difficult phy qns
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man 50kg stands in lift and take weight with weighing scale
what is weight of man when
a) lift travels up at constant velocity of 2m/s
b) when lift travel up and accelerate at 3m/s-2
2)
N
N S
S
state the neutral point. n and s are poles of different magnets
3) wall --- man --- wall
man points gun upwards and shoot ( in short, find spd of sound)
QN : name 2 factor that affect measurements
reaction time and ? ?
wind ? temperature of surrounding ?
apologise for slanted pic. last part need help please, the 2 advantages.
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dun really understand last part, what does '' in the measurement of resistance mean'' ? i did it the way i intrepreted and got 1 as my R.
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part iii ) would like to confirm the ans . 7.9n
btw is my cg correct? the X at the diagram.
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part iii ) would like to confirm the ans . 7.9n
btw is my cg correct? the X at the diagram.
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just guessing...
1a) 500N
b) 650 N
3) wind affects the measurement there
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f=ma huh. LOL.
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Originally posted by yiha093:
dun really understand last part, what does '' in the measurement of resistance mean'' ? i did it the way i intrepreted and got 1 as my R.
For series circuit:
Current (A) for both voltmeters (V) is the same whereas reading of voltmeters need to add up together.
For parallel circuit, vice versa.........
So 1ai) Current = voltmeter reading/ resistance
= 3V/ 2 ohms
= 1.5 A
1aii) p.d across the wire = 12V - 3V = 9V which I think you get the correct answer
1aiii) Resistance of test wire = Voltmeter reading/ current
= 9V/ 1.5 A (due to series circuit)
= 6 ohms
1bi) New current = volt reading/ resistance
= 9V/ 1 ohm
= 9A
1bii) Power dispitated = volt reading x current
= 9V x 9A
= 81W
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TIB, i do know those, i just wan the last part = )
thx anyway
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Continue from there:
1biii) Power for mass on nitrogen = Energy/ time
81W = Energy/ 300s
Energy needed = 81 x 300
= 24300J
Thus, mass of nitrigen needed = 24300/ 200
= 121.5g
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Originally posted by TIB657M:
Continue from there:
1biii) Power for mass on nitrogen = Energy/ time
81W = Energy/ 300s
Energy needed = 81 x 300
= 24300J
Thus, mass of nitrigen needed = 24300/ 200
= 121.5g
dun really understand last part, what does '' in the measurement of resistance mean'' ? i did it the way i intrepreted and got 1 as my R.
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Originally posted by yiha093:
dun really understand last part, what does '' in the measurement of resistance mean'' ? i did it the way i intrepreted and got 1 as my R.
It means 'In the measurement of resistance of test wire.' But the question asks you to find the mass of nitrogen vapour produced in 5 minutes.Don't be tricked by the measurement of resistance though.
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shouldn't the tension in that qn 3 be 9.79N?
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Originally posted by qdtimes2:
shouldn't the tension in that qn 3 be 9.79N?
ah ps. i trans ans wrongly. is 9.79.
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actually its qn 1 and 2
the rest i roughly know.
actually 1b im guesing is 350. . lolsxz
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Originally posted by yiha093:
dun really understand last part, what does '' in the measurement of resistance mean'' ? i did it the way i intrepreted and got 1 as my R.
1b (i)'s answer should be 4A of current passing through the test wire. instead of the 9A stated by TIB657M
Reason being: Since it is a series Circuit, there is only one path in which current can flow. Therefore each resistor in series has the same amount of current.
using this reason. and taking account that the voltage source and the standard resistor has not changed. The solution would be.
step 1 -- Total resistance: 2+1 ohms = 3 ohms
step 2 -- Total current: Emf (12V) divided by 3 ohm = 4A
1b(iI)'s answer should be changed too.
As we know the current passing through the whole circuit = 4A.
The Voltage for the standard resistor is. 4Ax2Ohm = 8V (ohms law, V= IR)
12V-8V = 4V.
P= VI
4Vx4A = 16W.
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Originally posted by kenn3th:
1b (i)'s answer should be 4A of current passing through the test wire. instead of the 9A stated by TIB657M
Reason being: Since it is a series Circuit, there is only one path in which current can flow. Therefore each resistor in series has the same amount of current.
using this reason. and taking account that the voltage source and the standard resistor has not changed. The solution would be.
step 1 -- Total resistance: 2+1 ohms = 3 ohms
step 2 -- Total current: Emf (12V) divided by 3 ohm = 4A
1b(iI)'s answer should be changed too.
As we know the current passing through the whole circuit = 4A.
The Voltage for the standard resistor is. 4Ax2Ohm = 8V (ohms law, V= IR)
12V-8V = 4V.
P= VI
4Vx4A = 16W.
Sorry for the mistake though.
So, the correct mass needed is: 24g
By following steps which I just posted.
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Originally posted by TIB657M:
Sorry for the mistake though.
So, the correct mass needed is: 24g
By following steps which I just posted.
dont need to say sorry (: we all learn. (:Just remember.
Series circuit: current constant throughout, Voltage drops.
Parallel Circuit: current drops, voltage constant throughout.
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so still bo rlpy to qn1 and 2 ?
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Originally posted by yiha093:
so still bo rlpy to qn1 and 2 ?
a) use 12 for g instead of 10 cos going up shld increase by 2 if not wrong.
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Originally posted by dkcx:
a) use 12 for g instead of 10 cos going up shld increase by 2 if not wrong.
g of 10 is acc.
so u add 2 = add spd to acc.
tehhhh ,
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Originally posted by yiha093:
g of 10 is acc.
so u add 2 = add spd to acc.
tehhhh ,
weight = mass X g
So when g increases, weight increase lo for a), giving u a weight of 60kg lo if i never remember wrong la. I haven't taught sec sch physics for ages.
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a) still 500n
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Originally posted by dkcx:
weight = mass X g
So when g increases, weight increase lo for a), giving u a weight of 60kg lo if i never remember wrong la. I haven't taught sec sch physics for ages.
Just taught yihao on MSN last night
Note: Constant velocity, so no change
Cuz net force is zero. -
eagle, that qn u taught me didnt even come out
no conversion of energy, moment, and f=ma. tat kind.
so sianz.
red bull didnt even use it man.
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Originally posted by yiha093:
eagle, that qn u taught me didnt even come out
no conversion of energy, moment, and f=ma. tat kind.
so sianz.
red bull didnt even use it man.
if u decide to go A levels, u will use it again
if u take poly engineering, or uni engineering, u will use again in foundation physics
also, u still have paper 1 coming up next week
so no harm learning it now ;)