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Physics: Post your doubts here!

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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
 
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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)
 
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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)
thnx

but how do u know that the skid marks appeared when the breaks were applied
 
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