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AS Physics P1 MCQs Preparation Thread.

rockerkunal

rockerkunal

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x-gamer-x said:
6Astarstudent rockerkunal

M SORRY
NOT QUESTION 4 BUT 9
Click to expand...

alright the answer is D to that, listen up, the horzontal velocity always remains the same and is the one which causes the horizontal distance moved, so from velocity = distance/time, you can use that, to find time use the formula s=0.5gt^2, after transformation to make t subject you will get t=root(2s/g) plug in the t formula in the velocity formula, mind you s is the vertical distance, and in the velocity formula its the horzontal, you willl get the answer
 
6

6Astarstudent

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x-gamer-x said:
6Astarstudent rockerkunal

M SORRY
NOT QUESTION 4 BUT 9
Click to expand...
you resolve the motion into horizontal and vertical components
the vertical initial velocity is 0 m/s with -9.81 m/s^2 acceleration and the distance is 1.25
so we apply the equation
s(displacement) = ut + (gt^2)/2
1.25 = 0 + (9.81 x t^2)/2 = 4.905 t^2
square root, so t= 0.505seconds

now we solve the horizontal component, since velocity is unchanged we can use the basic formula
speed = distance / time taken= 10/0.505=19.8 answer is D
 
x-gamer-x

x-gamer-x

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x-gamer-x

x-gamer-x

Messages
264
Reaction score
116
Points
53
rockerkunal said:
alright the answer is D to that, listen up, the horzontal velocity always remains the same and is the one which causes the horizontal distance moved, so from velocity = distance/time, you can use that, to find time use the formula s=0.5gt^2, after transformation to make t subject you will get t=root(2s/g) plug in the t formula in the velocity formula, mind you s is the vertical distance, and in the velocity formula its the horzontal, you willl get the answer
Click to expand...

6Astarstudent said:
you resolve the motion into horizontal and vertical components
the vertical initial velocity is 0 m/s with -9.81 m/s^2 acceleration and the distance is 1.25
so we apply the equation
s(displacement) = ut + (gt^2)/2
1.25 = 0 + (9.81 x t^2)/2 = 4.905 t^2
square root, so t= 0.505seconds

now we solve the horizontal component, since velocity is unchanged we can use the basic formula
speed = distance / time taken= 10/0.505=19.8 answer is D
Click to expand...

THANK U BOTH OF U
 
x-gamer-x

x-gamer-x

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6

6Astarstudent

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x-gamer-x said:
6Astarstudent
Click to expand...
for w11
#16: initial Ek = (mv^2)/2
since in a collision momentum must be conserved, initial momentum = mv . final = 2m x (v/2)
final Ek = (m (0.5v)^2)/2 = (mv^2)/4
so loses 50% of original Ek

for s11
#4
amplitude is 4x1.5= 6
period = 4 x 5ms = 20ms

#14
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2

#16
Power = Force x velocity
Force = weight = 80 x 9.81=784.8
Power = 784.8 x0.5 = 392.4W = 0.392kW

#25
ok, this one is hard to explain but
75Hz is the first harmonic, 1 node 1 antinode (1/4 wavelength)
for 2nd harmonic, it is (3/4 wavelength) so 75x3 = 225
3rd harmonic = (5/4 wavelength) so 75x5 = 375

#35
ratio of resistance is 1:2:5 and we know that ratio is current is inverse of that
so must be 1:1/2:1/5 = 10:5:2
so 5x 5/17 = 1.47

w10 qp 12

#12
torque = force x perpendicular distance to centre of pivot = 200 x 0.25 = 50Nm

#15
Ek = (mv^2)/2
since it is at steady speed, no net force so
Weight (mg) = retarding force (kv)
mg = kv
v = mg/ksubstitute this to Ek formula
Ek = (m(mg/f)^2)/2 = m^3 x g^2 / 2k^2
 
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