Physics: Post your doubts here!

[farhan141]

see 1033 at
http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-216.html

I got the primary coil thing but not the link/cut/thread secondary coil part

 

[nadeen64]

Can someone please explain this question? Answer is D

 

[Rizwan Javed]

Can someone please explain this question? Answer is D

View attachment 60213

When two objects have an elastic collision, the relative speed of approach is equal to the relative speed of separation.
When two objects are moving towards each others, the relative speed of one object relative to the other is given by sum of the speeds of two objects. So relative speed of appraoch here is u1 + u2.

When two objects are moving in a same direction, then the relative speed of an object is calculated by subtracting the individual speeds of the two objects. So the relative speed of separation here is given by: v2-v1

As stated above: "When two objects have an elastic collision, the relative speed of approach is equal to the relative speed of separation. ", the equation for this situation will become:

u1 + u2 = v2 - v1

So answer is D.

 

[Saad the Paki]

When two objects have an elastic collision, the relative speed of approach is equal to the relative speed of separation.
When two objects are moving towards each others, the relative speed of one object relative to the other is given by sum of the speeds of two objects. So relative speed of appraoch here is u1 + u2.

When two objects are moving in a same direction, then the relative speed of an object is calculated by subtracting the individual speeds of the two objects. So the relative speed of separation here is given by: v2-v1

As stated above: "When two objects have an elastic collision, the relative speed of approach is equal to the relative speed of separation. ", the equation for this situation will become:

u1 + u2 = v2 - v1

So answer is D.

Is this method valid only when the masses of the 2 objects are equal or whenever there is an elastic collision?

 

[nadeen64]

When two objects have an elastic collision, the relative speed of approach is equal to the relative speed of separation.
When two objects are moving towards each others, the relative speed of one object relative to the other is given by sum of the speeds of two objects. So relative speed of appraoch here is u1 + u2.

When two objects are moving in a same direction, then the relative speed of an object is calculated by subtracting the individual speeds of the two objects. So the relative speed of separation here is given by: v2-v1

As stated above: "When two objects have an elastic collision, the relative speed of approach is equal to the relative speed of separation. ", the equation for this situation will become:

u1 + u2 = v2 - v1

So answer is D.

Im still confused. When the objects are approaching, they are in different directions and when they're separating, they go in the same direction. so why not A?

 

[Rizwan Javed]

Is this method valid only when the masses of the 2 objects are equal or whenever there is an elastic collision?

It is applicable whenever objects have an elastic collision.

 

[Rizwan Javed]

Im still confused. When the objects are approaching, they are in different directions and when they're separating, they go in the same direction. so why not A?

Just remember a general rule. For finding the relative speed of two objects travelling in opposite directions, you add the speeds. When they are in same direction, we subtract their speeds.
There's a big explanation behind this why we do so. I can tell you that, if you wish.

 

[Saad the Paki]

Just remember a general rule. For finding the relative speed of two objects travelling in opposite directions, you add the speeds. When they are in same direction, we subtract their speeds.
There's a big explanation behind this why we do so. I can tell you that, if you wish.

I'd appreciate if you could

 

[Rizwan Javed]

I'd appreciate if you could

I explained this thing to another member here, so I'm just copy-pasting that explanation

Consider two objects: A & B. Now imagine that these two objects are moving towards each other with speeds v1 and v2 respectively. For relative motion of the objects, consider one object to be stationary. Let that be A. Now from the perspective of A, B will APPEAR to move towards A with a speed of v1 + v2. <--- this speed is not the actual speed of B, rather this is the speed by which it APPEARS to be moving towards A. Let me give you a practical example. Have you ever noticed that while you're sitting in a moving car and another car is coming from the opposite in front of you, the other car when crosses you, it seems like it is travelling with a 10000km/h speed? Have you ever noticed? This is what the relative speed is. This is not the actual speed of that car, rather this is what is observed by YOU when you are moving also! So in a nutshell, when two objects are moving towards each other, there speeds add up, to give a relative speed observed by any one of those two objects. (with the observer considering him/herself to be stationary!)

A similar things happens when two objects are travelling in opposite directions.

However, a totally different scenario is observed when two objects are moving in the same direction. In this case, the velocity of one of the objects is subtracted from the other objects velocity. Again take the example of the car thingy. Now imagine you are sitting in you car having a race with your class fellow ; just suppose this. You are driving at 80m/s and your friend's driving at 82m/s. Now even though your friend is travelling at such a fast speed, but when you'll see her, you'll think that she's just travelling at 2 m/s! And if she's travelling at the same speed as you, you'll think that she's not even moving! (this is what the magic relative velocity do!). But it just APPEARS to YOU (because you are the observer and you are considering yourself to be stationary) . It's not happening in the reality. So to sum up, when two objects are moving in the same direction, their speed are subtracted. v1 - v2.

Now relate this thing to momentum.
Two objects of equal masses are moving toward each other with speeds v1 & v2. They have a perfectly elastic collision, and move away from each other with speeds v3 & v4. So for this collision to be elastic, the relative speed of approach: v1 + v2, must be equal to relative speed of separation v3 + v4.

I hope I remained coherent enough, lol

 

[ashcull14]

Because the current in the primary coil is in phase with the flux in the core <-------- How do we know this??
But not the change in the flux. <-------- change is flux wont be generated as we using ac?

So why does change in the flux not in hase with flux in the core.

the magnetic force required to produce this changing magnetic flux (Φ) must be supplied by a changing current through the coil. this force generated by an electromagnet c is equal to the amount of current through that coil (in amps) multiplied by the number of turns of that coil around the core . Because the mathematical relationship between magnetic flux and magnetic force is directly proportional, and because the mathematical relationship between magnetic force and current is also directly proportional , the current through the coil will be in-phase with the flux wave.

the primary coil's induced voltage must remain at the same magnitude and phase in order to balance with the applied voltage, in accordance with Kirchhoff's voltage law. Consequently, the magnetic flux in the core cannot be affected by secondary coil current. However, what does change is the amount of magnetic force in the magnetic circuit.

this force is produced any time electrons move through a wire. Usually, it is accompanied by magnetic flux, ( "magnetic Ohm's Law" not in our course). In this case, though, additional flux is not permitted, so the only way the secondary coil's magnetic force may exist is if a counteracting magnetic force is generated by the primary coil, of equal magnitude and opposite phase. Indeed, this is what happens, an alternating current forming in the primary coil -- is 180 degree out of phase with the secondary coil's current -- to generate this counteracting magnetic force and prevent additional core flux.
it is very complicated
u dont need to know a lot of details as its not in our course
so dont worry about it
this is just a general idea

 

[ashcull14]

 

[The Sarcastic Retard]

View attachment 60226

0.18 :3

 

[The Sarcastic Retard]

the magnetic force required to produce this changing magnetic flux (Φ) must be supplied by a changing current through the coil. this force generated by an electromagnet c is equal to the amount of current through that coil (in amps) multiplied by the number of turns of that coil around the core . Because the mathematical relationship between magnetic flux and magnetic force is directly proportional, and because the mathematical relationship between magnetic force and current is also directly proportional , the current through the coil will be in-phase with the flux wave.

the primary coil's induced voltage must remain at the same magnitude and phase in order to balance with the applied voltage, in accordance with Kirchhoff's voltage law. Consequently, the magnetic flux in the core cannot be affected by secondary coil current. However, what does change is the amount of magnetic force in the magnetic circuit.

this force is produced any time electrons move through a wire. Usually, it is accompanied by magnetic flux, ( "magnetic Ohm's Law" not in our course). In this case, though, additional flux is not permitted, so the only way the secondary coil's magnetic force may exist is if a counteracting magnetic force is generated by the primary coil, of equal magnitude and opposite phase. Indeed, this is what happens, an alternating current forming in the primary coil -- is 180 degree out of phase with the secondary coil's current -- to generate this counteracting magnetic force and prevent additional core flux.
it is very complicated
u dont need to know a lot of details as its not in our course
so dont worry about it
this is just a general idea

Oh my!! I will read it later and ask doubts if I have in it. :3
Well thanks a lot for the time u took out for this... ^_^

 

[Xylferion]

View attachment 60226

You already know that the potential at A is 0.20. That means V in = 0.2

You're trying to find out what V out is, assuming there is a "linear" change in the voltmeter reading.



With a linear change, the feedback resistance at 15 degrees Celsius is 2500 ohms.

V out / V in = 2500 / 740
V out = ( 2500 / 740 ) * 0.20
V out = 0.68 V

Hope that helps!

 

[ashcull14]

0.18 :3

sorry didnt get wat u said

 

[ashcull14]

You already know that the potential at A is 0.20. That means V in = 0.2

You're trying to find out what V out is, assuming there is a "linear" change in the voltmeter reading.

View attachment 60227

With a linear change, the feedback resistance at 15 degrees Celsius is 2500 ohms.

V out / V in = 2500 / 740
V out = ( 2500 / 740 ) * 0.20
V out = 0.68 V

Hope that helps!

ure ans r always well explained thanks

 

[qwertypoiu]

I can't visualise this question
Which part of the needle is oscillating and how is it oscillating( which direction)? why is the amplitude 11mm? + in bi needle should be at its max height, i get it , but why is this max height 14mm below cloth or 8mm above cloth?
Can someone draw on the figure and explain
Please



biView attachment 60196

The whole needle is moving! Up and down. Like those knitting machines.

The position shown in the diagram is the highest position of the needle. They tell us that the needle moves down 22mm and comes up 22 mm. Since the gap between the sharp end of needle and the cloth below is only 8mm, it will poke through it! So to your question:

Why is amplitude 11mm? When they say it goes down 22 mm, they are quoting the distance from the maximum position to the minimum position:

Spoiler: image

Of course 'wave height' above does not refer to the amplitude, rather it is twice it.

So your amplitude is 22mm/2 = 11m

Now they already told us the maximum position of needle is 8.0mm above the cloth... So what's confusing? The other "maximum" is when the needle has moved 22mm down, so that it's 8 - 22 = -14mm above cloth (or simply 14mm below cloth)

Hope that makes sense

 

[princess Anu]

View attachment 60226

Which year is this

 

[princess Anu]

The whole needle is moving! Up and down. Like those knitting machines.

The position shown in the diagram is the highest position of the needle. They tell us that the needle moves down 22mm and comes up 22 mm. Since the gap between the sharp end of needle and the cloth below is only 8mm, it will poke through it! So to your question:

Why is amplitude 11mm? When they say it goes down 22 mm, they are quoting the distance from the maximum position to the minimum position:

Spoiler: image

View attachment 60233

Of course 'wave height' above does not refer to the amplitude, rather it is twice it.

So your amplitude is 22mm/2 = 11m

Now they already told us the maximum position of needle is 8.0mm above the cloth... So what's confusing? The other "maximum" is when the needle has moved 22mm down, so that it's 8 - 22 = -14mm above cloth (or simply 14mm below cloth)

Hope that makes sense

Yeah it sure did THANKS!!
One thing more, why have they shown 22mm label at that point?:/ that is what confused me in the first place

 

[qwertypoiu]

Yeah it sure did THANKS!!
One thing more, why have they shown 22mm label at that point?:/ that is what confused me in the first place

That is just to show the distance through which that block in the top moved as the needle oscillated. They could show it anywhere I guess. The point is just to show wave height.