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Initially it will have less speed as it falls from rest,after that it accelerates and falls quickly so distance fallen increases faster.C implies it already had speed at the startView attachment 54329
Why B? Why not C?
F is the total force applied and mgsina is the component of the force that actually works against gravity,i.e the useful force
Directed away so think of your face as positive end and paper as negative.Hence negative charge moves towards you.
Ammeters have a very small resistance,so most of the current will flow through the Ammeter as its separate from the resistors.Cant be B also as this combination in parallel has lowest overall resistanceView attachment 54333
How is it D?
Its inelastic so cant have the same velocity as before(4mv).Also it rebounds so it cant have zero velocity(2mv).Hence it can have momentum change like sowhy is it C?
In these situations it helps to assign values to the quantities.Suppose each has R as 2 Ohm.So total Resistance is 3Ohm.Hence total emf will be 6 v using P=V^2/r and I will be 2A.Hence at a junction it will split equally as both have identical resistance and power dissipated will be I2R as 1^2 x 2 as 2 W
can u send me the link for this
What year is this from?View attachment 54341 ans is D
can somebody post its solution pleaseee
oh lol.It is from an old paper.What year is this from?
This is related to A2.
Apply the formula F=QQ/4Pi*E*r^2
What is its ans?oh lol.It is from an old paper.
Can u solve the other one too please
What is its ans?
And are you sure velocity before collision isn't mentioned?
In elastic collision relative speed of approach is relative speed of separationCan't understand the question in physics p1 m/j 13 p13 Q#11
Resolve momentum in x and y directions.yeah.& It's B
In elastic collision relative speed of approach is relative speed of separation
B: 3-(-3) = 6-0
C: 3-(-2) = 6-1
D: 5-2 = 6-3
A 2-(-5) =/= -5-(-2)
hence A is inelastic.
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