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Physics: Post your doubts here!

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Momentum before=momentum after
Mv+-mv=2mv
Hence the velocity after momentum=0m/s
Kinetic energy before= 1/2*mv^2+1/2mv^2=mv^2
Kinetic energy after= 1/2*2mv^2 but since velocity after collision is 0 energy is 0j
Kinetic energy loss= energy before-energy after= mv^2-0=mv^2
Thankyou!
 
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Sorry to post it here but can someone explain why is option A right? for Q7 papers.gceguide.com/A%20Levels/Physics%20(9702)/9702_m16_qp_12.pdf

Note this is a Velocity distance graph.
The first part is obvious as velocity is constant.
The next is the slope part..Why isn't it a drastic decrease ? well because as the question states "uniform deceleration".. meaning it is a uniform decrease in velocity per unit time ; In the first graph the rate at which distance is being covered is also slower.
.velocity distance graph for deceleration   Google Search.png
 
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For q. 26:

equation for doppler effect:

[Speed of sound X Frequency of source]/[Speed of sound + or - speed of source]

so just put in the values and subtract when speed of sound - speed of source (approaching) from speed of sound + speed of source (Receding) ANS= C

for q. 25

for the length of wavelength: I used d = S X T

so wavelenght = 330 X 500 X 10^-6 = 0.165 m

so a translation of 0.165m means shift the graph by 1 lambda or 2 pi or 360 degrees

so basically it will be the same graph ANS= D

hope that helps :)
Thankyou!!
 
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que 34 anyone ?
Resistance of 1st Lamp = 960 (Power = 60W)
Resistance of 2nd Lamp = 40 (Power = 2.5W)

When Connected in series, PD across 1st lamp = (960/1000) * 250 = 240
PD across 2nd lamp = (40/1000) * 250 = 10
Since resistance is constant, Power Dissipated across 1st lamp = 60 W and Power Dissipated across 2nd lamp = 2.5 W.

Hence both lamps light normally.
 
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Can anyone give some advice regarding estimation and approximation in the mcqs?
How to prepare for it?
There were questions on estimating resistance of copper and on estimating young modulus of metals.. how would you approximate such a value without calculation..?
 
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A giant squid of length 20.0m is vertical in seawater, with the top of the squid at a depth of
8.00m. The density of seawater is 1050kgm–3.
What is the difference in pressure between the top and the bottom of the squid?
A 82000Pa B 206000Pa C 288000Pa D 389000Pa
Anyone?
 
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