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Physics MCQs thread.

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arlery said:
But how'd you solve it?
basically, u have
S = 1000 m ( 1 km )
a = 0.2 ms^-2
V = 0 m/s (as it stops after braking)
and u need to find (U) .. which is the initial speed ( the safe speed )
put the values in the equation
V^2 = U^2 + 2 a S
0 = U^2 - ( 2 x 0.2 x 1000 ) .................. ( the minus sign indicates the deceleration )
U^2 = 400
U = 20 m/s , so the answer is A
 
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arlery said:
how can I solve this? =$

It's just like any other speed, time ,etc. question! the distance between the yellow and red signs is 1000m , the train is decelerating at 0.2m s^-2 , and it stops when it reaches the red sign, so:

s = 1000
a = -0.2
v = 0
u = ? ( the maximum safe speed"

use the formula : v^2 = u^2 + 2as , and find "u"

hope that helped!
 
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Why is the answer A?
 

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arlery said:
Why is the answer A?
speed of approach = speed of separation
speed of approach = u - ( -u) = 2u
speed of approach = 5/3 u - (- 1/3 u ) = 2u ( this is the only choice which will give u the right answer ) .. A
 
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abdullah181994 said:
help plz ,explian it ans is B
when the weight is added. surface Z contracts (compression) and Y stretches (tension) as the upper rod is bent. similarly with X, the surface containing X stretches so its is tension.
 
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abdullah181994 said:
plzzzzzzzzzzzzzzzzzz expliannnnn plzzzzzzzzzzzzz :crazy: :%)
the phase difference between the molecules in a stationary wave is 180 degrees. so as u leave one node, the molecules should move in the opposite direction. now see the graph at A the longest arrow shows an antinode and the dot shows a node. in figure A, there are 2 antinodes ( 2 of the longest arrows ) and two dots ( nodes ) . so the answer is A
 
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see...F=qV/D...F=ma
qV/D=ma.....let V/D--->> A constant K....SO ma=Kq...rearranging....a=Kq/m....means a is proportional to q/m ratio....now chill out...apply this nd u ll get B
 
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