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Physics: Post your doubts here!

crazytaylorfanXD

crazytaylorfanXD

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sagar65265 said:
Q15)

We have to consider the two elements or conditions of equilibrium:

i) For translational equilibrium, no net force must act on the system;
ii)For rotational equilibrium, not net torque must act on the system about any point.

Let's take the crane itself to be the system. Then forces are exerted on the crane by the Earth (Gravity), the weight W (this is stationary so it has to be balanced), the weight L (same justification as previous one) and the ground (Normal force, maybe frictional force).

Now the first thing to note is that moving the weight L to the right, the crane feels a tendency to rotate in the clockwise direction, since the net moment about the center (Let's take that point to be at the top of the vertical rod, and at it's middle) will try to rotate it in that direction.

So if the crane tries to rotate clockwise, for it to be stationary there has to be some other torque that tries to rotate it anti-clockwise, right?
This is the second part of the conditions, and yes, there has to be a torque that tries to rotate the system anticlockwise.

This eliminates A - the force by the ground on A will only rotate it further clockwise, not resist the clockwise motion (imagine the crane is a door hinged at the top, and apply that force - the door will rotate clockwise, right? This is the opposite of what we want happening).

B is possible - the force by the ground attempts to rotate the "door" anti-clockwise and could thus aid in stabilizing the system, but we'll get back to that later.

C is wrong - suppose the normal force R moves to the left, then it will still try to rotate the system in a clockwise manner (since it is not perfectly aligned with that point we are taking moments about, it will try to rotate the system clockwise, which is not what we want).

D is possible - the reaction force R attempts to turn the crane anticlockwise and thus could stabilize the system, but there's a little more work to do.

We have to apply the first condition of equilibrium - that no net force acts on the system.
In B, there is a vertical force of gravity, a vertical force from W, a vertical force from L, a vertical force from the ground and a horizontal force from the ground.
Hang on. This isn't balanced! There is not horizontal force counteracting the horizontal force from the ground, so how can the system remain in equilibrium?

Let's confirm with D. In D, there is no horizontal force that upsets the equilibrium, but there is a variable vertical force that ensures equilibrium both, for translation and rotation. Therefore, D is our answer.

Q37)

Let's put aside the light factor and deal exclusively with the resistance.
First things first - the output voltage will be equal to the potential difference across the resistor R, since the output terminals are connected across R, and a voltmeter connected in the same manner would give the potential difference across R which is what we get in the output circuit.

Since the same current passes through both the resistors, the net resistance will be equal to

(Resistance of R) + (Resistance of LDR)

And since we can write V = IR, we can say

V = 10 - 0 = 10 Volts = I (Resistance of R + Resistance of LDR)

and therefore, I = 10/([Resistance of R] + [Resistance of LDR])

This is the current flowing through both the resistors. Now, we can find out the potential difference across resistance R by again using V = IR to get

V(across R) = Output Voltage = 10 * [Resistance of R]/([Resistance of R] + [Resistance of LDR])

Right now, I suggest you write this down if possible, in whatever manner is easiest for you to understand, since the calculation we can do now is going to be difficult to display here in a post.

Once you're done, note that we have to maximize this value of output voltage. We can do this with calculus, or we can do this with a neat math trick, which i''ll describe below:

We can divide both the numerator and denominator of the Output Voltage equation by [Resistance of R] - this is allowed since while doing this we are essentially multiplying the equation by (1/[Resistance of R])/(1/[Resistance of R]) - this is equal to 1, and multiplying an equation by 1 doesn't change it.

In this post, i'll do it step by step. First let's divide the top half by [Resistance of R] and we get

10 * ([Resistance of R]/[Resistance of R])/([Resistance of R] + [Resistance of LDR]) = 10/([Resistance of R] + [Resistance of LDR])

Now let's divide the bottom half by [Resistance of R] to get

10/([Resistance of R]/[Resistance of R] + [Resistance of LDR]/[Resistance of R]) = 10/(1 + [Resistance of LDR]/[Resistance of R])

So the output voltage is also equal to 10/(1 + [Resistance of LDR]/[Resistance of R]). To maximize this, we have to minimize the denominator, since the numerator is constant and cannot be changed.
Furthermore, since the denominator is equal to (1 + [Resistance of LDR]/[Resistance of R]) we basically have to minimize the second half, since the first half is a constant (1) and cannot be changed.

So, to minimize [Resistance of LDR]/[Resistance of R], we have to maximize [Resistance of R] and minimize [Resistance of LDR], since that will give us the smallest fraction possible.
To minimize the [Resistance of LDR], we note that we have to maximize the intensity of the light falling on the LDR, and so the only option that shows both these options is A.

Hope this helped!
Good Luck for all your exams!
Click to expand...

Thank you soo freaking much , this helped me out a lot :***
 
Thought blocker

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NIM said:
guyz need help :p
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf
8, 12, 17, 18, 24, 34, 35
Click to expand...
8)
okay, the equation we consider here is → s = 1/2 * g * t^2
When time = T, the dist = s.
However when the time = 0.5 T →time^2 = (0.5 T)^2 →0.25 T
So, the dist is proportional to the square of time.
Thus, when the time reduces to half its original value, dist reduces to 1/4th its original value.
So, the dist covered in time = 0.5T = 0.25L.
Ans = B

12)
we'll take this problem in terms of the forces acting on the barrel and the man
for the barrel: the force due to the barrel - the tension in the rope(T) = ma (because the barrel move's down, the force due to the barrel is greater than the tension)
(120*10) - T = 120a (we dont know the acceleration of the man and the barrel)
for the man: the tension (T) - the force due to the man = ma
T - (80*a) = 80a
find the value of a using simultaneous equations, which is 2 ms^-2
using v^2=u^2 + 2as
find the final velocity of the man at 9 m, taking u as 0 ms^-1

17)
F=α*V2
so
α=F÷(v2)
α=800÷20=2kg.m−1
P=W÷t=(F*d)÷t=F*v
F=(2kg.m−1)×(40m.s−1)2=3200N
and finally
P=40m.s−1*3200=128000Watts=128KW

18)
Efficiency is given by (Useful Output Work)/(Total Input Energy) =
(Useful Output Work per Second)/(Total Input Energy per Second)
The useful output work per second = useful power output = 150 * 10^3 Joules Per Second.
Therefore, Efficiency = (150 * 10^3 Joules/Second)/(Total Input Energy per second)
Every hour, 20 liters of fuel are consumed. Therefore, every second, 20/(60 * 60) liters of fuel are consumed.
The energy in these liters is equal to [20/(60 * 60)] * 40 * 10^6 Joules.
Therefore, the efficiency is equal to
(150 * 10^3)/[20/(60 * 60)] * 40 * 10^6 = (150 * 10^3 * 60 * 60)/(20 * 40 * 10^6) = B.

24)
http://www.s-cool.co.uk/a-level/phy...finitions-of-stress-strain-and-youngs-modulus
We know that E is inversely proportional to the area, and directly proportional to the length.
Now, when we increase the length to 3L, the extension also increases 3 times → 3x.
But becoz we increase the diameter to 2D, we have increased the area by 4 times. A = π(d/2)^2 → A is now 4A.
And since extension is INVERSELY proportional to area, the extension now falls → x/4

So, the new wire's extension = 3x/4
Ans = B

34)
img-20121122-00179-jpg.18658


35)
When the switch S is closed, a current is allowed to flow through the circuit. When the current flows through the circuit, a potential difference is produced across the resistor R. This means that energy is lost in the resistor.

Also, when current goes through the internal resistance of the battery, energy is lost in that internal resistance. Therefore, some of the battery's energy and potential difference is lost in the battery itself.

However, we have to remember the definition of e.m.f - it is the total work done by the battery in driving one coulomb of current around the complete circuit. This work includes the work done by the battery in pushing the current through the internal resistance also, and it is equal to the voltage across the battery when the circuit is not closed (if you put a voltmeter across the battery when the switch S is open, the voltage measured is the rating of the battery AND the e.m.f. value).

Suppose you take internal resistance and separate it from the cells in the battery, then the energy lost in the internal resistance + the energy lost in resistor R is still equal to the e.m.f.

Remember, e.m.f. is the work done by the cells in the battery to push one coulomb through the complete circuit. They will do the same work whether there is internal resistance or not, so the e.m.f. does not change in the circuit because the cells can be assumed to be separate from the internal resistance.

So, A and B are eliminated; e.m.f. does not change at all.

But because a battery = cells + internal resistance, when S is closed, current flows through the circuit and energy is lost in the internal resistance. Also, there is a loss in potential across the internal resistance, so there is a loss in potential across the battery. Therefore, the potential across the battery changes because energy is lost in the internal resistance = C.

Note: if energy is lost in any resistance, it is because work is done against that resistance (suppose you are pushing a box on the ground, friction is the resistance, and you need to do some work against that friction resistance. Your muscles are the potential pushing the current, and some of that energy is lost when you work against friction).
 
ZaqZainab

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Ahmed H. Al-Neel said:
question 16 please!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf
Click to expand...
ZaqZainab said:
the force 'wheel friction' is constant from the graph i can tell its 8kN
and so at speed 200 the wheel friction is still 8kN and the rest is wind resistance
so 40-8=32kN is wind resistance
32/8=4
Click to expand...
the resistive force at speed 0 is the wheel friction, from the graph it is seen that this is 8 kn, since wheel friction is constant at 8 kn, 32(40-8) kn must be the magnitude of wind resistance, thus ratio is 32/8 = 4
 
chishtyguy

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sagar65265

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Asad Moosvi said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf

Question 10 anyone?
Click to expand...

The instantaneous velocity (at any instant, instead of an average) of any moving object is equal to the rate of change of displacement.

In this case, the displacement is only along one axis, the up-down (vertical) axis. Therefore, the only measurements needed to qualify the displacement are the magnitude of the displacement and the sign (if the object moved in 3 dimensions, we'd need a vector arrow. Here, the object moves in only 1 dimension so we do not need that).

So, the instantaneous velocity of the object is simply equal to the gradient of the displacement-time graph.
Take a look at the graph they have given for displacement versus time - initially, at the very beginning, the gradient is zero, i.e. the gradient is a flat line. Therefore, the displacement is barely changing with time at t = 0. This means that the velocity of the object at t = 0 is also zero.

This narrows the answer down to A, C or D. Let's look closer at the curve.

After the initial moment, the displacement is negative but the gradient slowly becomes more and more positive - that means the object is moving upwards, which is a positive velocity (since upwards is arbitrarily given the positive direction). So, since the gradient is positive and the change in displacement is upwards, the speed just after t = 0 is also positive (and upwards). Let's move on.

At the second scale-bar on the time axis, you can see on the displacement graph that the gradient becomes zero; this means that the velocity also becomes zero, which means that at the second bar on the time axis the velocity must be zero. This still doesn't get us any further, so we'll go a last bit along the time axis.

Right after the flat-line at the second bar on the time axis, we see that the gradient on the displacement-time graph becomes negative, i.e. the displacement becomes more negative than before. So, the velocity right after the second bar must also be negative. The only option that has a negative velocity after the second bar (from A, C and D) is A, our final answer.

Hope this helped!
Good Luck for all your exams!
 
chishtyguy

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Asad Moosvi

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Hadi Murtaza said:
If der r 2 or more wires protruding from a single point, then da current is split. When current is split, da value of current is not da same everywhere in da circuit. Da only circuit arrangement, where dis is not da case is arrangement D. In D, a single wire is connecting every meter, so current is same throughout circuit
Click to expand...

I don't understand it.
 
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