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Physics: Post your doubts here!

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The equation relating the extension of a spring to the force that caused that extension is

F = -kx

Where F is the force vector, k is the spring constant of the spring (that concerns the stiffness of the spring) and x is the extension = (Length after application of force) - (original length). In terms of magnitude,

|F| = kx

So here, when a force of 25 Newtons is applied on the spring, if the spring constant is k = 150 Newtons/Meter, we can write

25 = 150 * x

x = 25/150 = (1/6) meters.

So, if the length of the spring after extension/after application of the force is 55 centimeters = 0.55 meters, then the extension = (Length after application of force) - (Original Length) = (1/6) meters.
So, (1/6) = 0.55 - (Original Length)

(Original length) = 0.55 - (1/6) = 0.55 - 0.1667 = 0.383333 meters.

The closest to this is A = 0.38 meters = our answer.

Hope this helped!
Good Luck for all your exams!
 
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Q17
since it's the bottom of the ball that hits the ground first, we have to consider the bottom of the ball
by law of conservation of energy,kinetic energy at bottom=potential energy at top
mx9.81x.72=0.75J
m=.11kg
second time,
.11x9.81x.37=ans
use the exact mass of the ball and you would end up with b as the answer
Q18
Power = Energy / time.
Power = ( 0.5 * m * v^2 ) / t
Power = ( 0.5 *(rho * volume)* v^2) / t
Power = ( 0.5 *(rho * area * length)* v^2) / t
Power = (0.5 *(rho * area) * v^3)
Power = 1300kW
Q24
We know that intensity indirectly proportional to area and directly proportional to the square of the amplitude.
Dividing area by 3 causes the power per unit area(intensity)to increase by a factor of 3 and increasing the amplitude by 2 causes the intensity to increase by a factor of 4
so, (4x4)P=12P
the other 2 u will have 2 give me some tym
Q35
v1 = 600 / (600+3000/7) * 3 = 1.75V
v2 = 3000/7 / ( 3000/7 +600) * 3 = 1.25V
v = v1 - v2 = 0.5V
35.png
 
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Q10
kinetic energy remains conserved,
initial = final
since its a stationary body, initial kinetic energy = zero
if velocity of m is v, then velocity of 2m has to be such that sum of momentum remains zero.
therefore,
so velocity of 2m is -v/2 ( negative 'cause its in the opposite direction)
you can check, mv + (2m)(-v/2) = 0
now, [ 1/2(m)(v^2) ] / [ 1/2(2m)(-v/2)^2]
it becomes 2/1.
Q15
if the depths of water are equal by the end then the gain in height of water in the second tank should be h/2.
since the tanks are identical each tank would have equal volumes so the mass of water that has moved to the other tank should be m/2. since mass is directly proportional to volume provided density is constant. And density IS constant since its water.
Now PE=( m/2)(h/2)g
=mgh/4
Q18
pressure = hdg. But here h is 2h since on one side liquid rises by h and another side liquid falls by h.
 
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Q26
Reflected distance = 2 times the orignal so, 300m
time = distance / speed
time = 300 / 3 * 10^8 = 1µs
Q30
Its a standing wave, so the 33cm is the distance between two nodes. The wavelength thus becomes 0.66m. F = 330/0.66 = 500, T = 1/500 = 0.002s = 2m.
so it means wave should be completed in 4 blockss as the time base is 0.50ms/cm isiliye B
 
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Q10
kinetic energy remains conserved,
initial = final
since its a stationary body, initial kinetic energy = zero
if velocity of m is v, then velocity of 2m has to be such that sum of momentum remains zero.
therefore,
so velocity of 2m is -v/2 ( negative 'cause its in the opposite direction)
you can check, mv + (2m)(-v/2) = 0
now, [ 1/2(m)(v^2) ] / [ 1/2(2m)(-v/2)^2]
it becomes 2/1.
Q15
if the depths of water are equal by the end then the gain in height of water in the second tank should be h/2.
since the tanks are identical each tank would have equal volumes so the mass of water that has moved to the other tank should be m/2. since mass is directly proportional to volume provided density is constant. And density IS constant since its water.
Now PE=( m/2)(h/2)g
=mgh/4
Q18
pressure = hdg. But here h is 2h since on one side liquid rises by h and another side liquid falls by h.
thank you sooo sooo much!
 
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