Physics: Post your doubts here!

[NIM]

need help
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w05_qp_1.pdf
5, 12, 13, 16, 23, 24, 25, 28, 29

 

[Thought blocker]

Stationary waves summery :​

Main concepts:

• do not appear to propagate
• are produced by the interference of two waves traveling in opposite directions with the same frequency and amplitude
• Positions on a standing wave:
  • node: a point where the amplitude is zero or a minimum
    (always form at fixed ends)
  • antinode: a point where the amplitude is a maximum
    (always form at free ends)

Another important point to note:

The distance between a node and an antinode is always 1/4 of the wavelength.
The distance between 2 adjacent nodes/antinodes is 1/2 of the wavelength.

Not in detail, but these are the important points that you should be knowing for MCQs

 

[Asad Moosvi]

You subtract the smaller value of k from the bigger one, then divide them by the bigger one, although I'm not entirely sure which value to divide. If its less than the percentage uncertainty its supported if its not then simply write its not supported rather than it has an inverse relation.

Which percentage uncertainty? Sometimes we calculate a percentage uncertainty earlier which isn't even part of the equation.

 

[Thought blocker]

i'm sooo sooo sorry i pasted the wrong link x_x i it was supposed to be w12 not s12 ugh!
For s12 my doubts are actually 7,10,17,28,30 ( and also 29,34,37 but you already answered them lol) soo sorry!

w12_12
29)
View attachment 44719
As you can see from the figure, there will be total difference b/w nodes = 3λ / 2 = L
Now we know the velocity, we can calculate λ so f = v / λ
λ = 0.2 m
f = 1650 Hz

34)
Let's go choice by choice shall we...:
B: It's said both are made from same material. So resistivity is same.
Reject B

A:
R = (resistivity * L)/A
Both are same material so we ignore resistivity for now.
So we have: R = L/A
We have to compare their cross-section areas.
A = L/R

For X:
When R = 20, L = 0.6
A = .6/20 = 0.03A

For Y:
When R = 10,
L = 0.6,
A = 0.6/10 = 0.06A

Compare 'em.
2X = Y
Is this what dear (NOT!) A is suggesting?
No. It says the opposite in fact.
So A's a no-no.

C.
Take both lengths equal. I took them 0.6m
For X, R = 20
For Y, R = 10

P = VI
SInce they're in series, I will be same for both.

For X,
P = 20I
For Y,
P = 10I

Compare 'em away.
Power of Y *2 = Power of X
WHich is exactly what C says.
So C's our choice.

But to be clear, let
s have a look at D

In parallel, current in one of the two branches, in equal to ratio of R/total of OTHER branch.

For X, (R=20) and R = 10 for Y. Total = 30
For current in X, therefore, it's 10/30 ie 1/3 I

For Y,
20/30 I
or 2/3 I

This is clearly NOT what's in D.
So D goes down.

36)
Here first we'll find the power at two points of current provided, and as its not a fixed current we would take the mean of power we get.
So it follows as :¬
Power at 2 ampere = 4 * 100 = 400W
Power at -1 ampere = -1 * 100 = -100W
Now as I said take mean of these powers : 500 / 2 = 250W

37)
Its not D as no current will pass if both switch are open. Then its not even A as both are closed so current will be I. In B S2 is open and S1 is closed so current just flows in series so current is still I. In C current flows through parallel circuit as S1 is open and S2 closed and current is not equal in parallel so it will not be I

38)
remember when two resistor are in series, more voltage drop is at resistor with more value, so their a flashing light then R = 5 mega ohm when dark, then 5M is very large than 1 K of other resistor, then almost all voltage drop at 5M, hope you get this, it is a very simple concept.

39)
Whats the big deal here ?
You know that nucleon number in helium is 4 (helium = alpha particle)
Then from original substance 100 alpha get removed every second, so in alpha, 4 * 100 = 400 nucleon each second.

s12_12
7)
Its A, because according to velocity time graph, we can see that its negative after half the time. That means that guy has changed his direction. this describes The scenario that a man run stratis gradually increasing hi speed and then after certain reducing speed and an he changes is direction(moves backward) with same speed pattern.
Now considering this troy with displacement, that guy reaches the farthest point and then comes back i.e max displacement after halftime.

10)
Keep this thing in mind when you are asked for projectile thingy :

• horizontal component of velocity = constant
• vertical component of acceleration constant = constant
• At top most, vertical velocity is zero

Now look it practically,
you throw a ball, when it reaches max height (top) it stops for a while
KE cannot be zero
KE = 0.5mv^2
v = root(square(vx) + square(vy))
vy = 0
but vx = constant
so v is nonzeror and hence KE canot be zero
similarly momentum canot be zero
P=mv

17)
Base has h = 0, i.e is block 1 = zero * mgh
Now we have other 3 blocks.
block 2 = at height h so m*g*h = mgh
block 3 = at height 2h = 2mgh
block 4 = height 3h = 3mgh
Sum all this P.E = 6mgh

28)
View attachment 44722

30)
Given
λ/2=33cm
so λ is 66cm or 0.66m
f=330/0.66=500
T=1/500=2ms
so it means wave should be completed in 4 blockss as the time base is 0.50ms/cm
the only one is B

 

[Kamihus]

Which percentage uncertainty? Sometimes we calculate a percentage uncertainty earlier which isn't even part of the equation.

The one you calculate in the earlier part like the one in w in the paper you gave above. That is used to calculate A which is in the equation. Alternatively you can just say its within a sensible limit like 10% so its supported.

 

[Thought blocker]

need help
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
5, 12, 13, 16, 23, 24, 25, 28, 29

5)
Vo = (5*2*1) = 10 (principle volume value)
Vu = (0.01/5) + (0.01/2) + (0.01/1) * 10 = 0.17

So volume = 10 +/- 0.17
And mass = 25 +/- 0.1

Uncertainty in density = (0.17/10) + (0.1/25) * 2.5 = 0.05

12)
Clockwise = 20 * 0.4 = 8 Nm
Anti-clockwise = 10 * 0.6 + 100 * 0.1 = 16 Nm (don't forget the weight of the beam!)

Therefore we need a clockwise moment of 8 more Nm

(20 * x) = 8, x = 0.4m from the pivot, so D

13)
Resultant torque = 45 N and resultant force = 60 N to the right

16)
Tension = mg sin θ = 10^3 sin 30 = 500 N (note: weight was already given, so need to multiply by 9.81)

Work = force * distance moved in direction of force = 500 * 5 = 2500 J

23)
In A and C, the amplitude is marked incorrectly. In D, λ is actually the time period.

24)
I α a^2 and I α f^2.

Rather than doing all the math to do this, compare the amplitude and frequency of the waves and use the formula to figure out this stuff:

If Q's amplitude is twice as much, the intensity will be four times as much.
If Q's frequency is half that of P, the intensity will be one-fourth.

Net change = 0, so the intensity remains the same.

25)
λ in water = 1500/150 = 10m and λ in air = 300/150 = 2m


28)
x=lambda*D/a
(Lambda/2*D)/2a
x=1/4*Lambda*D/a
so answer should be.. 0.75. A

29)
Since the adjacent 1st orders are 60 apart, that means one 1st order from the undeviated beam is 30 apart.

1.15 * 10^-6 sin 30 = 1 * λ
λ = 575 nm

 

[ZaqZainab]

need help
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w05_qp_1.pdf
5, 12, 13, 16, 23, 24, 25, 28, 29

Q5
Volume= (5*2*1) = 10
uncertainty V= (0.01/5) + (0.01/2) +( (0.01/1) * 10) = 0.17
So volume = 10 +- 0.17
And mass = 25 +- 0.1
using formula uncertainty of density/density= (uncertinity of volume/volume) +(uncertainty of mass/mass)
Uncertainty in density = (0.17/10) + (0.1/25) * 2.5 = 0.05

Q12
the 100N will act at the middle of the beam
that is at 0.5 m
and from pivot it is 0.1 m
so 100*0.1 +o.6*10 is the anticlockwise force =16
0.4*20 is the clock wise force =8
for the beam to be uniform anticlockwise=clockwise
but here
anticlockwise is more and clockwise less
so the 20 will be added in such a way that it acts clockwise
16=(8)+(20*x)
x=0.4
so 0.4 from the pivot D

Q13
torque=45 Non zero
and force=30+30=60 non zero

Q16
weight=mg=10^3
force = mg sin tetha
=10^3 sin 30 = 500 N
Work = force * distance moved in direction of force
=500 * 5 = 2500 J

Q23 amplitude is the displacement from MEAN position A and C are wrong
D is wrong because the wave length is marked wrong
that is the time period

Q24 information in the question
I proportional to a^2 and I proportional to f^2
I=a^2 I=f^2
For P
Io=x0^2
Io=(1/t0)^2
For Q
I1=(2x0)^2=4xo^2 =Io*4
I1=(1/2t0)^2=1/4to^2 =I0/4
finally Io=I1

Q25
v=f*lamda
In WATER
1500/150=lamda=10
In AIR
300/150=2

Q28
x=lamda*D/a
3*10^-3 = 700*10-9*D/a
a/D=7/30000
x=350*10^-9*D/2a
x=350*10^-9/((7/3000)*2)

Q29 d*sin(tetha)=n*lamda
tetha will be 60/2 as it is suppose to be the angle made by the normal to the granting
1.15 * 10^-6 sin 30 = 1 * λ
λ = 575 nm

 

[NIM]

help plzz
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf
6, 11, 17, 25

 

[crazytaylorfanXD]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w13_qp_13.pdf
someone please help me solve these question( QS.31,32.36.37,38)


http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_12.pdf
QS 9,19,

 

[sagar65265]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_11.pdf

I need your help in few (or many XP) questions. They are Q5,9,11,14 and 30.

Q5)

Okay, so what we need to find out is the percentage uncertainty in the value of the volume of the sphere with a radius measurement of r = 5.00 mm ± 0.01 mm.

The formula for volume of a sphere as a function of radius is 4πr³/3. Therefore, we are multiplying r into r into r to get r³, and this is the only value in the equation that has any uncertainty - there is no uncertainty in 4/3 and there is no uncertainty in π, so r is the only quantity we need to concern ourselves with.

The rule for uncertainties when any number of quantities are multiplied goes as follows:
"When one multiplies or divides several measurements together, one can often determine the fractional (or percentage) uncertainty in the final result simply by adding the uncertainties in the several quantities."

Let's do this by example. We are multiplying r by itself three times (to get r³) and so to get the percentage uncertainty in the final result, we resort to "adding the uncertainties in he several quantities". In other words, we add the (percentage uncertainty in r) to the (percentage uncertainty in r) to the
(percentage uncertainty in r).

Basically we multiply the percentage uncertainty in r by 3.
The percentage uncertainty in r = (0.01 mm/5.00 mm) * 100 = 0.2 %.
Multiplying this by 3, we get the uncertainty in the volume of the sphere with this radius to be 0.2 * 3 = 0.6% = C.

(more information here: http://spiff.rit.edu/classes/phys273/uncert/uncert.html)

Q9)

The mass of an object is an intrinsic property, i.e. it does not depend on any calculation or any phenomenon (at least at this level - yes, mass can be found from density but mass is not defined using density). So, it is one of the most basic quantities in nature.

Due to the above fact, we can say for sure that mass does not rely on any calculation to reveal it's nature - it is not dependent on momentum, it is not dependent on force or acceleration, it does not depend on weight - all those things are related using mass or defined using mass.

Therefore we can rule out A. Furthermore, looking at the facts, we can eliminate B and C as well (the definition of weight is that it is the gravitational force exerted on an object - on Earth, the weight of an object is practically equal to the gravitational force exerted on that object by the Earth itself). C is similar to B, but more specific, and it is still wrong - an object on Mars or even at a point infinity (where no gravitational influences exist) will have the same mass as it does on Earth. It's weight is the differing factor here.

Therefore, D is the most natural statement from nature - mass is that which opposes acceleration. Simple as that. No need for further definitions, this is the meaning of mass. Similarly, since the question says "weight of a body on Earth", D makes the most sense, because the weight of an object on Earth is equal to the magnitude of the pull the Earth exerts on it. So our final answer should be D.

Q11)

In 1 second, 5.0 × 10⁴ α-particles collide into the lead sample. Let's take these 50,000 particles as our system and work from there.

Imagine the collisions happening; initially, the particles each have a certain amount of momentum, by virtue of their mass and velocity. They keep moving closer and closer to the lead sample over time, and eventually they smash into the lead sheet. Their momentum seems to vanish since they come to a stop, but something does happen here, something very important - there is a force on our system (the particles) from the lead sheet, and that force is large enough to bring all these particles to a halt in one second.

So how large is this force that acts on our system? By Newton's Second Law, this force is equal to the rate of change of momentum of our system (again, the particles).
So they go from moving at a high speed to coming to a stop. Their speed changes, and that is why their momentum changes!

Let's find out this change in momentum. In 1 second, 50,000 α particles, each with a mass of 6.6 × 10⁻²⁷ kg and a speed of 1.5 × 10⁷ ms⁻¹ collide with the lead sheet over an area of 1.0 cm². After the collision, their velocity becomes zero. Therefore, we can write that the initial momentum is:

Initial total momentum (of all particles) = 50,000 particles * (6.6 × 10⁻²⁷ kg)(1.5 × 10⁷ ms⁻¹) = 4.95 * 10⁻¹⁵ kg ms⁻¹.
Final total momentum (of all particles) = 50,000 particles * (6.6 × 10⁻²⁷ kg)(0 ms⁻¹) = 0 kg ms⁻¹
Change in momentum (of system) = 0 - 4.95 * 10⁻¹⁵ = -4.95 * 10⁻¹⁵ kg ms⁻¹

This change in momentum takes 1 full second to occur, therefore the force on the system is

Force = (Change in Momentum)/(Time taken for change in momentum) = (Change in momentum)/1 = Change in momentum/second = -4.95 * 10⁻¹⁵ kg ms⁻².

By Newton's Third Law, this is the same force exerted by the system (all the particles) on the lead. Therefore, the pressure on the lead per second is equal to

Pressure = (Modulus of Force)/(Area) = (4.95 * 10⁻¹⁵ kg ms⁻²)/(1.0/10000) = 4.95 * 10⁻¹¹ kg m⁻¹s⁻² = 5.00 * 10⁻¹¹ Pascals = C.
Q14)

In the first case, there are two forces of tension on the arrow - the bowstring pulls the arrow along both sides, so the force component in the vertical direction is 0.
In the horizontal direction, the component of the tension force that pulls the arrow forward (to the left) due to the upper half of the string is

(100 Newtons) * cos(65) = 42.26 Newtons.

The lower half of the string exerts the same force in the forward direction, so the total force force in the first situation is

42.26 + 42.26 = 84.52 Newtons.

Onto the second situation. The bowstring has been drawn even farther back, and the tension in the string is 120 Newtons. That means that the upper part of the bowstring pulls the arrow with a force of 120 Newtons, and so the lower half. The components of this pull in the vertical directions cancel out (since both angles are the same) and the component in the horizontal direction of one half of the drawstring becomes

(120 Newtons) * cos(55) = 68.82 Newtons

Again, the lower half of the string exerts the same force in the horizontal (forward) direction, so the total force in this situation in the forward direction is

68.82 + 68.82 = 137.64 Newtons

The difference between the second and the first situation is

137.64 - 84.52 = 53.13 Newtons = D.

Q30)

Answered here, just ask in case you have any doubts.

Hope this helped!
Good Luck for all your exams!

 

[Snackbox86]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_13.pdf
hello, is there anyone who can give me explanations to the following questions
9, 10,15,34,
It is really appreciated

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w13_qp_13.pdf
someone please help me solve these question( QS.31,32.36.37,38)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_12.pdf
QS 9,19,

For the first link the text shows "s13_qp_13.pdf" while the pdf file that loads is "s13_qp_12.pdf". Which one is it that you have the doubts in?

For the second paper, here are the answers :

Q9) https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-498#post-806234

Q19)

When any sample of any material is at a particular temperature, the kinetic energy of 1 mole of that material will be the same as the kinetic energy of any other sample of any other material at the same temperature. This kinetic energy is related to the molecular movements - vibrational movement in a solid, sliding movement of molecules in liquid and linear movement of molecules in a gas.

The potential energy, on the average, of a molecule, is related to the average distance between it and the other molecules around it. Molecules very close together have low potential energies, while molecules that are far apart have large potential energies.

In the case of water molecules, they are actually farther apart in the solid state (ice) than in the liquid state (normal water) but this doesn't happen in any other common material. In other materials, the atoms/molecules are closer in the solid state than in the liquid and gas state.

But since the distance between molecules in the solid and gas state is practically independent of temperature, the mean potential energy does not depend on the temperature, but on the state. Therefore, the mean potential energy in a solid state is lower than the mean potential energy in a liquid state. So the potential energies are not the same, and we can eliminate A and B.

We have also established above (in the first paragraph of this question) that the mean kinetic energy only depends on the temperature of the sample concerned, so we can say that C is the correct answer, but to understand why D is wrong, we have to see that since the kinetic energy of both ice and water is the same but the potential energy of each sample is different, the sum can definitely not be the same. Therefore our answer, for sure, is C.

Hope this helped!

Good Luck for all your exams!

 

[ShreeyaBeatz]

Does it matter if k1 is the smaller one or the larger one?

No it doesnt. But most cases it is the larger one

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_13.pdf
hello, is there anyone who can give me explanations to the following questions
9, 10,15,34,
It is really appreciated

9)

If only one person is using the elevator, the mass of the person+elevator = 80 kg + 520 kg = 600 kg.
Suppose we take both the person and the elevator as our system. Let's take a look of at the forces acting on them.

i) The gravitational force of mg = (600 kg) * (9.81 ms⁻²) = 5886 kg ms⁻², acting downwards.
ii) The Tension force of magnitude T, acting upwards.

Since the set-up accelerates upwards, let's take upwards as the positive direction. Then tension is positive and weight is negative, and we can write

T - 5886 = ma
T - 5886 = 600a
where m is the mass of the system (600 kg) and a is the acceleration of the system.

Let's repeat for the weight. It will accelerate downwards, but the magnitude of the acceleration will be the same for the weight as for the elevator+person (because the rope is assumed to be unstretchable and this means any acceleration on one side will result in the same acceleration magnitude on the other side as well).

i) The gravitational force on the weight is (640 kg) * (9.81 ms⁻²) = 6278.4 kg ms⁻² downwards.
ii) Tension T (same magnitude as above) acting upwards.

Since the weight is accelerating downwards, we take downwards as the positive direction for the weight, and write

6278.4 - T = 640a

where a is the acceleration magnitude of the weight.

We can add both these equations as so, and this eliminates the variable of T:

T - 5886 + 6278.4 - T = 640a + 600a
392.4 = 1240a

So, a = 392.4/1240 = 0.3165 = 0.32 ms⁻² = B.

10)

Let's say the mass of a particle from the nucleus (also known as a nuclide) is "m". We'll keep this variable for use throughout this question.
Let's also select the atom as our system, since the atom on a whole is isolated, which allows us to apply the law of conservation of momentum.
Then, we can also write that the mass of the system is "Am", since m is the mass of 1 nuclide and A is the number of nucleons.

Initially, the atom has no velocity. Therefore, it has no momentum in any direction.

When the decay occurs, a proton of mass "m" is emitted with a velocity v in any direction. At the same time, the new nucleus is propelled in the opposite direction with a velocity u.
So, to conserve mass, the (mass of new nucleus) + (mass of proton) = (mass of original nucleus).
In other words, (mass of new nucleus) + m = Am. Rearranging, we get (mass of new nucleus) = Am - m = m(A - 1)
In this situation, momentum is conserved. Therefore, the momentum of proton is should be the opposite direction to the momentum of the new nucleus (check!) and the magnitude of this momentum should be the same as the magnitude of the new nucleus's momentum.

The momentum of the proton is mv. The momentum of the new nucleus is (A - 1)mu.
These have to be the same in magnitude, so we can set them equal to get

mv = (A - 1)mu.
Cancelling out the m's, we get

v = (A - 1)u = B.

15)

If the spring follows Hooke's Law, then we can write |F| = kx (Where |F| is the magnitude of the spring force, k is the spring constant and x is the extension of the spring, which is equal to [length of spring with load] - [length of spring without any load]).

If a load of 16 Newtons is applied, we can say that the spring exerts and equal and opposite force on it, so that 16 Newtons = kx.
If the final length is 5 times the initial length (40mm/1000 = 0.04 meters is the initial length) the extension =

(final length) - (initial length) = 5 * (initial length) - (initial length) = 4 * (initial length)

Since the initial length is 0.04 meters, the extension = 4 * 0.04 = 0.16 meters.

Putting this into the equation we get

16 Newtons = k * 0.16 = 0.16k so that k = 100 N/m.

The energy in a spring is given by the formula kx²/2 = (1/2)kx², where the symbols mean the same as in the earlier formula. Therefore, the energy stored is equal to

E = 0.5 * k * x² = 0.5 * 100 * 0.16² = 50 * 0.0256 = 1.28 Joules = 1.3 Joules = A.

34)

The wires are connected in parallel - that's the first thing you need to notice, and the best way of noticing this is looking at the arrangement - none of the wires are connected directly to each other, but each one of them has the same starting point and the same ending point. This usually means they have the same potential difference applied across their ends, which means they are connected in parallel.

So each cable has a resistance as a function of length, and for 1.0 km, we need to calculate the overall resistance.
The resistance formula for multiple resistances connected in parallel to each other is

1/R(equivalent) = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + 1/R₅ + 1/R₆.........

So we have 1 resistance of 100Ω (the steel core) and 6 resistances of 10Ω (the copper wires). Let's put this into the equation:

1/R(equivalent) = 1/100 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10
= 1/100 + 6/10
= 1/100 + 60/100
= 61/100
= 0.61
So since 1/R(equivalent) = 0.61, R(equivalent) = 1/0.61 = 1.63Ω = 1.6Ω = B.

Hope this helped!

Good Luck for all your exams!

 

[Hijab]

Can some one plz plz plz solve this question... I would really be grateful

 

[sagar65265]

Can some one plz plz plz solve this question... I would really be grateful

So the key thing to notice here is that when the circuit is working properly, the potential difference across the relay will be 16 Volts, and the current through it will be 0.6 Amperes (if the voltage isn't 16 Volts or the current isn't 0.6 Amperes, the relay circuit will not work properly).
However, since the wires used to connect the source to the relay also have a resistance, any current flowing through them will have to push against the resistance to keep moving, so they will lose some energy in the wires. In other words, there will be a drop in the potential across the wires.

Since one wire goes from the source to the relay, and the other wire goes from the other terminal of the relay back to the other terminal of the source, the wires and relay are all connected in series to each other. Therefore, the same current of 0.6 Amperes will go through the wires and relay.

Since the wires are 800 meters long each, and the resistance of each wire is 0.005 Ω per meter, we can say that the resistance of 800 meters of wire is equal to
0.005 Ω/m * 800 m = 4 Ω .
The potential difference across one wire is given by V = IR:

Potential Difference = V = 0.6 Amperes * 4 Ω = 2.4 Volts.

Since there are two wires of the same length, the total potential difference across them, i.e. the total volts lost in the wires, is equal to 2 * 2.4 Volts = 4.8 Volts.

Suppose the source has to supply at least another 16 Volts so that the relay can work properly, it totally has to supply 16 + 4.8 Volts to the circuit so that the potential difference across the relay is enough to run it properly even when the wires have some resistance and are wasting some potential. Therefore, the total minimum EMF of the circuit has to be 16 + 4.8 Volts = 20.8 Volts = C.

Hope this helped!
Good Luck for all your exams!

 

[Batguy]

please :'(

 

[Batguy]

Impulse = m(v - u)
Impulse = 0.5(8 - 12)
Impulse = -2 Ns

Force = Impulse/time
Force = -2/0.1
Force = -20 N
Since they only asked for da magnitude n not da direction, u can discard da negative sign.

Answer: C

naee answer D hai :'( maira bhee C tha :''(

 

[Hadi Murtaza]

naee answer D hai :'( maira bhee C tha :''(

ooops sorry v will b negative becuz of change in direction
Impulse = m(v - u)
Impulse = 0.5[ (8) - (-12) ]
Impulse = 0.5(8 + 12)
Impulse = 10 Ns

Force = Impulse/time
Force = 10/0.1
Force = 100 N

Answer: D

 

[crazytaylorfanXD]

For the first link the text shows "s13_qp_13.pdf" while the pdf file that loads is "s13_qp_12.pdf". Which one is it that you have the doubts in?

For the second paper, here are the answers :

Q9) https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-498#post-806234

Q19)

When any sample of any material is at a particular temperature, the kinetic energy of 1 mole of that material will be the same as the kinetic energy of any other sample of any other material at the same temperature. This kinetic energy is related to the molecular movements - vibrational movement in a solid, sliding movement of molecules in liquid and linear movement of molecules in a gas.

The potential energy, on the average, of a molecule, is related to the average distance between it and the other molecules around it. Molecules very close together have low potential energies, while molecules that are far apart have large potential energies.

In the case of water molecules, they are actually farther apart in the solid state (ice) than in the liquid state (normal water) but this doesn't happen in any other common material. In other materials, the atoms/molecules are closer in the solid state than in the liquid and gas state.

But since the distance between molecules in the solid and gas state is practically independent of temperature, the mean potential energy does not depend on the temperature, but on the state. Therefore, the mean potential energy in a solid state is lower than the mean potential energy in a liquid state. So the potential energies are not the same, and we can eliminate A and B.

We have also established above (in the first paragraph of this question) that the mean kinetic energy only depends on the temperature of the sample concerned, so we can say that C is the correct answer, but to understand why D is wrong, we have to see that since the kinetic energy of both ice and water is the same but the potential energy of each sample is different, the sum can definitely not be the same. Therefore our answer, for sure, is C.

Hope this helped!

Good Luck for all your exams!

Thank you so much
your help is much appreciated
Also the paper that i was referencing in the first part was actually NOV 2013 P13