Physics: Post your doubts here!

[Thought blocker]

Uhm... Can someone provide me the solution to my question please?

I am so sorry I am late. Did you found out the answer or you want me to do?
If you still havent got it.
Here it is :¬
At node B: Ih = Ia + Ib
For Loop BCDE using KVL: 0 = 1.2*Ih - 12 + 0.01Ib
For Loop ABEF using KVL: 0 = 0.1Ia - 0.01Ib + 12 - 14

3 equations for 3 unknowns - the rest is math.

I would rarely be seen in threads. Coz now I am here and get busy in conversations of studies. Nothing confidential. So I am sorry if I cannot. And it would be better if you inbox me questions coz I prefer checking inbox first. I rarely click on alerts. As today. I totally forgot to reply you, but as I got alert, its here. Sorry again.

 

[Thought blocker]

Can someone explain why ans is B. plz detail each of the options

June 2011 P12 Q24

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_ms_12.pdf


This paper is not on the yearly solutions that were available in another post.

Thanks.

What are you not getting in this question?

 

[Llamas]

The intensity I of a sound at a point P is inversely proportional to the square of the distance x of P
from the source of the sound. That is
I ∝ 1/x^2

Air molecules at P, a distance r from S, oscillate with amplitude 8.0µm.
Point Q is situated a distance 2r from S.
What is the amplitude of oscillation of air molecules at Q?

Question 26, May 2008.
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s08_qp_1.pdf

 

[Llamas]

The intensity I of a sound at a point P is inversely proportional to the square of the distance x of P
from the source of the sound. That is
I ∝ 1/x^2
Air molecules at P, a distance r from S, oscillate with amplitude 8.0µm.
Point Q is situated a distance 2r from S.
What is the amplitude of oscillation of air molecules at Q?

May 2008, question 26.

 

[Thought blocker]

actually, the only formula that come to me is F = kx. how to apply it here? I can't understand any of the answers provided, the calculation involved???

As well as telling me why B is correct, could you tell me why the others are wrong???


Thanks

Probability A and B. I think you obviously know why the other two are rejected.
B because, no matter how many springs you add the total compression will be unchanged as each of the springs have the same spring constant.
A is not the answer coz, i think that should make it harder because each layer does not have the same force applied to it.

 

[Thought blocker]

The intensity I of a sound at a point P is inversely proportional to the square of the distance x of P
from the source of the sound. That is
I ∝ 1/x^2

Air molecules at P, a distance r from S, oscillate with amplitude 8.0µm.
Point Q is situated a distance 2r from S.
What is the amplitude of oscillation of air molecules at Q?

Question 26, May 2008.
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s08_qp_1.pdf

The intensity I of a sound at a point P is inversely proportional to the square of the distance x of P
from the source of the sound. That is
I ∝ 1/x^2
Air molecules at P, a distance r from S, oscillate with amplitude 8.0µm.
Point Q is situated a distance 2r from S.
What is the amplitude of oscillation of air molecules at Q?

May 2008, question 26.

26)
I ¹/α x² so, doubling x means I would be divided by 4.
I α A² so, √(8²/4) = √16 = 4.0 μm

This can also be done like this :¬

I1/ I2 = A1² / A2²
1/(1/4) = 8² /x²
4/8² = 1/x²
x²= 16
x= √16
x= 4.0 μm

 

[Llamas]

Which operation involves the greatest mean power?
A a car moving against a resistive force of 0.4kN at a constant speed of 20ms–1
B a crane lifting a weight of 3kN at a speed of 2ms–1
C a crane lifting a weight of 5kN at a speed of 1ms–1
D a weight being pulled across a horizontal surface at a speed of 6 m s–1 against a frictional
force of 1.5kN

 

[Thought blocker]

Which operation involves the greatest mean power?
A a car moving against a resistive force of 0.4kN at a constant speed of 20ms–1
B a crane lifting a weight of 3kN at a speed of 2ms–1
C a crane lifting a weight of 5kN at a speed of 1ms–1
D a weight being pulled across a horizontal surface at a speed of 6 m s–1 against a frictional
force of 1.5kN

Power is same as force times velocity.
A) 8Kpa
B) 6Kpa
C) 5Kpa
D) 9Kpa = Answer.

An electric power cable consists of six copper wires c surrounding a steel core s.

1.0 km of one of the copper wires has a resistance of 10 Ω and 1.0 km of the steel core has a
resistance of 100Ω.
What is the approximate resistance of a 1.0km length of the power cable?

Question 32, October 2008.

32)
All the wires are connected in parallel.
So 1/R = (1/10)X6 + 1/100
Therefore, 1/R = 61/100
R = 100/61 = 1.6 ohms.

 

[Llamas]

A particle has a charge of 4.8 × 10–19 C. The particle remains at rest between a pair of horizontal,
parallel plates having a separation of 15 mm. The potential difference between the plates is
660V.
What is the weight of the particle?
A 2.1 × 10−14N
B 2.1 × 10−15N
C 2.1 × 10−17N
D 1.1 × 10−23N

Question 30, October 2008

 

[Llamas]

Thank you!

Power is same as force times velocity.
A) 8Kpa
B) 6Kpa
C) 5Kpa
D) 9Kpa = Answer.

32)
All the wires are connected in parallel.
So 1/R = (1/10)X6 + 1/100
Therefore, 1/R = 61/100
R = 100/61 = 1.6 ohms.

 

[Thought blocker]

A particle has a charge of 4.8 × 10–19 C. The particle remains at rest between a pair of horizontal,
parallel plates having a separation of 15 mm. The potential difference between the plates is
660V.
What is the weight of the particle?
A 2.1 × 10−14N
B 2.1 × 10−15N
C 2.1 × 10−17N
D 1.1 × 10−23N

Question 30, October 2008

This question is based on derivation of equations.
We know that F = mg = W
We also know that W = EQ Where E = v/d
So we got an expression related to the question we have here that is W = v/d * Q
Substituting the values you have in question final answer will be on your calculator display as "2.1 x 10^-14" that is A.

 

[Llamas]

This question is based on derivation of equations.
We know that F = mg = W
We also know that W = EQ Where E = v/d
So we got an expression related to the question we have here that is W = v/d * Q
Substituting the values you have in question final answer will be on your calculator display as "2.1 x 10^-14" that is A.

Thank you again!

 

[ZaqZainab]

Sorry, but I`m still not getting something. Maybe I'm having a confusion but I don't know what it is.

I'm thinking this way.

F = ke. e = F/k. For smaller k, shouldn't e be greater. So, why is C wrong.

Also, Young modulus E = stress / strain = (F/A) / (e/L) = FL / Ae. So, e = FL / AE. If E is smaller, e would be bigger. So, why D wrong?

Also, can you show me mathematically (using the formulae) how B is the answer. And what am I having wrong. The above logic shows that both C and D are good?

For smaller k, e WILL be greater but if you look at the question it says 'Which change will not have the desired effect?'
and e is extension which in this case is the compression. you haven't seen the not in the question. the e shouldn't be greater. If it is, it does not support the question they asked. both C and D are bad in that case.

 

[Zepudee]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w03_qp_4.pdf

Can anyone give me the answer of 1(a) and 4(c)

 

[Zepudee]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_4.pdf

And this hehe, Q5 (b) and Q7 (a) thank you so muchhhh

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_4.pdf

Can anyone give me the answer of 1(a) and 4(c)

Hello.
For 1a)
You need a textbook and a fresh mind to understand whats written in your book.
4c)
Capacitor linearise the voltage signal. the signal given in the ques is already linearised once. if it is linearised again with the second capacitor the signal's ripple is further reduced but the time period and freq remains the same

 

[The Godfather]

Zepudee
5B : This I saw somewhere, IDK but I had doubt too. Anyways here he showed this :¬

7A : At t=0, the mass of iron will be zero grams. After 2.6 hours, mass of iron will be 0.7 micrograms. At t=5.2 h, mass of iron will be (0.7+0.35) micrograms. When m grams of manganese decays, it produces n grams of iron because both of them have same Ar.

 

[Zepudee]

Hello.
For 1a)
You need a textbook and a fresh mind to understand whats written in your book.
4c)
Capacitor linearise the voltage signal. the signal given in the ques is already linearised once. if it is linearised again with the second capacitor the signal's ripple is further reduced but the time period and freq remains the same

for the diagram q4(c) im not sure if my diagram is right or wrong, just needed some clarification thanks thought blocker

 

[Zepudee]

Zepudee
5B : This I saw somewhere, IDK but I had doubt too. Anyways here he showed this :¬

7A : At t=0, the mass of iron will be zero grams. After 2.6 hours, mass of iron will be 0.7 micrograms. At t=5.2 h, mass of iron will be (0.7+0.35) micrograms. When m grams of manganese decays, it produces n grams of iron because both of them have same Ar.

Thanks ! :3

 

[Thought blocker]

for the diagram q4(c) im not sure if my diagram is right or wrong, just needed some clarification thanks thought blocker

Dikhao!