• 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!

A*(a*)

A*(a*)

Messages
434
Reaction score
237
Points
53
Trisha Graal Emile said:
A 120 kg crate is dragged along the horizontal ground by a 200 N force acting at an angle of 30° to tswhe horizontal. The crate moves along the surface with a constant velocity of 0.5 m s–1. The 200 N force is applied for a time of 16 s.

a Calculate the work done on the crate by:
i the 200 N force
ii the weight of the crate
iii the normal contact force R.
b Calculate the rate of work done against the frictional force FR.
Click to expand...

Are these answers correct and is it from past papers ?
ai 200cos(30) * 8 J
aii 120*9.81 N
aiii 120*9.81 - 200sin(30) N
b 1600 cos(30) / 16 W
 
U

umair999

Messages
2
Reaction score
0
Points
11
XPFMember said:
Hi everyone, AsSalamoAlaikum Wr Wb...

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your PHYSICS doubts in this thread. InshaAllah other people here will help me and you all. :D ;)

NOTE: If any doubts in the pastpapers, please post the link! You can find links here!

Any Physics related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites! :)
Thanks!
Jazak Allah Khair!

Physics Notes:

Some links & Notes - by destined007

Physics Practical Tips - by arlery

Notes for A2 Direct Sensing (Applications) - shared by sweetiepie

Physics Summarised Notes (Click to download)

AS and A-Level Physics Definitions

A2: Physics Revision notes - by smzimran

Paper:5 Finding uncerainty in log - by XPFMember

Physics Paper 5 tips - by arlery

Physics Compiled Pastpapers: <Credits to CaptainDanger for sharing this..>

Here are the compiled A level topical Physics questions in PDF form...

Paper 1 : http://www.mediafire.com/?tocg6ha6ihkwd

Paper 2 & Paper 4 : http://www.mediafire.com/?g65j51stacmy33c

(Source : http://www.alevelforum.com/viewtopic.php?f=15&t=14)
Click to expand...
where are the answers of mcq's(paper 1)
 
Adeel12346

Adeel12346

Messages
23
Reaction score
84
Points
23
I think u will have to see mark scheme from papers section... U can see which year u are doing at the moment and then u can see it or download it from papers section :)... Hope it helps u :)
 
emkay

emkay

Messages
259
Reaction score
162
Points
53
Yousif Mukkhtar said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_11.pdf
Can someone explain q5?
Click to expand...

As it says, 3000 revolutions/min (apply unitary method)
3000 = 60sec
1 = x
x = 60/3000
x= 0.02s(one wave) in the 10cm wide display of c.r.o
looking at the options, the second one
showing 10ms means 0.01
and 0.01*2 = 0.02 means one wave and when we multiply 0.02*5 (as there could be 5 waves made in the 10cm wide display of c.r.o) or (10*10^-2/0.02 = 5) = it gives 0.1 i.e the option B.
(as 10ms-1 = 0.01)
 
Y

Yousif Mukkhtar

Messages
373
Reaction score
125
Points
53
emkay said:
As it says, 3000 revolutions/min (apply unitary method)
3000 = 60sec
1 = x
x = 60/3000
x= 0.02s(one wave) in the 10cm wide display of c.r.o
looking at the options, the second one
showing 10ms means 0.01
and 0.01*2 = 0.02 means one wave and when we multiply 0.02*5 (as there could be 5 waves made in the 10cm wide display of c.r.o) or (10*10^-2/0.02 = 5) = it gives 0.1 i.e the option B.
(as 10ms-1 = 0.01)
Click to expand...
Thanks again for saving another lost soul :)
 
emkay

emkay

Messages
259
Reaction score
162
Points
53
Yousif Mukkhtar said:
Guys one last question:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_12.pdf
q13 and 15
Click to expand...
for 13,
Forward force - Frictional Force = Mass*Acceleration
Forward Force = 2.0*9.81
Frictional Force = 6N
Mass (net)= 8.0 + 2.0
Acceleration = F-f/mass
a=19.62-6/10
a=1.36 or 1.4 i.e A is the answer.
---
for 15,
since tourqe is equal to force*distance,
the distance moved by the spindle will be the torque,
i.e L/4 means, WL/4 and will always be clockwise in this condition
ps. this was a stupid question just remember the fact that the distance moved by
the spindle will be multiplied by the weight of the cube W.
 
You must log in or register to reply here.
Top