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I don't understand what happens at Y in question 10. The total mass of what remains the same? The sand falls away, so the total mass should be less.
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Physics: Post your doubts here!
- Thread starter XPFMember
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I don't understand what happens at Y in question 10. The total mass of what remains the same? The sand falls away, so the total mass should be less.
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14: Moment is calculated by force*distance so they are inversely proportional. B is the correct graph for this.
17 and 35
37. When resistance is zero, p.d. will be zero and current passing thorugh the circuit will be maximum due to low resistance.
When resistance is maximum, p.d. will be maximum but current will not be zero as some current passes through.
39. Few particles will bounce back as most of the mass is concentrated at a point so a head-on collision only will result in this which has a low probability.
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The mass of the truck is less but total mass of the system is same. The sand will also leave with speed and will not be stationary when thrown out.
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Convert the height into meters which will be 0.8m and apply the formula Pressure=p*g*h
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf
Q9. How to solve it? If using scale factor, how to convert percentage to scale factor?
Q9. How to solve it? If using scale factor, how to convert percentage to scale factor?
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Speed is v first and then increased to v+(20/100*x)=1.2v
Distance is proportional to square of v when acceleration is same so 1.2^2=1.44x
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Q9
First find the time, using S = Ut + 0.5t^(2) ; S = 1.25 , U = zero, t = ?
That is 1.25= 0.5 x 9.81 x t^(2), t = 0.5 seconds.
Now for the Velocity, we know it equals Displacement / time taken ----> V = 10 / 0.5 = 20 m/s so in this type of question either of 2 quantities are given, we just need to find 3rd and input them all in second eqn to get the final result we are asked.
Q18
pressure = hdg. But here h is 2h since on one side liquid rises by h and another side liquid falls by h.
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use
s=ut+0.5aT^2 and find accelration as per using T and L and then using this a and putting T as 0.5T solve for s and ans will be B.
hope this will help Bro,easy.
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At point S and P both have highest accelerationCan someone tell me what each of these letters represent coz a lot of questions with the same graphs come inn different papers and I keep getting them wrong plz
and Q wont be zero all the time as it is oscillating.
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20)
Take the leftmost column of liquid.
There is a rule in fluid-statics that "the pressure in a sample of liquid is the same at any vertical height, as long as the fluid at that height is part of the whole."
What that says is that in any single sample of liquid, the pressure at any height is the same. Here, it means that the pressure at the top of the leftmost column (connected to the bulb) is the same as the pressure in the next column on the right at the same depth/height. (Neat info here)
So, take the unbroken fluid column on the left. At the top of the leftmost tube (which exposes the fluid to the pressure of the bulb), the pressure at the surface is simply 16,000 Pascals, because that is the pressure in the bulb and is also the pressure with which the gas pushes on any section of the surface around it. The liquid forms part of this surface, so the pressure on the liquid is also 16,000 Pascals.
Now let's move to the next column on the right, where the liquid column is exposed to pressure P. Where it is exposed to pressure P, the surface faces a pressure of P Pascals. This is to be expected. However, as we descend down the liquid, the pressure increases by (ρgh) at a depth h, and continues decreasing with depth.
When we reach depth h1, we note that this level coincides with the vertical level of the fluid in the leftmost column. The pressure on the right side column is
P + ρgh1, while the pressure on the left hand column is 16,000 Pascals. Since these are equal, we can write
P + ρgh1 = 16,000.
Repeating that calculation on the right side, we get
P + ρgh2 = 8,000
Eliminating P,
16,000 - ρgh1 = 8,000 - ρgh2
13,600 * 9.8 * (h1-h2) = 8000
h1 - h2 = 0.06
So the difference is 0.06 meters = 6 centimeters. The only option that agrees is D.
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14)
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm
16)
80*28/100=112/5
112/5 =potential energy
112/5=mgh
112/5=(o.12)(9.81)h
P.Sonly 120 g taken into account as only the arrow has a movement.
17)
mgsinteta*speed=40% of total power mgsinteta=force
force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans
18)
Show me your working, as its easy!
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
q 17! please helppp its simple i know but i m not gettin it!
also q 18 22 25
thanx
q 17! please helppp its simple i know but i m not gettin it!
also q 18 22 25
thanx
Last edited:
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14,15,21, and 31!!
14)
Moment = force times distance from the pivote
----------> F * 2R happens in B
15)
( 10cos30 + 10cos30 ) - 10 = 7.3
21)
17.5 *10^6 = 830*g*h + 1000*g*h
17.5 * 10^6 = 8142.3x + 9810(2000 - x)
17.5 *10^6 = 8142.3x + 19 620 000 - 9810x
-2 120 000= -1667.7x
1271.2 = x
~ 1270 m (3 s.f.)
31)
Use Q = It
so I = Q/t
here Q = 4Q as for 1 disc
and f = f so t = 1/f so 4Q/1/f = 4Qf
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Oh I took g as 9.8114)
(0.35*3*9.81)+(0.1*1.4*9.81=(0.15*6*9.81)+ans
ans=2.8Nm
16)
80*28/100=112/5
112/5 =potential energy
112/5=mgh
112/5=(o.12)(9.81)h
P.Sonly 120 g taken into account as only the arrow has a movement.
17)
mgsinteta*speed=40% of total power mgsinteta=force
force*velocity=power
(total m)(9.81)sin30)==40%
100%=ans
18)
Show me your working, as its easy!
I actually don't get 18 work den=force x distance. Pressure = force|area so pressure x area x distance should be the work done right? That's pressure x volume but that much is given in all the choices
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
Q5, 11, 22, 26 and 34. Please don't ignore this post at least XP
Q5, 11, 22, 26 and 34. Please don't ignore this post at least XP
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28)
Weight acts downwards so since it's in equilibrium the electric force has to act upwards. Since the plate up is positive and attracting the drop, that drop has to have a negative charge.