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K.E. gained = P.E. lost - Work done against friction
K.E. gained = 50 - 10
K.E. gained = 40 J
Final K.E. = Initial K.E. + K.E. gained
Final K.E. = 5 + 40
Final K.E. = 45 J
Answer: B
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K.E. gained = P.E. lost - Work done against friction
Which formula could be correct for the speed v of the ocean waves in terms of the density p of sea water, the acceleration of free fall g, the depth of the ocean h and the wavelength, lambda?
31)
I still don't get 31 and 35...why is the electron deflected out of the field? In 35 why is B wrong...I mean I got those set of values - power decreases initially and then increases according to I2R=P31)
electron would be attracted by the positive charge, so B and D are fundamentally wrong.
now since the electron is moving against the electric field, it will be deflected out of the field by the negative charge.
line of field represents the direction of the field .
so A
35)
answer is A because in tthe question it said there are 2 resistors so as one increases in resistance it requires more voltage so voltage across other one decreases
using formula V/E=R1/R1+R2 to test values you will get something similar to graph A
37)
the contact X would be adjusted around the resistor with 4 kΩ, so the max. n min. limits of p.d. across PQ wud be when its complete resistance is considered and when none is considered. at R = 4kΩ, the resistance across it is 20V
[ 4 /(4+1)] * 25 = 20V
when there is no 4kΩ resistor, p.d. across PQ becomes zero. therefore the answer is B.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Question 3, can someone please explain it for me?
q.35
Q.39
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Question 3, can someone please explain it for me?
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s10_qp_11.pdf
quest.16 ans is D ...
The Efficiency = (Useful Output energy)/(Total Input Energy).
Let's look at the changes in energy in the process.
Since the kinetic energy of the car doesn't change and remains at v, there is no change in the kinetic energy of the car.
Since friction does not do any work (we can assume this, and even otherwise it is not useful output so it doesn't count) there is no energy lost to internal energy.
However, potential energy does change - the car moves up a height of mgs * sin(α) (Since ΔP.E. = mgΔh, and sin(α) = Δh/s, Δh = s * sin(α) and ΔP.E. = mgs * sin(α)).
Therefore, the useful energy output = mgs * sin(α)
The work we put into the system is the only input to the system; no other influence is exerted by us, so the only input to the system is the work done by our force = F.
This work = F.s = Fs.
So, (Useful Output Energy)/(Total Input Energy) = mgs * sin(α)/Fs = mgsin(α)/F = D.
Hope this helped!
Good Luck for all your exams!
sagar65265 I just said na wo concept thing... I was talking about this types of quest onlyhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf
Guys please question 5 COMPLETE EXPLANATion !!!
Thank u
sagar65265 I just said na wo concept thing... I was talking about this types of quest only
I am sorry idk
idkwhats the answer of this question :/?
The answers Bwhats the answer of this question :/?
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