact with the plate momentarily when the plate is at its maximum displacement from rest position.
This happens because the plate is oscillating, and at its maximum displacement, the plate will instantly reverse direction, which causes the cube to lose contact with the plate for a very short time. Remember that the cube is not glued to it, it's simply resting on it.
ii) we have to find this amplitude. We know accMax=Aw^2
And at this instant, the cube has an acceleration of 9.81, the acceleration due to gravity.
We're given in first part that f= 4.5 Hz, and w=2pie/T or 2pief
So substitute
9.81=A(2pie4.5)^2
A= 0.012 m or 1.2 x10^-2 m

 

[Xylferion]

And isn't absolute uncertainity with
Max - Min / 2 =Absolute uncertainity ?

you've got 3 readings... and those readings are not proper either. Like I've already said twice, have your values differ by 1.00 from each other.

 

[The Sarcastic Retard]

you've got 3 readings... and those readings are not proper either. Like I've already said twice, have your values differ by 1.00 from each other.

I face lot of troubles in practical papers.
I generally do it like 0.2/(reading) * 100 for %UC :/ though idk if its correct or not acc to above post but i get marks in my exams.. Recently I scored 33/40

 

[TheJDOG]

The number of the page we are on right now ain't cool man xD

 

[Midnight dream]

I face lot of troubles in practical papers.
I generally do it like 0.2/(reading) * 100 for %UC :/ though idk if its correct or not acc to above post but i get marks in my exams.. Recently I scored 33/40

and unluckily i am even worse than you..i got 29/40

how do you get the last part done ..Which wants limitations..?

 

[amal sharkawi]

Can anyone answer this question please???

 

[Mayarzawaydeh]

View attachment 52289
Can anyone answer this question please???

Answer is B.
Stress = force per unit area = F / A
Let the force (weight) and the (cross-sectional) area in the full-size crane be F and A respectively.
Stress in cable of full-size crane = F / A
As mentioned in the question, the model is one-tenth full-size in all linear dimensions.
For the model,
The load, which has a cube-shaped load, is in 3-dimension. So, for each of the 3 dimensions, the length of the model is reduced by a factor of 1/10. Therefore, the volume of the load is reduced by a factor of (1/10)3 = 1 / 1000.
Weight = mg and Mass = density x volume. Since the same material is used, the density is the same. So, the mass is proportional to the volume. A reduction in the volume causes the mass to be reduced by the same factor. The weight, which depends on the mass (g is constant), is also reduced by the same factor.
Force (weight) in model = F / 1000
Similarly, the cross-sectional area (which depends on (diameter)2) will be reduced by a factor of (1/10)2 = 1 / 100.
(Cross-sectional) Area in model = A / 100
Stress in cable of model crane = (F/1000) / (A/100) = 0.1 (F/A)
Ratio = (F/A) / 0.1(F/A) = 10