Physics: Post your doubts here!

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_11.pdf

Q16,21,23,24 help!!!

16)
The vertical components of both H and W are cancelled by the upwards vertical component of T. Furthermore, T balances out the horizontal component of H too. Thus T has to be the largest of the three. C is the only option.

21)
As Menu answered

23)
F = kx so x = F/k
For A : x = 4/k
For B : x = 3/k
For C : x = 3k
For D : x = 8/3k
So here constant k is either being multiplied or divide, hence we take k as 1 if we would have addition or subtraction included, we take k as zero
Now put 1 instead k .. A = 4, B = 3, C = 3, D = 2.5 so A is the answer

24)
As Menu answered

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_13.pdf

Q4

 

[MYLORD]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf q 11

 

[MYLORD]

it's A because v=w-u
w would be in the same direction and u would be upwards

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf q 11

 

[MYLORD]

thnx

F=ma
10-4=[20/10]xa
a=3ms^-2

 

[ShreeyaBeatz]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
question no 8 and 4

 

[ShreeyaBeatz]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf
qno. 8 !

 

[MYLORD]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf q15

 

[Hadi Murtaza]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf q15

Power = Force * velocity
Power = (1000 + 1000*10) * 0.5
Power = 11000 * 0.5
Power = 5500 W = 5.5 kW
Answer: B

 

[zem]

Can someone help me with this question?

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
qno. 8 !

using s=ut+0.5gt^2
for distance x,
x=0*t1+0.5gt1^2 (since initial velocity =0)
x=1/2gt1^2 (eqn 1)

for distance x+h,
x+h=0*t2+1/2gt2^2
x+h=1/2gt2^2
making x the subject gives x=1/2gt2^2-h (eqn 2)
since x=x
eqn 2=eqn 1

1/2gt2^2-h=1/2gt1^2
2(1/2gt2^2-h)=gt1^2
gt2^2-2h=gt1^2
gt2^2-gt1^2=2h
g(t2^2-t1^2)=2h
g=2h/(t2^2-t1^2)

 

[Hijab]

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

 

[_Ahmad]

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

check the 11th post of the previous page Menu Mendz has already solve it.

 

[sagar65265]

Can someone help me with this question?View attachment 43858

Wow, is this a good question! The trick here is to simply work through all the steps, and see what you get.
First step is in find the extension from the values we know.

The Young's Modulus Formula is

Y = Fl/Ae (where F is the tension, l is the original, unladen length of the sample, e is the extension, and A is the cross section perpendicular to the extension)

So, by multiplying both sides by e, we get

eY = Fl/A From here, suppose we divide both sides by Y, we get

e = Fl/AY which is the extension formula that we need. What we need to do is find out the values of this ratio for both the scale model and the full size version, and divide them to get the final answer.

A very important point to note here is that the question says that all linear dimensions are in a ratio of 1:10; this means that quantities such as length, radius, diameter, height, and so on are 10 times greater in the real version than in the scale model. Quantities such as Area, Volume, and Density are non-linear and so do not follow this ratio.

On reading the question again, there is something crazily interesting there - they have clearly mentioned that the load is cubic. Why? WHY?
It's important, that's why:

Suppose the material has some density ρ and it has a side length a. Then, for the scale model, the mass = ρV = ρa^3.
For the full size model, the same follows - however, the length has increased to 10 * a, so the mass of the load in the full-size crane = ρ(10a)^3 = 1000ρa.
Major importance there!
I'm pretty sure there is no more to be taken in consideration, since the extension of a wire requires you to consider only the dimension of the wire (going to do it), the load (done by above discussion) and the kind of material being stretched (Young's Modulus is the same since the material used is the same. Done!)

So, for the scale model:
i) The force can be written as "F" (let's say). This is also equal to ρa^3, as discussed above.
ii) The length of the cable without any strain applied can be written as "l".
iii) The Cross Sectional Area of the wire should be written as πr^2, where r is the radius of the cable.
iv) Finally, the Young Modulus can just be written as "Y".

Therefore, we can write

(extension of cable on the model crane) = [(ρa^3)gl]/Yπr^2

For the real crane:
i) The force can be written as "1000ρa^3", again as discussed above.
ii) The length of the cable without any strain can be written as "L". We know from the question that this is equal to 10 * l, so the length = "10l".
iii) The Cross Sectional Area of the wire should be written as "πR^2" where R is the radius of the full scale wire. We know this is equal to 10*r, so we can write that
the cross-sectional area of the real cable = π * (10r)^2 = 100πr^2.
iv) Again, the Young Modulus is just Y, since the material of the real cable is the same as the material of the scale model cable.

Therefore, we can write

(extension of cable on the full-size crane) = [(1000ρa^3 ) * 10l]/[100Yπr^2] = [(100ρa^3)gl]/Yπr^2

Dividing those two, we get (extension of cable on the full-size crane)/(extension of cable on the model crane) = 100 = 10^2 = C.

Hope this helped!
Good Luck for all your exams!

 

[Thought blocker]

Well what the heck is Doppler Effect ._. ?

 

[Wolfgangs]

Guys, about negative acceleration. If an object is decelerating, what will the direction of the resultant force and acceleration be? Will both face backward?

 

[amal sharkawi]

anyone answer this question please

 

[sagar65265]

Well what the heck is Doppler Effect ._. ?

Suppose you are standing x meters away from a sound source (turned off), and the speed of the waves = v. Also suppose wavelength = λ.

If it is turned off, you will hear a crest of the wave in time x/v seconds.
The wave will take a time of λ/v to completely register in your ear, so you will hear the next crest at a time of x/v + λ/v.

So, the time period will appear to be T = λ/v, and so frequency = 1/T = v/λ.

But suppose you are moving towards the source with velocity u (or the source is moving towards you). In that case, imagine you hear the first crest at time t.
Since you are moving along with the sound wave (or the source is moving along with the sound wave, or maybe both of you are moving towards each other) then the wave will cover distance vΔt in time Δt, and you will cover uΔt in same interval.

You will hear the next crest when vΔt + uΔt = λ, which means that Δt = λ/(u+v).
But Δt = period T, so this time, T = λ/(u+v), which means that frequency in this situation = 1/T = (u+v)/λ

But that is not the frequency of the source; this frequency is greater than the earlier frequency (when both were stationary) of v/λ. Strange, right?
This is called the Doppler effect - the change in apparent frequency of a source when relative motion occurs between the source and listener.
(But this isn't needed for CIE AS or A levels).

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

View attachment 43912

anyone answer this question please

The diving board is in equilibrium - it is not accelerating, it is not revolving.
If an object is not accelerating, there is no net force on the object.
If an object is not rotating, there is no net torque on the object.

The boy is stationary, so the force exerted on him by the board that holds him up is equal in magnitude to his weight. By Newton's Third Law, he exerts the same force downwards on the board. Since his weight is mg, the force he exerts on the board is equal to mg, and acts downwards. This force acts 5.0 meters away from the hinge.

The spring, if deformed by a distance "x", will exert a force of magnitude "kx" upwards on the board. This force contacts and pushes the board 2.0 meters from the hinge.

In this case, we can see that the board isn't rotating, so the net torque on it is zero. So, to find out our answer, we take moments about the hinge:

Moment due to the boy = mg * 5.0 meters = 5mg in the anti-clockwise direction.
Moment due to the spring = kx * 2.0 meters = 2kx in the clockwise direction.

So the net torque is zero, therefore 5mg = 2kx. Therefore, x = 5mg/2k
Putting in the values, we get x = 5 * 50 * 9.81/(2 * 10,000) = 1962/20000 = 0.0981 meters
This is equal to 0.0981 * 100 = 9.81 centimeters = C.

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

View attachment 43911

someone answer this question please

Alright, so here Q represents the actual voltage (since it is prorperly calibrated) and P represents a skewed voltage. The problem is that P has a zero error.

The meaning of a zero error is that the meter will not actually give a reading of exactly zero when the voltage through it is zero; it might give 0.3, -0.4, etc. In this case, P is anyways not calibrated, so it will not give the right reading; however, this means that in the given graph, P and Q only coincide at a Voltage of Zero because of the zero error.

Another example is that P says the voltage is zero even when the real voltage is 0.3 Volts. Therefore, it is giving us a reading of 0.3 lower than it's uncalibrated reading.

What we can do to correct this on the graph is to either shift the line to the left, or shift the line up, until the zero error is compensated for (we shouldn't do this in real life, but for this problem we can afford to do that since the gradient doesn't change, as is given in the question). And in reality, both are the same.

Now, then. Suppose we do that, the graph then gets an equation of y = mx or (Reading on P) = m(Reading on Q) [since y = mx + c, and c is adjusted to zero, y = mx).
All we have to do is figure out the gradient, which is the same as the gradient for the graph given, so we can calculate it from the given graph.
Taking the topmost point and the lowest point, we have:
Topmost point = (8, 5.7)
Lowest point = (0, 0.3)

So m = y(2) - y(1)/x(2) - x(1) = (8 - 0)/(5.7 - 0.3) = 8/5.4 = 1.48
So, in the final graph, (Reading on P) = 1.48 * (Reading on Q)
Since the reading on Q is the real voltage, and we want to know the reading for the real voltage of 5 volts, we have
(Reading on P for 5 volts) = 1.48 * (Reading on Q for 5 volts) = 1.48 * 5 = 7.4 = D.

Hope this helped!
Good Luck for all your exams!