Physics: Post your doubts here!

[JD REBORN]

Q 10 and Q31 with proper explanation.

 

[elbeyon]

Can someone please explain to me how to do question 15, 18 and 20?
I kinda understand 18 my reasoning is that the pressure already in the flask is already hdg so in the increase would also be hdg giving 2hdg (correct me if i am wrong)

For question no. 15
Since the water will be equally divided in both the vessels after the tap is opened so vessel X will now have h/2 height of water and m/2 mass of water.
Due to which potential energy lost= (m/2)*(g)*(h/2).
So, final answer will be (mgh)/4. Hence no. B

 

[SkyPilotage]

Q 10 and Q31 with proper explanation.

Question 10:-
When sand is added, the mass increases. Therefore, its speed decreases since energy is nor created nor destroyed so k.e=0.5mv^2 has to be constant
When it is removed, some k.e is lost when it is dropeed out. Therefore, k.e decreases when its mass is less. Its speed will remain constant because it needs a force to give it a higher speed forward i.e Answer is B.

 

[SkyPilotage]

Q 10 and Q31 with proper explanation.

Question 31:-
The direction of the field lies represent the direction of the force acting on a POSITIVE test charge from its defiition.
Therefore, the force on a proton at any point is tangential to the field line.
If its an electron, then just flip the arrow 180 Degrees. So the answer is A

 

[SkyPilotage]

For question no. 15
Since the water will be equally divided in both the vessels after the tap is opened so vessel X will now have h/2 height of water and m/2 mass of water.
Due to which potential energy lost= (m/2)*(g)*(h/2).
So, final answer will be (mgh)/4. Hence no. B

Woops Sorry, I didnt see Number 15 in there Thanks for solving that

 

[elbeyon]

Woops Sorry, I didnt see Number 15 in there Thanks for solving that

WC

 

[JD REBORN]

Question 10:-
When sand is added, the mass increases. Therefore, its speed decreases since energy is nor created nor destroyed so k.e=0.5mv^2 has to be constant
When it is removed, some k.e is lost when it is dropeed out. Therefore, k.e decreases when its mass is less. Its speed will remain constant because it needs a force to give it a higher speed forward i.e Answer is B.

I dont understand why speed will remain constant when sand is removed???

 

[SkyPilotage]

I dont understand why speed will remain constant when sand is removed???

Because by Newtons First Law. An object will remain in constant motion with steady speed when no force is acting on it. There is no friction to slow it down as stated in the question "frictionless"
and there is no force pushing it either. Hope that cleared it up,

 

[JD REBORN]

Because by Newtons First Law. An object will remain in constant motion with steady speed when no force is acting on it. There is no friction to slow it down as stated in the question "frictionless"
and there is no force pushing it either. Hope that cleared it up,

OK but why wont we consider it when sand is added?Shouldnt the speed be constant then?

 

[SkyPilotage]

OK but why wont we consider it when sand is added?Shouldnt the speed be constant then?

Who said the speed wasnt constant? When sand is added, speed decreases but continues in constant speed.
As I said in my previous posts, k.e is conserved! so when we add still sand with no energy then speed has to decrease.
But when sand was dropped, it was moving with the car so it had some k.e so the car lost some k.e but continues to move in lower constant speed.

 

[SkyPilotage]

Who said the speed wasnt constant? When sand is added, speed decreases but continues in constant speed.
As I said in my previous posts, k.e is conserved! so when we add still sand with no energy then speed has to decrease.
But when sand was dropped, it was moving with the car so it had some k.e so the car lost some k.e but continues to move in lower constant speed.

*******Alright let me show it to you using conservation of momentum as the students were trying to investigate.
Take the mass of sand and car to be the same to make calculations easier
-At X
Initial Momentum = mcar x v + msand x 0 ----------> speed of sand is zero since it was dropped from rest
Final Momentum = 2m___ --->
Initial = Final Momentum so the new speed is v/2 ms^-1
-At Y
Initial Momentum = 2mv/2
Final Momentum = ( mssand x v/2 ) + (mcar x ___) ------> the speed of sand dropped is v/2 since it was moving with the car and dropped from it
Initial = Final Momentum ----> speed of car should be v/2
I hope now it clears your confusion

***** Now let me show it to you as conservation of kinetic energy
At X:-
Initial K.e = 1/2 mcar v^2 + 1/2 msand x 0
Final K.e = 1/2 (2m) ___^2 so speed must have decreased!
At Y:-
Initial K.e = 1/2 (2m) v^2
Final K.e = 1/2 msand (v)^2 + 1/2 mcar (v)^2

If momentum hasnt satisfied you, I hope kinetic energy does the job

 

[SkyPilotage]

Why do you think that is the case? Why does the frequency of the shock wave have to be much higher than the natural frequency of the seismometer?
Thanks in advance.

 

[JD REBORN]

*******Alright let me show it to you using conservation of momentum as the students were trying to investigate.
Take the mass of sand and car to be the same to make calculations easier
-At X
Initial Momentum = mcar x v + msand x 0 ----------> speed of sand is zero since it was dropped from rest
Final Momentum = 2m___ --->
Initial = Final Momentum so the new speed is v/2 ms^-1
-At Y
Initial Momentum = 2mv/2
Final Momentum = ( mssand x v/2 ) + (mcar x ___) ------> the speed of sand dropped is v/2 since it was moving with the car and dropped from it
Initial = Final Momentum ----> speed of car should be v/2
I hope now it clears your confusion

***** Now let me show it to you as conservation of kinetic energy
At X:-
Initial K.e = 1/2 mcar v^2 + 1/2 msand x 0
Final K.e = 1/2 (2m) ___^2 so speed must have decreased!
At Y:-
Initial K.e = 1/2 (2m) v^2
Final K.e = 1/2 msand (v)^2 + 1/2 mcar (v)^2

If momentum hasnt satisfied you, I hope kinetic energy does the job

U r a offcialy a genius,Thnx ur explanation was brilliant.Love u bro

 

[SkyPilotage]

U r a offcialy a genius,Thnx ur explanation was brilliant.Love u bro

Nah, Im no genius. Im just A2 infected
I am very much glad to help Feel free to post any more doubts and Ill try best to reply

 

[XPFMember]

Nah, Im no genius. Im just A2 infected
I am very much glad to help Feel free to post any more doubts and Ill try best to reply

Assalamoalaikum wr wb!

I've so many...

Can I flood them all at once?

P.S. do u take biology?

 

[XPFMember]

Aoa,
Q2:
In (i) you have found the amount of total hydrogen which was found out to be 415 mol H2
Now in (ii), they are asking about how many balloons may be filled by that 415 mol of H2
The problem is that they have given the amount of H2 in one balloon not at a pressure of 2.50 × 10^7 Pa but they have given for 1.85 × 10^5 Pa.
So, we need to find volume at pressure of gas at 1.85 × 10^5 Pa.
Volume of gas at 1.85 × 10^5 Pa = V2
The formula is p1V1 = p2V2
V2 = p1V1 / p2
V2 = (2.5 × 10^7× 4.00 × 10^4) / (1.85 × 10^5)
V2 = 5.41 × 10^6 cm3

Now, the gas that remains in the cylinder is :
V = (5.41 × 10^6) - ( 4.00 × 10^4) = 5.37 × 10^6 cm3
1 balloon contains (7.24 × 10^3) cm3 of H2
1 cm3 H2 fills 1 / (7.24 × 10^3) balloons
(5.37 × 10^6) cm3 fills (5.37 × 10^6) / (7.24 × 10^3) balloons
= 741 balloons

If we had solved this question by dividing no of moles of H2 in cylinder by no of moles of H2 in one balloon, the answer would be a little high value because we would have ignored the amount of gas that remains in the cylinder.

I hope u got it!

Waalaikumassalam wr wb!!

seriously...thanks a lotttttttttttttt jazakAllahu khairen!!!!!!

umm the last part doesnt makes sense to me, and i;m still kinda confused
like y do u find the gas remaining....i dont seem to get this one...

Q4
Attachment

JazakAllah...can u explain, how did u know which was the point of max displacement, and that of zero displacement?

Q6(a)
The field due to current in wire PQ will be circular at point Q
But they are asking about field due to WIRE XY AT POINT Q
I think you did not read the question correctly...


Q6(c)
Any values eg I = 10, d= 0.1
I = 11, d= 0.1
etc

oh ok...just to confirm, the force there is going to be the arrow we get from the field produced by XY? XY has a.c.w field...so Q will have field downwards cuz of that?

and mind telling me what exactly to write in Q:6c..the whole answer please...

I just wanted to point out,
Magnetic Flux is the number of magnetic field lines which is Phi.
Magnetic Flux Linkage is the product of the magnetic flux and the number of turns of the coil. which is N x Phi.
"Magnetic flux density is no. of field lines per unit area (Symbol:B )" like you said.
Thank you

aoa wr wb!!!!
jazakAlllah!!!!!!! that's exactly what i needed to know....!!!

Yes, i admit a little mistake of mine!

it's alright..no problem but thnx loads for the great help....


JazakAllah khair 2 both of u...i'll surely pray for u ppl!!

 

[SkyPilotage]

Assalamoalaikum wr wb!

I've so many...

Can I flood them all at once?

P.S. do u take biology?

Well you should ask all your questions, so flood them all and I will try my best to solve the ones I know

for Q4 smzimran, look at the piston . When it is at the middle, this means we can consider it as zero dispacement since the piston moves up and down.
So when the piston is at the middle, check the position which is at S.
When the piston is at the top or bottom, check position A which is Max displacement

Q6 a) yes the force is tangential to the field produced by XY

P.S:- No i dont take biology, sorry!

 

[XPFMember]

Well you should ask all your questions, so flood them all and I will try my best to solve the ones I know

for Q4 smzimran, look at the piston . When it is at the middle, this means we can consider it as zero dispacement since the piston moves up and down.
So when the piston is at the middle, check the position which is at S.
When the piston is at the top or bottom, check position A which is Max displacement

Q6 a) yes the force is tangential to the field produced by XY

P.S:- No i dont take biology, sorry!

oh ok...jazakAllah..i'll post them ... may Allah bless you..

i still dont get Q:4....how did u know that at those particular positions, that'd be the piston position...? :s

 

[Pasindu]

Re: Physics Help here! Stuck somewhere?? Ask here!

J04/P2/Q.1 (B), Q.2 (B)
J05/p1/Q.3,4,11,20,22,24,32,33,36
and thanks for the previous solutions

J04 P2

1 B - To get the resultant it is ( 8 cos 20 ) + ( 6 cos 20 ) = 13.2 error is 0.2

J 05 /P1

Q 3 - The velocity is 10 ms per second therefore when you calculate KE = 0.5 x 80 x 10 ² = Ans is 4000
which is answer B

Q 11 - This is a perfect inelastic collision only momentum is conserved therefore
60 - 40 = 2v which is 10 answer is B, - 40 because velocity is a vector and has direction 2v because
v +v because they are stick together they both are going in the same direction.

Q 20 - I just used some imaginary value like 5 and got the answer. X = 2.5 , Y = 5 , Z = 7.5.
hope you know the equations for strings in parallel and series

Q 32 - Charge = current x time therefore the current = 10, Q =10 x 1 second ,

10 / 1.6 x 10 power -19 = answer is 6.3 x 10 power 19. answer is C
[ is don't know why they have give area of the cross section ]


Hope this is helpful and my methods are correct

 

[Pasindu]

Re: Physics Help here! Stuck somewhere?? Ask here!

I'm having a little doubt in vector components. Consider this: A trolley of weight, W, 600 N is going down a slope, which is at an angle of 30* to the horizontal. They say that the component of W down the slope is = W sin 30*. How is that? The way I'm thinking it should be W/sin 30*. I'm obviously wrong. Can anyone explain why?
Thank you.

When resolving forces in any direction remember

resolving towards the angle is - COS angle
resolving away from the angle is - SIN angle

This makes every thing easy !
Hope this is helpful