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Physics: Post your doubts here!

maryam fatima

maryam fatima

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Shaoli Hassan

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Snowysangel said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf8:B
8: B
9: D
11:B
12:A
Could someone please explain why!
Click to expand...

Q-8: We know, s=ut+0.5gt2
So, L=0.5gT2
since initial speed is 0
Now, for 0.5T,
s=0.5(0.5T)2
=0.5(0.25)T2
=0.25L

Q-9: For projectile motion, the vertical acceleration is the same throughout and is equal to g. There is no component of acceleration in the horizontal direction, so horizontal velocity is unchanged. But vertical speed changes due to which the resultant velocity changes.

Q-11: Since the collision is elastic, relative velocity of approach is equal to relative velocity of separation. So, the speed of each mass after collision will be v. Kinetic energy after collision is mv2.

Q-12: For the barrel,
F=ma
120g-T=120a where T is the tension in the string
T=120g-120a

For the stake,
F=ma
T-80g=80a
T=80g+80a

Equate them to obtain,
120g-120a=80g+80a
a=0.5g

v2=u2+2as
v2=0+2(0.5g)(9)
v=6
 
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Snowysangel

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Shaoli Hassan said:
Q-8: We know, s=ut+0.5gt2
So, L=0.5gT2
since initial speed is 0
Now, for 0.5T,
s=0.5(0.5T)2
=0.5(0.25)T2
=0.25L

Q-9: For projectile motion, the vertical acceleration is the same throughout and is equal to g. There is no component of acceleration in the horizontal direction, so horizontal velocity is unchanged. But vertical speed changes due to which the resultant velocity changes.

Q-11: Since the collision is elastic, relative velocity of approach is equal to relative velocity of separation. So, the speed of each mass after collision will be v. Kinetic energy after collision is mv2.

Q-12: For the barrel,
F=ma
120g-T=120a where T is the tension in the string
T=120g-120a

For the stake,
F=ma
T-80g=80a
T=80g+80a

Equate them to obtain,
120g-120a=80g+80a
a=0.5g

v2=u2+2as
v2=0+2(0.5g)(9)
v=6
Click to expand...
How do know s is 9m? Won't the distance moved by the heavier object he greater? Oh and also, what happens to the acceleration of an object in both the presence and the absence if air resistance?
 
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Shaoli Hassan

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Snowysangel said:
How do know s is 9m? Won't the distance moved by the heavier object he greater? Oh and also, what happens to the acceleration of an object in both the presence and the absence if air resistance?
Click to expand...
No, the distance travelled won't be different since both of them are connected by the same string. The question says what will be the speed when the man's head is level with the bottom of the barrel. This means, the distance travelled by both the objects must be half of the length of the string=(18/2).
If an object if falling vertically (not connected to a string), in the absence of air resistance, the acceleration remains unchanged. But, in the presence of air resistance the acceleration decreases. If connected by a string as in question 12, just remember that the acceleration is the same for the whole system.
 
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Shaoli Hassan

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sagar65265

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DeViL gURl B) said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w04_qp_1.pdf

COMPLETE EXPLANATION NEEDED FOR QUESTION 14 pleaseplease!!
Click to expand...

The concept here is pretty straightforward - at any point in an electric field, the force on a charged particle acts along the tangent to the electric field line at that point.

In other words, the force on a particle placed on a curved field line will act along the direction of the tangent of the field line at that point.

Also, the arrows on electric field lines (in diagrams only) tell you the direction of the force a positively charged particle would experience in the field - a negatively charged particle will experience a force along the opposite direction.
So the positive particle faces a force along the arrow, the negatively charged particle experiences a force in the direction opposite the arrow.

Electric Field.JPG

On the diagram above, the blue arrows represent forces on a negative charge (D,E,F) and the red arrows represent force on a positive charge(A,B,C). You'll note that both the forces at any point act along the tangent to that point, just in opposite directions (NOTE: The arrows here are not drawn to scale, so do not concern yourself with the length of the arrows and the magnitude of the field - here, just focus on the directions relative to the field lines).

The tangent represents the direction of the electric field at that point, so the electric field direction can be said to be in the direction of a force on a positive particle at that point in the field, with the force on a negative particle in the opposite direction.

So, in this question, the electric field arrows points to the right, and the lines are all straight and evenly spaced. Therefore, this is a uniform electric field, with the forces on a proton placed anywhere in this field acting to the right.

Therefore, the force on an electron (as in the question) should be in the opposite direction, i.e. to the left. The correct answer is, therefore, B.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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meerul264 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf

Q33 please. Thanks
Click to expand...

Suppose the resistance of one of the resistors is R. Then the combined resistance of Q and R is given as follows:

1/R(Q + R) = 1/R(Q) + 1/R(R) = 1/R + 1/R = 2/R
Therefore, R(Q + R) = R/2.

This resistance is in series with R(P), and since they are in series we can add them straight away:

R(P + Q + R) = R + R/2 = 3R/2.

So the total resistance of the circuit is equal to 3R/2. Suppose the voltage is V, then we can see that the current = I = V/R:

I(circuit) = V/(3R/2) = 2V/3R Amperes

From this, since we know that the total power delivered is 12 Watts and the power given is equal to P = IV, we can write

P = 12 Watts = 2V/3R * V = 2V²/(3R)
So, we can write that

V²/R = 3/2 * 12 = 18
So V²/R = 18. Let's move on.

Since we know that the resistance of network Q + R = R(Q + R) = R/2, we can say that the potential difference across Q + R section is equal to V(Q + R) =
I * R(Q + R) =

V(Q + R) = 2V/3R * R/2 = V/3.

So the potential difference across the resistor called R is equal to V/3. Therefore, since P(resistor) = V(across resistor)²/R, we can write

P(R) = (V/3)²/R = V²/3²R = V²/(9R)

Since we know that V²/R = 18, we can write

P(R) = (18)/9 = 2 Watts = A.

Hope this helped!
Good Luck for all your exams!
 
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sagar65265

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ShreeyaBeatz said:
Which property of a metal wire depends on its Young modulus?
A ductility
B elastic limit
C spring constant
D ultimate tensile stress
Click to expand...

The main point to understand is this - the Young's Modulus is basically a calculated value that we can find out from measurements and use later. It is akin to density - there's very few things that depend on density, but density depends on many other factors and density is also a constant for a material that we regularly use in calculations.

The ductility of a material should have nothing to do with the Young's Modulus - it reflects in the Young Modulus, but is definitely not dependent on it. The ductility of a material is difficult to quantitatively measure and is not calculated using Young's Modulus. The Young's Modulus value just tells us how well the object will stretch, not whether it will do so efficiently without wasting too much energy or whether it will spring back. A wire that springs back is not very useful as far as I can imagine (though who knows how wrong I could be!).

The elastic limit of a material cannot be calculated with the Young's modulus - think about it! The Young's Modulus simply tells us that if you have a sample of this cross section and that length, and you apply so-and-so force to it, it will have such-and-such extension - what it tells us is that with a large enough force, a large enough length, and a small enough cross sectional area, we can cause a near-infinite extension of any sample! This is obviously not true, since the Young Modulus formula implies a linear stress-strain relationship, which does not predict anything that happens after the limit of proportionality - the elastic limit, the plastic stage, the ultimate tensile strength or the fracture point. We only assume that the Young's Modulus holds in our problems, otherwise the equations to describe plastic deformation would be a very tricky set to solve, indeed!

All this only leaves us with C - the spring constant is relevant to the stretchiness of the material, and is a constant that is very close to the Young Modulus. Both make the assumption that the limit of proportionality is not reached, and neither are the plastic stage or the fracture point. This makes sense, since wires that stretch easily will make sensitive springs (but probably not always the best of wires).

It is a rather intuitive question, but I hope this at least contributed to your thoughts on the topic.

Good Luck for all your exams!
 
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