# Physics: Post your doubts here!

Discussion in 'International A And AS Level' started by XPFMember, Sep 29, 2011.

1. ### Raheel327

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Do we always use the resultant force only to calculate power, in the formula p=fv?
And, in the first question the car later accelerates. Wouldnt that change its speed so it's no longer 10?

3. ### Hamnah Zahoor

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When the train is at its' maximum speed the acceleration is to be considered zero. Which means the resultant force acting on the train is zero. In order to calculate the force acting:
P = Fv
2*10^6 = F (40)
F = 50,000 N
Forward force is 50,000 and the resistive force is also 50,000 N which makes the total resultant force equal to zero.
It is said in the question statement that the resistive force is constant at all speeds thus at 10 ms^-1 Rf= 50,000 N
P = Fv
2*10^6 = F(10)
F = 2*10^5
Resultant force equals to forward force minus resistive force
(2*10^5)-(50,000)
1.5*10^5 N

4. ### Hamnah Zahoor

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No the forward force is used and also in the question I have used F = 400 N not the resultant which is 400-160= 240 N

At the time when acceleration begins the speed is 10 ms^-1 .

5. ### Raheel327

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How can the acceleration be 0.20ms^-2 when the speed is still 10ms-1?

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6. ### Ibrahim Yaqoob

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I just gave my phy paper2 in oct/nov session . It had a non rectilinear motion question in which we had to calculate max vertical displacement. I made a blunder and took the value of accelaration positive (9.81). It was of three marks i guess. the next question asked KE/GPE. so it subsequently used the height calculated in previous question . I took the other things completely fine.this was of 4 marks. Can anyone guess how much marks will I be losing?

7. ### Hamnah Zahoor

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Raheel327
The philosophy of reaching a maximum, any maximum, means that philosophically one is climbing and climbing higher and higher to reach the maximum peak, as long as there is a higher peak one keeps on climbing the slope. Once one reaches the top, the top must be a horizontal plane, even if it is a sharp peak, it must be a plane otherwise one has not reached the maximum.
Then one goes down the peak.

So, it is the same with velocity, if the velocity reached is a maximum, one can only go down the slopes around it, after retaining that maximum velocity for some time.

When acceleration is equal to zero than, F =ma a= 0 than resultant force is equal to zero.

Last edited: Nov 5, 2018
8. ### Hamnah Zahoor

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As I have shown in the graph above although the acceleration begins to be 0.20 but the initial will remain 10 ms^-1 . the point where it begins it will change after wards.

9. ### kfemknwnsx

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has anyone got model diagrams for q1 of p5?

10. ### Lemniscates

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Yes.

11. ### Maheen S

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Could anyone help me with q36 november 2017 variant 11

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12. ### Hamnah Zahoor

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13. ### Wâlèé Atèéq

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I missed my lecture of electromagnetism, is there any helpful detailed notes for this topic? Thanks

14. ### Mr."S"

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Is urdu,islamiyat and pak studies required for getting admission at nust
Because I have e-mailed those guys like 10 times and they don't give a proper answer.Also i have done my olevels from uae and doing a levels in pak

15. ### Thought blocker

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16. ### kfemknwnsx

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You will require those 3 subjects. Check the IBCC website though since they're the ones that create the Equivalency Form(you may get some leeway if you are a dual national).

17. ### Alis alish

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how momentum = impulse?

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18. ### Hamnah Zahoor

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momentum= mass*velocity
kg(ms^-1)
impulse = force*time
Newton*second
(kgms^-2) s
kgms^-1

units are same thus, both are equal.

19. ### Navin K

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Moemntum is not equal to Impulse

Change in Momentum = Impulse

Just like: Transfer of Energy = Work

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20. ### Clark20

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Need help with this.

#### Attached Files:

• ###### Screenshot_20190211-185855__01.jpg
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21. ### Hamnah Zahoor

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Mg - T = Ma -------(1)
T - mg = ma ------(2)

Make T subject in the second equation
T = ma + mg
substitute in first equation
Mg - (ma+mg) = Ma
Mg - ma -mg = Ma
Mg - mg = Ma + ma
(M - m)g = (M + m)a
a=(M-m)g/(M+m)