IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: Phosu on February 01, 2010, 04:33:57 pm
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could someone plz explain how to solve these questions
cie
Jun 02 pp1 question 14, 21 and 23
Thanks : )
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The answer is attatched.
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thanks dude
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im kinda confused with the concept of momentum
for eg ques 12 pp1 may/june 06,
the trucks collide and stick together and then the question asks to find its speed after collision, how can there be a resultant speed if they stick together on impact?
aside from that i have a doubt on solving 2 more questions from the same paper, question 33 and 34
Thanks :)
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They stick together and move togther - a single particle with mass m1 +m2
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33)4v
for the top route the 12v is divided in ratio 1@ ie 4v:8v so midpoint is at 4v
but for the bottom bridge it is 2:1 ie 8v:4v so midpoint is 8v
difference between midpoints is 4v
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34)the voltage accros the resistor is 12* 3/(3+1)=9v
p=V^2/R=9^2 /3=27
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nov 08
paper 1 ques 11, 21, 31
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11. Resloving for 2kg mass 2g-T=2a
resolving for 8kg mass T-6=8a
add these two 2g-6=10a so a=(2g-6)/10=1.4 c
21)c think of resistors in paralell and series. Maximise the resistance or extrension.
31)R=pl/A so halving the diameter means the resistance =pi*l/(d^2/4) is 4 times bigger and the current is divided by 4. d
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plz explain how to solve
Nov 09 paper 1, ques 8 and 12,
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When I get home
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Can you post a link to the paper
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here you go
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Thanks. Home in 2 hours
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33)4v
for the top route the 12v is divided in ratio 1@ ie 4v:8v so midpoint is at 4v
but for the bottom bridge it is 2:1 ie 8v:4v so midpoint is 8v
difference between midpoints is 4v
could u explain this in furthur details, like how did u get the voltages first?
and could explain question 12 indetail 2.
thankyou
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Can you upload the file again it may be corrupted
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i hope it works this time
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Thanks
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8)only A momentum and Kinetic energy both conserved
before
momentum2mu-mu=mu
kinetic enery =1/2*2m*u^2 +1/2*m*y^2=3/2mu^2
after
momentum 2m*-u/3+m*5u.3=mu
kinetic energy 1/2*2m*(u/3)^2+1/2*m*(5u/3)^2=3/2mu^2
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B resolve if the tension is not vertical it must increse so the vertial component remains constant.
If the angle to the vertical increases the tension must increase further
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M1 doubt
A car is towing a trailer along a straight horizontal road by means of a horizontal towrope.
The mass of the car is 1400 kg. The mass of the trailer is 700 kg. The car and the trailer are
modelled as particles and the towrope
as a light inextensible string. The resistances to motion
of the car and the trailer are assumed to be constant and of magnitude 630 N and 280 N
respectively. The driving force on the car, due to its engine, is 2380 N. Find
(a) the acceleration of the car,(3)
(b) the tension in the towrope.
could someone plz explain the B part, Thanks :)
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When I get up
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https://studentforums.biz/index.php/topic,5802.0.html
on this page
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M1 doubt
A car is towing a trailer along a straight horizontal road by means of a horizontal towrope.
The mass of the car is 1400 kg. The mass of the trailer is 700 kg. The car and the trailer are
modelled as particles and the towrope
as a light inextensible string. The resistances to motion
of the car and the trailer are assumed to be constant and of magnitude 630 N and 280 N
respectively. The driving force on the car, due to its engine, is 2380 N. Find
(a) the acceleration of the car,(3)
(b) the tension in the towrope.
could someone plz explain the B part, Thanks :)
tension is the force...its g0nna b...mass times acceleration..
F= MAss x Acceleration !!...
!!!!!!..(unsure)..bt i think itz this wayyy
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c part
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2 hours
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See this
http://www.astarmathsandphysics.com/a_level_maths_notes/M1/a_level_maths_notes_m1_particles_on_slopes_%282%29.html
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plz show how to solve
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plz show how to solve
whats the correct answer PHOSU :P
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its A
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its A
i think its short circuiting ??? so voltage is zero
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If you look v closely they are the same circuit
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If you look v closely they are the same circuit
alright , in the question they say voltage is zero
how come????? is it liek each resistor in parallel to the other??
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Wow, that was a fun question :P
took a while:
Heres how you do it,
You need to firstly work out the Resistance of the resistor from the first circuit...
BE CAREFUL WHERE THEY HAVE PLACED THE AMMETER...ITS ON THE PARALLEL CIRCUIT... so the total current in the circuit is 2 times the ammeter reading i.e. 2A
V = I R-total
6 = 2 x R-total
R-total = 3 ohms
Now use the resistance in parallel combination:
1 / R-total = 1/2R + 1/2R
you will get:
R-total = R
therefore,
each resistor has a value of 3 ohms
in the second circuit,
V = I R
6 = I ( 3 + 3)
I = 1.0 Amps
and voltmeter must read 0 Volts because there is no emf in the path across which the voltmeter is connected
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Wow, that was a fun question :P
took a while:
Heres how you do it,
You need to firstly work out the Resistance of the resistor from the first circuit...
BE CAREFUL WHERE THEY HAVE PLACED THE AMMETER...ITS ON THE PARALLEL CIRCUIT... so the total current in the circuit is 2 times the ammeter reading i.e. 2A
V = I R-total
6 = 2 x R-total
R-total = 3 ohms
Now use the resistance in parallel combination:
1 / R-total = 1/2R + 1/2R
you will get:
R-total = R
therefore,
each resistor has a value of 3 ohms
in the second circuit,
V = I R
6 = I ( 3 + 3)
I = 1.0 Amps
and voltmeter must read 0 Volts because there is no emf in the path across which the voltmeter is connected
THANKYOU ALOOOOTTT ;D ;D
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10 Two spheres approach each other along the same straight line. Their speeds are u1 and u2
before collision, and v1 and v2 after collision, in the directions shown below.
before collision
-> <-
O O
u1 u2
after collision
-> ->
O O
v1 v2
Which equation is correct if the collision is perfectly elastic?
A u1 – u2 = v2 + v1
B u1 – u2 = v2 – v1
C u1 + u2 = v2 + v1
D u1 + u2 = v2 – v1
please explain, i ticked A, but the ans is D
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For perfectly elastic collision approach speed u1+u2=separations speed v2-v1
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For perfectly elastic collision approach speed u1+u2=separations speed v2-v1
is this ALWAYS the formula ????
and could you please explain when do we use these formulas exactly?
very urgent plz and wat s speed of approach and separation anywway
thxx aloot
VERY URGENT PLZ
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Use the formula if you know a collision is perfectly elastic.
Speed of approach is the closing speed - add if they are moving towards each other, subtract if they are moving in the same direction.
Separation speed is the speed at which they are parting. Add if they are moving away from each other, subtract if they are moving in the same direction
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The interference patterns from a diffraction grating and a double slit are compared.
Using the diffraction grating, yellow light of the first order is seen at 30° to the normal to the
grating.
The same light produces interference fringes on a screen 1.0 m from the double slit. The slit
separation is 500 times greater than the line spacing of the grating.
What is the fringe separation on the screen?
A 2.5 × 10^–7m
B 1.0 × 10^–5m
C 1.0 × 10^–3m
D 1.0 × 10^–1m
plz show how to solve,
i know the two formulas, but i dont know how to use them
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The interference patterns from a diffraction grating and a double slit are compared.
Using the diffraction grating, yellow light of the first order is seen at 30° to the normal to the
grating.
The same light produces interference fringes on a screen 1.0 m from the double slit. The slit
separation is 500 times greater than the line spacing of the grating.
What is the fringe separation on the screen?
A 2.5 × 10^–7m
B 1.0 × 10^–5m
C 1.0 × 10^–3m
D 1.0 × 10^–1m
plz show how to solve,
i know the two formulas, but i dont know how to use them
is it C?
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When I get home
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is it C?
yah, how did you find it out?
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yah, how did you find it out?
ok so first we have these 2 formulas:
d sin thita= n (lambda) equ 1
(lambda) = ax/D equ 2
the information we got for equation one is :
thita=30
n=1 (first order as mentioned)
d= unknown
wavelength of yellow light approximately = 500nm
the information we got for equ 2 is:
D=1 m
a = 500 d ( as they say its 500 times)
therefore substitute values in equ 1 :
d sin thita= n lambda
rearrange
d= (1 * (500 *10^-9) )/ sin 30 >>>>> (mind u we multiply 500nm * 10^-9 to change it to "m" )
d= 1 * 10^-6 m
now substitute in equ 2 :
lambda= ax/D
rearrange
lambda/aD =x
now substitue values:
( 500 * 10^-9)/(500 * (1*10^-6) *1)
answer:
1 *10^-3 which is C
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Thanks so much for the explanation
a few doubts, do we need to know the wavelength of each light?
i know n stands for the order, but what does "d" stand for?
and you got the last rearrangement rong,
if (lambda)= ax/D
then (lambda)D/a=x
in this case since D is equal to 1, it wouldn't affect the answer. but in other cases be careful.
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A particle is acted upon by two forces F1 and F2 , given by
F1 = (i – 3j) N,
F2 = ( pi + 2pj) N, where p is a positive constant.
(a) Find the angle between F2 and j
how do we find the angle?
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Cos x =u.v/|u||v|
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Cos x =u.v/|u||v|
could you expalin in details plz?
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F1 = (i – 3j) N,
F2 = ( pi + 2pj) N, where p is a positive constant.
F1.F2=1*p+-3*2p=-5p
|F1|=sqrt(1^2+(-3)^2)=sqrt(10)
|F2|=sqrt(p^2+(2p)^2)=sqrt(5p^2)
cos theta=-5p/sqrt(10)sqrt(5p^2)=5over sqrt(50)=1/sqrt(2)
theta=cos^-1 1/sqrt()=45