Physics: Post your doubts here!

Tkp

Tkp

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m nt getting it.i dnt understand these questions.same question in mj07 p1 23.pls cn any1 clear the concept by drawing it?
 
littlecloud11

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Tkp said:
m nt getting it.i dnt understand these questions.same question in mj07 p1 23.pls cn any1 clear the concept by drawing it?
Click to expand...

If i drew the Graph it would be same as the one in the question so let me just try explaining using that. You know that when an object is moving through the air it's Kinetic energy will be zero at it's maximum height, right? So naturally, it's velocity will also be zero at maximum height. When am object is at it's equilibrium position the KE is max. hence the speed is also maximum. When you consider the velocity time graph like the one in m/j 06 Find the points where the velocity is zero because they represent the max and min points. The first point (B) shows zero velocity, when you throw an object upwards it reaches it highest point first before it goes down, so B would be the max height. The next point where the velocity is o is D, so that would be the min. height.

For m/j 07 it's a displacement time graph. Again, the same principal applies, at the max and min height the velocity is o. Also, you should know that at the max and min height the acceleration is maximum.This can be proved by the graph below. At point B and D where the velocity is zero the gradient of the curve is actually the steepest, and since the gradient represents acceleration, it's value is max. So your answer to 23 should be D, where the velocity is 0 and acc. max.

phy.png

Hope this helps!
 
asadalam

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In nov12 p51 q1 they have asked us to vary the speed of the bar magnet as it leaves the coil...is it ok if we explain the use of electromagnets placed above and below the coil and each connected individually to a power supply?
 
Norton

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Can someone explain to me what would happen if white light is incident on a diffraction gratinginstead of a monochromatic light?

PLEASE can u explain it thanx a lot!!!!
 
littlecloud11

littlecloud11

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Norton said:
Can someone explain to me what would happen if white light is incident on a diffraction gratinginstead of a monochromatic light?

PLEASE can u explain it thanx a lot!!!!
Click to expand...

You know how monochromatic light gives dark and white bands, the same happens for white light, except that the white region splits to form 7 different colors because white light is composed of 7 colors. The dark band remains as it is.
 
Ahmedraza73

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Can any solve this:
The frequency of the fundamental mode of transverse vibration of stretched string wire 1 m long is 250 Hz.when the wire is shortened to 0.4 m at the same tension the fundamental frequency is?

A: 102 Hz
B: 162 Hz
C: 312 Hz
D: 416 Hz
E: 640 Hz
 
Norton

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littlecloud11 said:
You know how monochromatic light gives dark and white bands, the same happens for white light, except that the white region splits to form 7 different colors because white light is composed of 7 colors. The dark band remains as it is.
Click to expand...
thanx alot again!!!
 
snowbrood

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Ahmedraza73 said:
Can any solve this:
The frequency of the fundamental mode of transverse vibration of stretched string wire 1 m long is 250 Hz.when the wire is shortened to 0.4 m at the same tension the fundamental frequency is?

A: 102 Hz
B: 162 Hz
C: 312 Hz
D: 416 Hz
E: 640 Hz
Click to expand...
fundamental wavelength =2L
L = 1/2 lambda
lambda = 2L
V is constant both cases, so f1(lambda1) = f2(lambda2)
(250)(2) = 0.8f
f = 625
answer is 625 i had contacted many phds uni teachers and students they all agree with this answer
 
snowbrood

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littlecloud11 said:
You know how monochromatic light gives dark and white bands, the same happens for white light, except that the white region splits to form 7 different colors because white light is composed of 7 colors. The dark band remains as it is.
Click to expand...
do u agree?
with my answer as 625 ??
 
Ahmedraza73

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snowbrood said:
fundamental wavelength =2L
L = 1/2 lambda
lambda = 2L
V is constant both cases, so f1(lambda1) = f2(lambda2)
(250)(2) = 0.8f
f = 625
answer is 625 i had contacted many phds uni teachers and students they all agree with this answer
Click to expand...
I even got the answer as 625 but there was a difference of 15 in the mcQ's answer and my answer so i might think its wrong:
Anyways thnx brood:)
 
snowbrood

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Nazib Intesar said:
What is the detailed solution of:-
View attachment 22028
Click to expand...

let energy before rebounding E1 and energy after rebounding E2
energy is proportional to mgh
m and g are constant here they arent changed

since m and g are constants dividing E2 by E1 we get 0.5mgh/mgh=1/2
kinetic energy=(1/2)mv^2
now lets go kinetic energy part (1/2)m is constant here


kinetic energy=1/2mu^2 before rebounding
kinetic energy=1/2mv^2 after rebounding
ratio of k.e after rebounding/ k.e before rebounding=v^2/u^2
before rebounding potiential energy lose and kinectic energy gain
after rebounding potiential energy gain and kinetic energy lose but remember energy is conserved
p.e is converted into k.e
and k.e into p.e
so we can say kinetic energy before rebounding equals E1
and k.e before rebounding equals E2

since we want ratio of u/v
so
v^2/u^2=1/2
square rooting both sides we get
v/u=1/sqrt(2)
 
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