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Physics: Post your doubts here!

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sagar65265

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Snowysangel said:
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?
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This question needs the use of dimensional analysis - there is no theory in the syllabus that can be used to derive this, so we use dimensional analysis.

The units of speed are m/s = Meters/Second. These are the dimensions of Length and Time respectively, so we write their dimensions as LT^-1.
Density has units kg/m^3 = Kilograms/Meters^3. These are the dimensions of Mass and Length, so we write their dimension as ML^-3.
Acceleration (g) has units m/s^2 = Meters/Second^2. The dimension of this are written as LT^-2.
Depth and Wavelength both have units of Meters, so their dimensions are L.

On the left side, with speed, we have LT^-1 as the dimensions. Therefore, we must have a similar combination on the right side.

We note that density has the dimension M of mass; no other quantity given to us has this dimension, so if we introduce the density ρ on the right side, there will be no way to eliminate the dimension of mass on the right side (we want to eliminate mass, since there is no dimension M on the left-hand side in the speed of the wave).

So we can eliminate the need to use ρ.

Let's move on to the next elimination. We have acceleration g, depth and wavelength left.
The only one of these that has a dimension of time T (which we need on the left-hand side in speed) is acceleration. But this has dimensions of LT^-2. We want T^-1, as it is in speed. Therefore the only choice we have is to take the square root of acceleration g. So, we then have units on the right side of (L^0.5)*(T^-1).

We have to introduce (L^0.5) on the right side to ensure that we get LT^-1. Therefore, we can either multiply (g)^0.5 into (h)^0.5 or (λ)^0.5.
However, we can estimate that the speed depends on λ and say (gλ)^0.5, but there is little certainty in this. Are there any options provided with the question?
If so, you can carry out dimensional analysis on each one of those options, and you will obtain the right answer when the dimensions turn out to be LT^-1.

Hope this helped!
Good Luck for all your exams!
 
Thought blocker

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persandkesh
14)This question is a tricky one indeed. First of all you need to decide at which point will you take the moment. If you look at the diagram, you'll find that you need to avoid the upper force "F" since you will not know the perpendicular distance from this force at any point on the line except at the lowest point, and at the lowest point's moment will be FH. Moreover, if we take the lowest point and calculate the moment about it, it'd be: FH = WA which is not in the given answers. The only alternative to avoid the upper F is to take the moment about the highest point on the ladder, because the moment of the upper F would be zero, and the equation will have 3 components.

By calculating moments about the highest point on the ladder. F x H + W x a = W x 2a
Thus the answer is A.

16)We know that work is the force multiplied by distance along the direction of the force. The direction of the force is obviously the field lines. So the change in work ( the change in potential energy is FS ). To make it more clear, the diagram below explains that all points on the red line have the same potential energy and all points on the orange line has the same potential energy as well because there is no change in the distance along the direction of the force. This is quiet similar to the Gravitational potential energy. If we consider the blue line to be the earth surface and we put equal masses on the pink horizontal line which has the same vertical height from Earth. This means that all the equal masses have the same potential energy.
4-jpg.12447

Ok to know whether the charge lost or gained potential energy. Think of it as a gravitational force as well... If you move an object away from Earth ( There is an attractive force towards the Earth ) then it gains energy (mgh). The field is from the left to right meaning that it is negative to the right and the charge is positive. Thus think of the right side as the surface of the Earth, the object is becomes nearer to the Earth thus it loses energy.

In short :
for potential energy, vertical height is always taken, so distance would be s.
The field is towards the right side, so left is positive and the right negative, which means that the potential energy will decrease. energy = workdone, therefore, w.d. = F x s

24)
Young Modulus is a property of the material itself not it's length or cross section. Since the two wires are both made of Steel, then they have the same Young Modulus.
 
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Snowysangel said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf 31, 35 and 37
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31)
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.
 
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Thought blocker said:
31)
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.
Click to expand...
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=P
 
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Hadi Murtaza

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_Ahmad said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_13.pdf

Q35,39
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Q.39
Al (27)(13) + α(4)(2) -------> P(30)(15) + _____

The sum of atomic masses of left side is 31, while dat of right side is 30, so either a proton or neutron is missing
THe sum of atomic numbers of left side is 15, n dat of right side is same. Atomic number is the number of protons, n becuz these r both same on either side of da equation, da by product can only b a neutron.
full equation:
Al (27)(13) + α(4)(2) -------> P(30)(15) + N(1)(0)

hence Answer: D
 
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sagar65265

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RoyalPurple said:
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s10_qp_11.pdf

quest.16 ans is D ...:confused:
Click to expand...

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!
 
RoyalPurple

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sagar65265 said:
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!
Click to expand...

thank you so much :) i got it :)
all the best to you too :)
 
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