in the question thy mention The road has marker posts
every 100 m. When the car passes one post, it has a speed of 10 m s–1 and, when it passes the
next one, its speed is 20 m s–1.
so it passes 2 post 2*100=200 !
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AS Physics P1 MCQs Preparation Thread.
- Thread starter mujtabashahnawaz
- Start date
every 100 m. When the car passes one post, it has a speed of 10 m s–1 and, when it passes the
next one, its speed is 20 m s–1.
so it passes 2 post 2*100=200 !
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hannn ur ryt!!
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really...??? amazing.... life is so easy now a day
...so if i say the distance between by house to urs is 100m.. and if i start from my house to urs at a speed of 10ms-1 and reach at 20ms-1 then the distance i think i travelled must be 100m... to solve the numerical if you are using 10ms-1 as initial speed then you are considering that the person is on the first post... to reach the next post he has to cover up a distance of 100m... ... ... ...
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oh ... fine mate..hannn ur ryt!!
yeh walla u r rytreally...??? amazing.... life is so easy now a day
so if i say the distance between by house to urs is 100m.. and if i start from my house to urs at a speed of 10ms-1 and reach at 20ms-1 then the distance i think i travelled must be 100m... to solve the numerical if you are using 10ms-1 as initial speed then you are considering that the person is on the first post... to reach the next post he has to cover up a distance of 100m... ... ... ...
anyway thanks man
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solve q.8 !!!!!
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yeh walla u r ryt
han han.. pta tha mjhe
im arabian i dont speak urdu
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explanation? plz
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i dont have the marking scheme
but the ans is D...we donot know the initial speed .... so we have to form two simultaneous equations to solve this.. s=ut+.5at^2...
1. 40=12u+1/2a(144)
2. 40=6u+1/2a(36)... solve them.. and you will numerical value equal to 1.1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
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the ans in A!!!i dont have the marking scheme
we donot know the initial speed .... so we have to form two simultaneous equations to solve this.. s=ut+.5at^2...
1. 40=12u+1/2a(144)
2. 40=6u+1/2a(36)... solve them.. and you will numerical value equal to 1.1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
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can u solve the 2 eq.. i may have made calc. errorz...
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OKii lemmmeeeeeeeecan u solve the 2 eq.. i may have made calc. errorz...
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OKii lemmmeeeeeeee
no calulation errors, its 1.11111!!!can u solve the 2 eq.. i may have made calc. errorz...
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oh.... let me do it all again..
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yes pleaseeee!!
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n10.
32.. battery will always provide constant emf if not damaged.. so option A and B is wrong... since the voltmeter is connected across the the battery, it will only tell the pd drop across battery not the lamp.. so option D also go wrong.. in addition to that.. lamp is connected in series, so voltage should have been same as before.. but high resistance voltmeter dropped the potential...
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HELP ME ON THESE QUESTIONS PLS
15 22 32 of n10 variant 12
16 19 34 of J11 variant 12
waiting for quick reply pls with details on this and my previous asked questions on page 38
n10
22... sir option B and D will produce the extension on 1/2 ratio for every weight.. so forget about them... consider A.. spring with high K is difficult to be expanded.... by placing a high k spring down will enable it to work both ways around.. when a heavy mass is to be suspended, rigid box will not allow the lower k spring in the box to move.. this will produces little extension... so low sensitivity with higher mass... when lighter mass is to be suspended, it will produce by rapid extension by release of the box.. option C cannot work with this phenomenon....
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Average speed between XY=40m/12s=10/3 ms^-1 . The average speed occurs at t/2 for uniform acceleration, so at t=6s.
Average speed between YZ=40m/6s=20/3 ms^-1. The average speed occurs at t/2 for uniform acceleration, so at t=3s but for total time it occurs at T=12s(from X to Y) +3s=15s.
So a=Δv/Δt=(20/3-10/3)/(15-6)=10/27=answer A.