physics mcq help needed! urgent!

[superduper2009]

The answer is B
The maxima and minima would on the graph i.e. A and C would occur at the middle of the motion... where speed is maximum..
therefore u can cross out A and C,
deciding between B and D ---
u know that at both points B and D the velocit is zero...
but u have to think... for the minimum point on the movement... the gradient at the point of the graph has to be negative.. (decreasing) ... and at point B.. the gradient (of tangent) is negative

hope it helps

[halosh92]

The answer is B
The maxima and minima would on the graph i.e. A and C would occur at the middle of the motion... where speed is maximum..
therefore u can cross out A and C,
deciding between B and D ---
u know that at both points B and D the velocit is zero...
but u have to think... for the minimum point on the movement... the gradient at the point of the graph has to be negative.. (decreasing) ... and at point B.. the gradient (of tangent) is negative

hope it helps

 but the answer is D

[superduper2009]

oops... sorry
ur right...
i gt it....between 0 and B the mass is moving upwards due to the positive velocity then it starts moving downwards.... C is the middle of the motion... and the mass is moving downwards...and then at D the mass has zero velocity and is at the lowest point of motion

[ifraha18]

q8 n 10..june 2007..??plz someone.

[superduper2009]

Q 8 --->
s=0.5g t^2
therfore -->
t = (2s/g)^.5

time to fall 40m = (2(40)/9.81)^0.5 = 2.85
time to fall 30 m = (2(30)/9.81)^0.5 = 2.47
2.85 - 2.47 = 0.38... answer is A

[halosh92]

q8 n 10..june 2007..??plz someone.

Q8) first find for the 40m
s=40
u=0
a=9.8
s=ut +1/2at^2
s=1/2at^2
sqrt(40*2/9.= t
t1= 2.857


now find for the first 30m
same thing except substitute "s" with 30m
u will get=2.47
t2= 2.47

t1-t2= 2.857 - 2.47= 0.38

[halosh92]

Q10)
IMPULSE
Ft=mv-mu
F=mv-mu/t
F= p2 -p1/t2-t1

[ifraha18]

fr 10 the answers b.

thks fr de eighth

[ifraha18]

q33 10 june 2007???

[superduper2009]

 10 -Momentum = mv
also momentum = Ft

mv=Ft
F=(mv)/t
Average F = change in momentum / change in time
t2-t1 is the change in time...
change in momentum is P2-P1
therfore the answer is
p2-p1/t2-t1

the answer is B

[ifraha18]

but b is p1-p2/t2-t1

nd q 33 of june 2007 as well..plz.?

[halosh92]

q33 10 june 2007???

the voltage in parallel is same...
now first find V1
2/10 x 5 = 1
and V2
2/5 x 3 = 1.2

and now V1-V2=1-1.2 = -0.2

[ifraha18]

thnkyou wht about q10.??

i am nt good wid circuits how can i solve em properli!!!...??wht should i do wid circuits

[superduper2009]

it's already answered...

[astarmathsandphysics]

nd q 33 of june 2007 as well..plz.?
V_1 =pd between S and P=1/2*2=1V
V_2 =PD between S and Q= 3/(2+3)*2=1.2V
1-1.2=-0.2V