As physics p1 MCQS YEARLY ONLY.

[crys123]

how is that possible? if F and a are in the same direction, shouldn't v be in that direction too?
and one more question:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Q1

how is that possible? if F and a are in the same direction, shouldn't v be in that direction too?
and one more question:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Q1

ok i am help u with this one.

SI base SI base units
Electric current Ampere (A)
Time Seconds (s)
Mass Kilogram (kg)
Length metre (m)
Temperature Kelvin (K)
Amount of substance Mole (mol)
Luminous intensity Candela (Cd)


Coulomb, joule or any other quantity is not SI base. Just the above 7.

 

[IGCSE O/L student]

ok i am help u with this one.

SI base SI base units
Electric current Ampere (A)
Time Seconds (s)
Mass Kilogram (kg)
Length metre (m)
Temperature Kelvin (K)
Amount of substance Mole (mol)
Luminous intensity Candela (Cd)


Coulomb, joule or any other quantity is not SI base. Just the above 7.

sorry, I meant Q10 and Q12

 

[crys123]

sorry, I meant Q10 and Q12

ok, no problem

Force = change in momentum (division sign) change in time.
This is newton's second law. The net force acting on an object is directly proportional to the rate of change linear momentum. the net force and the change in momentum are in the same direction.

since this is a graph, imagine the upper portion as positive and the lower portion as negative. so its +p1 and -p2

p1-p2/ t2-t1 or we could say -p2+p1/t2-t1

u try to understand this while i try to solve no 12.

 

[IGCSE O/L student]

ok, no problem

Force = change in momentum (division sign) change in time.
This is newton's second law. The net force acting on an object is directly proportional to the rate of change linear momentum. the net force and the change in momentum are in the same direction.

since this is a graph, imagine the upper portion as positive and the lower portion as negative. so its +p1 and -p2

p1-p2/ t2-t1 or we could say -p2+p1/t2-t1

u try to understand this while i try to solve no 12.

isn't change in momentum equal to the final momentum minus the initial momentum, ie; p2-p1/t2-t1

 

[crys123]

yeah u r right. u got it. but its a graph. and in a graph we consider the upper part (+) and the lower part as negative. so the value of p1 (coz its on the upper portion) is +p1. and for p2 its on the lower portion so its -p2.

 

[crys123]

no 12: i dont know whats wrong but im getting the wrong ans everytime! sorry couldnt help with this one too

 

[IGCSE O/L student]

yeah u r right. u got it. but its a graph. and in a graph we consider the upper part (+) and the lower part as negative. so the value of p1 (coz its on the upper portion) is +p1. and for p2 its on the lower portion so its -p2.

ok, thanks a bunch

 

[IGCSE O/L student]

no 12: i dont know whats wrong but im getting the wrong ans everytime! sorry couldnt help with this one too

I got it finally. they say that the two objects move off together, which means they stuck together on impact

 

[MathsFan]

how is that possible? if F and a are in the same direction, shouldn't v be in that direction too?

Let's say we have an object moving to the right, i.e. the vector v is pointing to the right. Then, we pull the object to the left, i,e. F is pointing to the left. This causes the object to have an acceleration in the opposite direction of the initial velocity.

EDIT: Sorry for the late reply.

 

[IGCSE O/L student]

Let's say we have an object moving to the right, i.e. the vector v is pointing to the right. Then, we pull the object to the left, i,e. F is pointing to the left. This causes the object to have an acceleration in the opposite direction of the initial velocity.

EDIT: Sorry for the late reply.

oh thanks alot. no worries. it didn't come in my paper. plus, your explanation's very beneficial

 

[IGCSE O/L student]

btw crys123 the momentum question came in my paper, and you have no idea how good I felt because I knew why my answer was right.
many thanks

 

[crys123]

btw crys123 the momentum question came in my paper, and you have no idea how good I felt because I knew why my answer was right.
many thanks

now i feel really good.

 

[ramiemajed96]

thank you

 

[IGCSE O/L student]

now i feel really good.

I know right? May God bless you sis

 

[crys123]

can anyone help me with uncertainty and error? Oct/nov 2005. Q: 5

 

[crys123]

the ans in C. why is D wrong?

 

[hino]

the ans in C. why is D wrong?

at the beginning of the fall the object has maximum accl. but with the passage of time the air resistance increases causing the acceleration to decrease
the air resistance keeps increasing until a point comes where it becomes equal to the weight of the object,at this point the object has terminal velocity or constant velocity which means acceleration becomes zero since constant velocity means zero acceleration so the answerz C

 

[hino]

can anyone help me with uncertainty and error? Oct/nov 2005. Q: 5

first find the uncertainity in mass
uncer.in mass=0.1/25=0.25
now find the uncert. in volume
volume=length*breadth*height
uncert. in vol.=0.01/5+0.01/2+0.01/1
=0.017
finally add both the uncertainities in mass and volume and multiply it with the density
=(0.25+0.017)*2.5
uncert. in the result=0.05 Ans

 

[crys123]

first find the uncertainity in mass
uncer.in mass=0.1/25=0.25
now find the uncert. in volume
volume=length*breadth*height
uncert. in vol.=0.01/5+0.01/2+0.01/1
=0.017
finally add both the uncertainities in mass and volume and multiply it with the density
=(0.25+0.017)*2.5
uncert. in the result=0.05 Ans

thank you so much!!!

 

[hino]

thank you so much!!!

np