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Physics: Post your doubts here!

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View attachment 55367
Wkhan860 here comes another two questions for you. :(
14-
Horizontal component of velocity = v cos(45)
Vertical component of velocity = v sin(45)

initial kinetic energy E = 1/2 m v^2
now see the question says KE at highest position which means there wont be vertical velocity as at the highest point vertical v is zero
we only hve horizontal so final KE = 1/2 m (cos 45)^2 v^2
final kinetic energy is 1/2 m 0.5 v^2......this is half of E
hope u got it!
 
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14-
Horizontal component of velocity = v cos(45)
Vertical component of velocity = v sin(45)

initial kinetic energy E = 1/2 m v^2
now see the question says KE at highest position which means there wont be vertical velocity as at the highest point vertical v is zero
we only hve horizontal so final KE = 1/2 m (cos 45)^2 v^2
final kinetic energy is 1/2 m 0.5 v^2......this is half of E
hope u got it!
Thankyouuuu:D
 
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upload_2015-7-5_16-51-31.png


The problem is for c(ii)
i use Vout/V in=1 +Rf/R1
Replacing the values Vin=+0.4 , Rf =120 R1=5
i get answer :10 V
But the answer from marking scheme is 9 V . Can someone tell me the Reason ? is it linked to the diagram in part B ?
 
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View attachment 55423


The problem is for c(ii)
i use Vout/V in=1 +Rf/R1
Replacing the values Vin=+0.4 , Rf =120 R1=5
i get answer :10 V
But the answer from marking scheme is 9 V . Can someone tell me the Reason ? is it linked to the diagram in part B ?
When the gain is more than the supply voltage 10>9 V, the op-amp will go into saturation. So, normally you get back the supply voltage.
 

tdk

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20. Approximately how long would it take an electron to travel from the battery of a car to a head light and back (complete loop)?

a. seconds


b. hours

c. years


d. one-millionth of a second

e. one-tenth of a second
 
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How is this answer correct? can someone explain?
its q7 from oct/nov 2006 paper 4 the answer is from mark Untitled.png scheme.
 
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