Physics: Post your doubts here!

[Thought blocker]

Pg 187
Practical circuits!

tysm!

 

[Mohammed salik]

tysm!

Welcome

 

[Suchal Riaz]

Mark scheme: http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_ms_22.pdf
Question paper: http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf
Question number 2(b) my attempt:

I am not sure i made it correct. Please tell me how to draw it in a better way. I couldn't find a better scale.
It's my first question i asked in As level even though i m a member for 3 years. So i hope someone will answer me.
Thought blocker asma tareen ZaqZainab

 

[Thought blocker]

Mark scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_ms_22.pdf
Question paper: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_22.pdf
Question number 2(b) my attempt:
View attachment 38924
I am not sure i made it correct. Please tell me how to draw it in a better way. I couldn't find a better scale.
It's my first question i asked in As level even though i m a member for 3 years. So i hope someone will answer me.
Thought blocker asma tareen ZaqZainab

While moving downward from A to the ground the velocity changes from 8.4 to 13 m/s. The time elapsed is .47 sec. There is a time delay of .02 sec during which the ball is in contact with the ground and the velocity drops from 13 m/s to 4.2 m/s this is shown by the sharp fall in the graph. As the ball begins to move upwards the velocity is now in the opposite direction or negative. You can calculate the time taken for the ball to come to rest by using v = u + at
0 = 4.2 - 9.8t
t =.42 sec.
So the ball moves for the further .42 sec before it comes to rest.

ZaqZainab

 

[Suchal Riaz]

View attachment 38933
While moving downward from A to the ground the velocity changes from 8.4 to 13 m/s. The time elapsed is .47 sec. There is a time delay of .02 sec during which the ball is in contact with the ground and the velocity drops from 13 m/s to 4.2 m/s this is shown by the sharp fall in the graph. As the ball begins to move upwards the velocity is now in the opposite direction or negative. You can calculate the time taken for the ball to come to rest by using v = u + at
0 = 4.2 - 9.8t
t =.42 sec.
So the ball moves for the further .42 sec before it comes to rest.

ZaqZainab

so the problem was the scale which was too small or there was some other errors? and if you chose 4 m/s as one block each small block will be of 4/5 m/s = 0.8m/s while the values are given correct to 0.1 m/s
the graph was small for the vales. as the values were given to one decimal place there was no way to make the graph accurate to one decimal place.

 

[Suchal Riaz]

thanks by the way

 

[Thought blocker]

so the problem was the scale which was too small or there was some other errors? and if you chose 4 m/s as one block each small block will be of 4/5 m/s = 0.8m/s while the values are given correct to 0.1 m/s
the graph was small for the vales. as the values were given to one decimal place there was no way to make the graph accurate to one decimal place.

So the graph I posted is incorrect ?

 

[Suchal Riaz]

So the graph I posted is incorrect ?

no it's not incorrect. i meant that there is no way to draw the graph to the full accuracy because the graph is small.

 

[Thought blocker]

no it's not incorrect. i meant that there is no way to draw the graph to the full accuracy because the graph is small.

OK

 

[mehria]

y not C ?? :/

 

[Thought blocker]

y not C ?? :/

Is answer B ?

 

[mehria]

Is answer B ?

yea

 

[Thought blocker]

yea

Thanks to mark P
R^2 = F(1)^2 + F(2)^2 + 2*F(1)*F(2)*cos(theta)
The above is the Law of Cosines used to find the resultant of two given forces with angle theta between them. Notice the "+" sign in the above equation, NOT a "-" like in the law of cosine applied to a triangle.

Back to the solution: R^2 = 10^2 + 10^2 + 2(10)(10)cos(120) = 10^2 + 10^2 + 2(10)(10)(-1/2) = 10^2
Final answer: Resultant = 10 N

 

[mehria]

Thanks to mark P
R^2 = F(1)^2 + F(2)^2 + 2*F(1)*F(2)*cos(theta)
The above is the Law of Cosines used to find the resultant of two given forces with angle theta between them. Notice the "+" sign in the above equation, NOT a "-" like in the law of cosine applied to a triangle.

Back to the solution: R^2 = 10^2 + 10^2 + 2(10)(10)cos(120) = 10^2 + 10^2 + 2(10)(10)(-1/2) = 10^2
Final answer: Resultant = 10 N

ohhh thanks..... instead of adding i was subtracting 2(10)(10)cos120
but y r we adding it?

 

[Thought blocker]

ohhh thanks..... instead of adding i was subtracting 2(10)(10)cos120
but y r we adding it?

It's not a triangle.

 

[mehria]

It's not a triangle.

then wat is it?? sorry m confuseddddd...........

 

[Thought blocker]

then wat is it?? sorry m confuseddddd...........

I mean its not a right angle triangle ZaqZainab M I CORRECT ?

 

[usama321]

It's not a triangle.

then wat is it?? sorry m confuseddddd...........

Well, i don't know how that worked out but the easier way to do it would be to resolve the forces. The horizontal forces are 10 to the right and 10sin30 to the left. The resultant is 5 N to the right. The vertical force is 10cos 30 = 8.66
Now just use pythagoras theorem to get the resultant.
resultant square = 8.66square + 5 square
resultant = 10

 

[ZaqZainab]

I mean its not a right angle triangle ZaqZainab M I CORRECT ?

Using cosine rule: c^2 = a^2 + b^2 +2abcosC
c^2 = 10^2 + 10^2 +2(100)cos 120
C = 10 N
What are you guys talking about

 

[usama321]

Using cosine rule: c^2 = a^2 + b^2 +2abcosC
c^2 = 10^2 + 10^2 +2(100)cos 120
C = 10 N
What are you guys talking about

the law of cosine is a square + b square -2ab cos angle c = c square