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Physics: Post your doubts here!

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Q1 A box of mass 8.0 kg rests on a horizontal, rough surface. A string attached to the box passes
over a smooth pulley and supports a 2.0 kg mass at its other end.
When the box is released, a friction force of 6.0 N acts on it.
What is the acceleration of the box?
answer is 1.4ms-2
can u solve it via newton second law plz?

Q2--

An electric power cable consists of six copper wires c surrounding a steel core s
1.0 km of one of the copper wires has a resistance of 10 Ω and 1.0 km of the steel core has a
resistance of 100 Ω.
What is the approximate resistance of a 1.0 km length of the power cable?
B 1.6 Ω

1. Draw a free-body diagram on the 8 kg mass. The force to the right will be the weight of the 2 kg mass. The force backwards will be 6 N. a=F/m, and m=8+2 kg.
 
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people can anyone please explain how calculate uncertainties in wich the values are dividing multiplying and the percentage uncertainty....?
thanks!
 
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For 17. click attachment

for 31. Its B because it is a positive charge repel means it goes opposite to arrow for repel and it says X to Y, but also increasing force which means lines become concentrated rather than uniform lines. Hope you get it

Thank you :)
 
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Can you help me out then on June 2010 paper 11 question 33

here u have to assume a value of p.d=eg:6 V
so current in circuit =power/voltage=2 amp
nw voltage across P would be 4V and across Q and R would be 2V since they are in parallel
see hw itS 2V(ASSUMING EACH RESISTANCE TO BE 3,SO TOTAL R OF RESISTORS IN PARALLEL=1.5 OHM
NW THERE SHARE OFBATTERYS VOLTAGE=1.5/4.5(TOTAL R) x 6=2V)
anyway nw,
nw since P=VI
power in P=2 x 1(since divided between Q and R=2W

hope u got it,i tried my best
 
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In question no. 5, fractional uncertainty is being asked.... so:
as speed = distance traveled / time taken
here, distance traveled is 385 - 115 = 270, and its uncertainty is, 1+1= 2mm (its not 1mm bcoz we have to take the uncertainties of both the initial and final values of distance travelled)
again for time taken, 3.5 - 1.5 = 2.0 and its uncertainty is .02 + .02 =0.04mm (same reason as above)
hence, there fractional uncertainties should be, 2/270 + 0.04/2.00
So the key is A.

Hope it helped... :)
 
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In question no. 5, fractional uncertainty is being asked.... so:
as speed = distance traveled / time taken
here, distance traveled is 385 - 115 = 270, and its uncertainty is, 1+1= 2mm (its not 1mm bcoz we have to take the uncertainties of both the initial and final values of distance travelled)
again for time taken, 3.5 - 1.5 = 2.0 and its uncertainty is .02 + .02 =0.04mm (same reason as above)
hence, there fractional uncertainties should be, 2/270 + 0.04/2.00
So the key is A.

Hope it helped... :)

Thankyou very very much :LOL:
 
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here u have to assume a value of p.d=eg:6 V
so current in circuit =power/voltage=2 amp
nw voltage across P would be 4V and across Q and R would be 2V since they are in parallel
see hw itS 2V(ASSUMING EACH RESISTANCE TO BE 3,SO TOTAL R OF RESISTORS IN PARALLEL=1.5 OHM
NW THERE SHARE OFBATTERYS VOLTAGE=1.5/4.5(TOTAL R) x 6=2V)
anyway nw,
nw since P=VI
power in P=2 x 1(since divided between Q and R=2W

hope u got it,i tried my best

It's ok I found it out already, thank you for the effort just ask me if you have doubt
 
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