Physics: Post your doubts here!

[hino]

oct/nov 2010 pap 12 Q8 A moving body undergoes uniform acceleration while travelling in a straight line between points
X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and
from Y to Z is 6.0 s.
What is the acceleration of the body?
A 0.37 m s–2 B 0.49 m s–2 C 0.56 m s–2 D 1.1 m s–2
one more question that i need help with oct/nov 2010 pap12 Q13
wud appreciate some help

 

[wasifsamad]

give me pressure insolids and liqiuds notes.........2 or 3 pages maximum

 

[unseen95]

please help me with detailed explanation on question number 5 and 14 of 9702/01/M/J/03. Help!!

no one willing to help

 

[VicSalty]

any work sheets for physics?

 

[zephyr86]

Hi.can anyone help me with these physics questions?

 

[hino]

no one willing to help

Q 14 since they asked for the vertical component of the resultant force we ll simply subtract weight from the upthrust that is 10000-9000 to get 1000N which is the vertical component,as for that 500N we wont consider it since itz the horizontal component and we have nothing to do with it
hope you get it

 

[hino]

oct/nov 2010 pap 12 Q8 A moving body undergoes uniform acceleration while travelling in a straight line between points
X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and
from Y to Z is 6.0 s.
What is the acceleration of the body?
A 0.37 m s–2 B 0.49 m s–2 C 0.56 m s–2 D 1.1 m s–2
one more question that i need help with oct/nov 2010 pap12 Q13
wud appreciate some help

would someone please kindly help with these questions

 

[snoonono]

May june 2010, question 3 part A, in the question, it didnt say cm x10^-2, but they used x10^-2 in ms, help?

 

[snowbrood]

http://postimage.org/image/8xv0e6tmx/ can anyone solve this question

 

[snowbrood]

for first question
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37

for the second question well lets see option a
according to Archimedes principle upthrust = weight of liquid displaced
water displaced depends upon the volume of the object do u think that the cube's volume will increase if it is furthur lowered into water i hope u get it

for option b spring shows the tension which equals T=W-U
CORRESPONDS MEANS IS SIMILAR TENSION WONT BE SIMILAR TO UPTHRUST

FOR OPTION C UPTHRUST CHANGES THE READING WHICH ACTS TO HORIZANTAL SURFACE NOT VERTICAL.

FOR OPTION D GRAVITIONAL PULL DEPENDS ON MASS. MASS WONT CHANGE BEFORE AND AFTER IMMERSION HOPE I WAS ABLE TO HELP

 

[snowbrood]

May june 2010, question 3 part A, in the question, it didnt say cm x10^-2, but they used x10^-2 in ms, help?

THERE U GO I SOLVED UR QUESTION HOPE THAT HELPS

 

[Yousif Mukkhtar]

Can some explain q25
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s04_qp_1.pdf

 

[RadzMau]

SOMEONE HELP PLS!!!
A projectile leaves the ground at an angle of 60 to the horizontal. Its initial kinetic energy is E. Neglecting air resistance, find in terms of E its kinetic energy at the highest point of motion.

 

[hino]

for first question
lets see the motion in XY
initial velocity at x=v1
final velocity at y=v2
average velocity=distance/time
(V1+V2)/2=40/12 equation 1
simplifying it we get V1+V2=6.67

now consider motion in XZ
initial velocity at x=v1
final velocity=v3
V1+V2/2=(40+40)/(12+6)
simplifying it we get V1+V3=8.89 equation 2
subtract equation 1 from 2
i mean equation 2-equation 1
V1 is cancelled and we get V3-V2=2.22
since acceleration is constant so
acceleration in yz = acceleration in xz
= change in velocity/time
=2.22/6
=0.37

for the second question well lets see option a
according to Archimedes principle upthrust = weight of liquid displaced
water displaced depends upon the volume of the object do u think that the cube's volume will increase if it is furthur lowered into water i hope u get it

for option b spring shows the tension which equals T=W-U
CORRESPONDS MEANS IS SIMILAR TENSION WONT BE SIMILAR TO UPTHRUST

FOR OPTION C UPTHRUST CHANGES THE READING WHICH ACTS TO HORIZANTAL SURFACE NOT VERTICAL.

FOR OPTION D GRAVITIONAL PULL DEPENDS ON MASS. MASS WONT CHANGE BEFORE AND AFTER IMMERSION HOPE I WAS ABLE TO HELP

didnt get the second one but anyway thankyou for your help

 

[hino]

SOMEONE HELP PLS!!!
A projectile leaves the ground at an angle of 60 to the horizontal. Its initial kinetic energy is E. Neglecting air resistance, find in terms of E its kinetic energy at the highest point of motion.

Velocity at launch
E=1/2 mv^2
v=√2Em
horizontal component of velocity at launch (u)
cos60=u/√2Em
u=0.5√2Em

at the highest point, the object has no vertical velocity, but has the same horizontal velocity it had at launch (if we ignore air resistance, horizontal velocity does not vary through the flight since there are no horizontal forces acting on the projectile)
so
vertical component=0 horizontal component=0.5√2Em
now since we have both the components of velocity at the highest point we ll calculate the resultant velocity using pythagoras theorom which is going to be the same as the horizontal component of velocity
using this resultant velocity we ll calculate the K.E at the highest point
K.E=1/2 mv^2
K.E=1/2 m(0.5√2Em)^2
K.E=0.25E (answer)
hope i was able to help

 

[Yousif Mukkhtar]

Can some one explain this urgently: q 4) b) i) ii)
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf

 

[snowbrood]

 

[snowbrood]

View attachment 21134

can anyone please help its the part b that i need help in

 

[Dukeofwin]

Velocity at launch
E=1/2 mv^2
v=√2Em
horizontal component of velocity at launch (u)
cos60=u/√2Em
u=0.5√2Em

at the highest point, the object has no vertical velocity, but has the same horizontal velocity it had at launch (if we ignore air resistance, horizontal velocity does not vary through the flight since there are no horizontal forces acting on the projectile)
so
vertical component=0 horizontal component=0.5√2Em
now since we have both the components of velocity at the highest point we ll calculate the resultant velocity using pythagoras theorom which is going to be the same as the horizontal component of velocity
using this resultant velocity we ll calculate the K.E at the highest point
K.E=1/2 mv^2
K.E=1/2 m(0.5√2Em)^2
K.E=0.25E (answer)
hope i was able to help

Your informations seems perfect but I guess you messed the formula up! If its E=1/2 mv^2 then should V=√2E(divided by)M. Because M is multiplying with V and would go and divide there?

 

[hino]

Your informations seems perfect but I guess you messed the formula up! If its E=1/2 mv^2 then should V=√2E(divided by)M. Because M is multiplying with V and would go and divide there?

yes i guess i made a little mistake there,i mistakenly multilpied m rather than dividing it
anyway thankyou for pointing out