Physics: Post your doubts here!

[biscuitbiscuit]

Oct/Nov 2010 /21 6.b . Please give me the solution with complete explanation. URGENT

Okay, this is how i did this :
Power=V^2/R, therefore resistance of a heater is, R= V^2/P
R= 230^2/1000= 52.9 ohms

(i) Since the heaters are in parallel therefore voltage across them would be 230V
Total resistance of the circuit= 52.9/2= 26.45 Ohms
therefore power diss...= 230^2/26.45= 2000W= 2KW

(ii) Series combination, voltage across the heaters would be 115V
Total resistance of the circuit= 52.9*2= 105.8
therefore power diss...= 230^2/105.8= 500W= 0.5KW

(iii) bottom are parallel to each other, therefore their effective resistance= 26.45ohms
therefore the total resistance of the circuit= 26.45+52.9= 79.35 ohms

power diss..= 230^2/79.35= 666.6666...= 0.67KW

Hope it helps

 

[biscuitbiscuit]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_23.pdf
"Q8(b)" someone please

 

[Suchal Riaz]

I didn't get this thing, did u mean 1.5 to 3.5 ohms if not how did you get volts?

from graph

 

[biscuitbiscuit]

from graph

well the on the graph y-axis is labelled in ohms!

 

[Suchal Riaz]

Suchal Riaz My teacher had not teach me anytime to half. Do we have to half ? :'(

idk what you are talking about. i guess it's error?
powers multiply:
V=r²
so dV/V = 2dr/R
dR/R means fractional error in R(convention). like in R= 3 +- 0.1 R=3 and dR=0.1 so fractional error is 0.1/3=dR/R
when multiplying or dividing the fractional error add and if the power multiply.
such as :
M=AZⁿ
dM/M = dA/A + n(dZ/Z)
and dM = M[ dA/A + n(dZ/Z) ]

 

[Suchal Riaz]

well the on the graph y-axis is labelled in ohms!

oops.
i meant to say that the deflection will be small. if you work out the change in reading of volt meter as temperature is changed from 0 to 25 C it will be very small. if the change is small it means that the precision will be less.

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_23.pdf
"Q8(b)" someone please

Calculate the current in the circuit by adding the resistances and V = IR. Then just use P = I^2R... bhai tumhari current electricty kuch zada hi weak hai ye to asan tha

 

[Thought blocker]

Calculate the current in the circuit by adding the resistances and V = IR. Then just use P = I^2R... bhai tumhari current electricty kuch zada hi weak hai ye to asan tha

I=V/R What is R ? 44.1 but in ms it is 44.6 y ?

 

[usama321]

I=V/R What is R ? 44.1 but in ms it is 44.6 y ?

The resistances of the wires

 

[Thought blocker]

The resistances of the wires

but how 44.6 ?

 

[usama321]

but how 44.6 ?

They just rounded the value of .492 to .5. You will use the actual value of 44.1 and .492. Round the answer to 3 sig figures at the end

 

[Thought blocker]

They just rounded the value of .492 to .5. You will use the actual value of 44.1 and .492. Round the answer to 3 sig figures at the end

What are you talking about ?????
Mark scheme has :¬
current = 230 /44.6<---------------------------------------- How 44.6 ? ( this is the doubt)
power = (230 /44.6)2 * 44.1
= 1170 W

 

[usama321]

What are you talking about ?????
Mark scheme has :¬
current = 230 /44.6<---------------------------------------- How 44.6 ? ( this is the doubt)
power = (230 /44.6)2 * 44.1
= 1170 W

Read my previous post again. Check the link below -.-

http://www.mathsisfun.com/numbers/addition.html

 

[Snow Angel]

please can someone explain the questions 2a) and 6bi) from the paper

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_2.pdf

in q 2a), why do we use 12.8 and 29.3 m in the equations to find the velocity. please explain. Thanks in advance

 

[usama321]

please can someone explain the questions 2a) and 6bi) from the paper

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf

in q 2a), why do we use 12.8 and 29.3 m in the equations to find the velocity. please explain. Thanks in advance

We know that the breaks were applied when the skid marks appeard. We also know the final speed would be zero. Use
2as = vsquare - u square
for second part he is asking the time interval before the brakes were applied

Just use s = ut + 1/2atsquare ( no acceleration)

 

[Thought blocker]

Read my previous post again. Check the link below -.-

http://www.mathsisfun.com/numbers/addition.html

It was so rude of you! Leave it I still dont get you I'll ask from some one else, ty.

 

[usama321]

It was so rude of you! Leave it I still dont get you I'll ask from some one else, ty.

My dear friend, one does get frustrated when the other ignores the simple answer again and again. I am really sorry, and won't do that again.
44.1 + 0.492 = 44.592. He has just rounded that off to 44.6 thats all

 

[Thought blocker]

My dear friend, one does get frustrated when the other ignores the simple answer again and again. I am really sorry, and won't do that again.
44.1 + 0.492 = 44.592. He has just rounded that off to 44.6 thats all

Thanks, and sorry.

 

[biscuitbiscuit]

Calculate the current in the circuit by adding the resistances and V = IR. Then just use P = I^2R... bhai tumhari current electricty kuch zada hi weak hai ye to asan tha

Ohhhhh! formula mey current ko square karna bhol gaya aur tum to na kaho phy weak hay. Ghar mey sab nay tang kar rakha hai, chor do phy ko nahi hoti tum sey

 

[ashcull14]

help!