• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Physics: Post your doubts here!

Thought blocker

Thought blocker

Messages
8,477
Reaction score
34,837
Points
698
Asad Moosvi said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_13.pdf

question 17
Click to expand...
17)
The question is asking what is the rate does the energy provided to the system by the motor. So this shows that the motor is actually supplying some of its power. You must also know that the mass of the elevator is greater that the mass of the heavy weight because the question already stated "the elevator is partly counterbalanced by a heavy weight." So (m1 > m2) but if that is the case then how can the elevator move up? Yes. The motor and the heavy weight(m2) are providing the energy required to move the elevator up. So I came up with a formula: Power by m2 + power by motor = Power required to move m1. And we will have to use a formula P=FV where F can be the weight and V can be the velocity/speed. So substitute this formula into the earlier formula and we will get: m2gv + Power by motor = m1gv. Rearrange the formula: Power by motor = m1gv - m2gv. Simplify further and we will get: Power by motor = (m1 - m2)gv
 
Asad Moosvi

Asad Moosvi

Messages
129
Reaction score
306
Points
73
Thought blocker said:
17)
The question is asking what is the rate does the energy provided to the system by the motor. So this shows that the motor is actually supplying some of its power. You must also know that the mass of the elevator is greater that the mass of the heavy weight because the question already stated "the elevator is partly counterbalanced by a heavy weight." So (m1 > m2) but if that is the case then how can the elevator move up? Yes. The motor and the heavy weight(m2) are providing the energy required to move the elevator up. So I came up with a formula: Power by m2 + power by motor = Power required to move m1. And we will have to use a formula P=FV where F can be the weight and V can be the velocity/speed. So substitute this formula into the earlier formula and we will get: m2gv + Power by motor = m1gv. Rearrange the formula: Power by motor = m1gv - m2gv. Simplify further and we will get: Power by motor = (m1 - m2)gv
Click to expand...

Oh. I see. Thanks. :)
 
crazytaylorfanXD

crazytaylorfanXD

Messages
54
Reaction score
83
Points
28
Thought blocker said:
No..
case i)
You had 3 strings with load W
so imagine if you have 3 students and 1 chocolate
What is the ratio of getting chocolate to each child ?
Obviously 1/3
so same over here
3 strings = 3 students
W load = 1 chocolate
so load (Force) on each spring will be W/3

case ii)
You have 2 strings with load 2W
so imagine now you have 2 students and 2 chocolates
What is the ratio of getting chocolate to each child ?
Obviously 1
so same over here
2 strings = 2 students
2W load = 2 chocolates
so load (Force) on each spring will be W

Got it ? :)
Click to expand...

Ohh i get it now thanks :***
 
princessnoor

princessnoor

Messages
239
Reaction score
341
Points
73
chishtyguy

chishtyguy

Messages
473
Reaction score
1,511
Points
153
Thought blocker said:
5)
IDK

7)
At R jumper will be at rest, coz no velocity
At Q accn is zero as Q is the maximum height gained by jumper.

8)
Short answer, because the time taken with each speed is different.

Let's do the math behind this - suppose the distance traveled with both speeds is denoted by "s".
Suppose the time taken at 600 kmph = t(1).

Then, since Speed = Distance/Time, we can write

600 = s/t(1)
So that
t(1) = s/600

Similarly, let's do the same for the other speed. The distance traveled is still "s", the speed = 400 kmph, and the time taken to traverse this distance = t(2). So:

400 = s/t(2)
So that
t(2) = s/400

The average speed is the total distance traveled dividing by the time taken (the average velocity is different - it is equal to total displacement divided by time taken).
= (Total Distance traveled)/(Time taken to cover that distance).

Since the plane travels a distance "s" in one direction and returns the same distance "s", the total distance traveled = s+s = 2s.

The time taken = t(1) + t(2) = s/400 + s/600 = 3s/1200 + 2s/1200 = 5s/1200 = s/240

Therefore, the average speed = 2s/(s/240) = 480 kmph = C.

11)
Mass of α = 6.6 x 10⁻²⁷
Speed of α = 1.5 x 10⁷
No. of α per sec = 50000 in area of 0.0001 m²
Use P = f / A
P = ma / A
mass of 50000 = 50000 * 6.6 x 10⁻²⁷
acceleration = V / t Where V = 1.5 x 10⁷ and t = 1
P = [ (50000 * 6.6 x 10⁻²⁷) x 1.5 x 10⁷ ] / 0.0001
P = 4.95 x 10⁻¹¹ ≈ 5 x 10⁻¹¹

39)
using a double thickness foil
Theory based answer, come on
  1. After voltage passes through some resistance it will decrease
  2. The higher the resistance the greater the decrease in voltage
-->So it will decrease less after passing through 2 ohm than when passing through 4 ohm
Hence it is D
Click to expand...
Q5 is .6% as Volume 4/3 (Pie) r^3
so cube will become the co-efficient as 3( l'/l)
Percentage uncertainity in Volume is v'/v x100 = 3(l'/l) x 100
that is .6%
 
Asad Moosvi

Asad Moosvi

Messages
129
Reaction score
306
Points
73
crazytaylorfanXD said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

qs 5????
Click to expand...

When you add or subtract quantities, you add their absolute uncertainties.

d2 = 385
d1 = 115
uncertainty in d2 = 1
uncertainty in d1 = 1
delta d = 385 - 115 = 270
absolute uncertainty in delta d = uncertainty in d1 + uncertainty in d2 = 2

t2 = 3.5
t1 = 1.5
uncertainty in t2 = 0.02
uncertainty in t1 = 0.02
delta t = 3.5 - 1.5 = 2.0
absolute uncertainty in delta t = 0.02 + 0.02 = 0.04

fractional uncertainty in delta t = 0.04 / 2
fractional uncertainty in delta d = 2 / 270

You just add the fractional uncertainties.
 
chishtyguy

chishtyguy

Messages
473
Reaction score
1,511
Points
153
Wolfgangs said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_1.pdf
Q1, 5 and 10 pleaSE
Click to expand...
Q1 answer is C
as c= Wavelenght/ time
t=wavelenght/c= 2x10^15
waves in one sec is frequency so f= 1/2x10^15= 5x10^14
Q5
Well I got it wrong at first... But it's just
look at the actual true current .8 and look at faulty ammeter its also .8, so both the readings are closest
Q10
M not completely sure of the formula but in A's We use the formula
initial + final velocity of sphere 1= initial +final velocity of sphere 2
by arranging them we will have
u1+v1= -u2 + v2
rearrange them
u1+u2=v2- v1
About the formula Thought blocker M I correct???
 
chishtyguy

chishtyguy

Messages
473
Reaction score
1,511
Points
153
Menu Mendz said:
Could someone help me in Q's 10,38 and 39 of this paper
Oct 2012 paper 12
Thanks in advance!!!
Click to expand...
For question no 10 it's D
its not B, As velocity vertical component is continuously changing while horizontal component is Constant
Its not A, Veloicity can't be zero because object has not yet landed
Its not C, Although at maximum height vertical component is zero but there is a horizontal component so Velocity can't be zero..
D correctly represents it

For q 38 it's B
use the formula R/R+5000 x 1.5
At R = 1000, V= .25V
R=1000000, V= 1.49~1.5
look at graph B's max and Min Val

Q 39 its C
C as helium nuclei has only neutron and proton
Nucleon has Proton and neutron
so as Alpha particles increases number of nucleons decreases
 
You must log in or register to reply here.
Top