Rod's Physics 3/4 Questions Thread

[Rod]

http://i.imgur.com/DMWBmRl.jpg
Isn't the best example due to time constraints sorry.
Clipping occurs when the output voltage of the amplifier is constant even though the input voltage is varying.
In the transfer characteristic graph, this would occur when the input voltage is below 3V or above 5V. We can see that even if we input 1V, we still get a 4V output, same with 2V and 3V.

Comparing the difference between the two left hand graphs, we can see that parts of the output graph have been 'clipped', with their tops and bottoms missing. Examining the top of the first peak, we can see that it goes above 5V all the way to 6V.
Back to the transfer graph, 5V will go to 10v but 6V will still only go to 10V.
Thus when we draw our output graph, we see a flat line.
Anything above the flat line has been 'clipped', in that we have lost the original signal, akin to distortion in speakers when they are too loud.

Essentially when doing clipping questions you do nothing different from the rest, simply take the input voltage, find the output voltage based on the amplifier and you are finished. There aren't any special steps etc.

Ahh that was an awesome example Steven. I finally understand now haha!

Thanks so much

So to summarise:
We basically use the same rules to sketch a clipped output graph from an input and transfer characteristic. Except, we know this time the graph will be clipped because the transfer characteristic shows that the output voltage will remain constant in certain times. So we just graph it like we did before, except there will be flat lines indicating the clipping.

I have my SAC on this in three weeks. It's going to be an exam style test. I've going through all the hard stuff like plotting the graphs and modulation. Would you recommend doing anything else? Topics/questions that tripped you up in your SAC or the exam (unlikely though haha)

Thanks

[knightrider]

how would you do these questions

Calculate the energy in electronvolts of
a)an alpha particle with 8.5*10^-12 J of energy
b)a beta particle with 6.4*10^-11 J of energy
c)a gamma ray with 4.7*10^-11 J of energy

[lzxnl]

Remember, the electron volt is a much smaller unit of energy than the joule, so you need more electron volts for the same amount of energy.
1 eV = 1.6 * 10^-19 J
So 1 J = 6.25 * 10^18 eV
See if that helps.

[Rod]

Hi everyone, just struggling to understand the physics behind this.

It's about the conservation of momentum.

So two roller bladers and rolling, one, who is moving faster than the one at front catches up, they both hold hands and move at the same speed (although the speed will be less than what they were travelling before, but momentum before will equal to the momentum after).

After they let go of their hands, they both continue to move at the same speed, if there is no external forces acting on them.

My answer was correct, but I just don't understand why 'After they let go of their hands, they both continue to move at the same speed, if there is no external forces acting on them'. Can someone please explain?

Thanks!

[Rishi97]

Hi everyone, just struggling to understand the physics behind this.

It's about the conservation of momentum.

So two roller bladers and rolling, one, who is moving faster than the one at front catches up, they both hold hands and move at the same speed (although the speed will be less than what they were travelling before, but momentum before will equal to the momentum after).

After they let go of their hands, they both continue to move at the same speed, if there is no external forces acting on them.

My answer was correct, but I just don't understand why 'After they let go of their hands, they both continue to move at the same speed, if there is no external forces acting on them'. Can someone please explain?

Thanks!

I'm no expert in physics but I assume it means that since there are no forces such as friction, air resistance acting specifically on the roller bladers, they will continue to move off in a constant speed. This also satisifes Newton's 1st law. Every object will continue to move in a straight line or at a constant speed unless acted on by a force.
Hope this is right

[Rod]

I'm no expert in physics but I assume it means that since there are no forces such as friction, air resistance acting specifically on the roller bladers, they will continue to move off in a constant speed. This also satisifes Newton's 1st law. Every object will continue to move in a straight line or at a constant speed unless acted on by a force.
Hope this is right

Thanks Rishi

But I was just wondering why they both remain at the same speeds and how that still applies to the laws of conservation of momentum

thanks

[Thorium]

Thanks Rishi

But I was just wondering why they both remain at the same speeds and how that still applies to the laws of conservation of momentum

thanks

When together: total p=(m1+m2)v
When separated, due to their inertia, they keep moving with the same speed (as Rishi said). This means that when separated: total p=m1v+m2v=(m1+m2)v again. Thus the momentum in an isolated system is conserved.

Hope that helps =)

[Rod]

When together: total p=(m1+m2)v
When separated, due to their inertia, they keep moving with the same speed (as Rishi said). This means that when separated: total p=m1v+m2v=(m1+m2)v again. Thus the momentum in an isolated system is conserved.

Hope that helps =)

Thanks, lets test it out

Say one weights 50 kg, and the other 20 kg, and are travelling at 5 m/s-1

so:
When together: total p=(m1+m2)v : p= (50+20)5 = 350 kg m/s

When separated, due to their inertia, they keep moving with the same speed (as Rishi said). This means that when separated: total p=m1v+m2v=(m1+m2)v again. Thus the momentum in an isolated system is conserved.

so: p= m1v + m2v should equal 350
= 50x5 + 20x5 = 350 kg m/s

Yep you guys are right, thanks Rishi and Thorium

So now I understand the maths behind it, still don't understand why it happens.

[Thorium]

Thanks, lets test it out

Say one weights 50 kg, and the other 20 kg, and are travelling at 5 m/s-1

so:
When together: total p=(m1+m2)v : p= (50+20)5 = 350 kg m/s

When separated, due to their inertia, they keep moving with the same speed (as Rishi said). This means that when separated: total p=m1v+m2v=(m1+m2)v again. Thus the momentum in an isolated system is conserved.

so: p= m1v + m2v should equal 350
= 50x5 + 20x5 = 350 kg m/s

Yep you guys are right, thanks Rishi and Thorium

So now I understand the maths behind it, still don't understand why it happens.

When they separate, they do not push each other away, they only stop holding hands. So They have the same speed and same momentum because they have no external force applied to them. And as you know, when no other force is applied, an object remains as it is.

[Rod]

When they separate, they do not push each other away, they only stop holding hands. So They have the same speed and same momentum because they have no external force applied to them. And as you know, when no other force is applied, an object remains as it is.

Thanks so much man. I fully understand.

You too Rishi thank you

[Rod]

Another question has popped up, I still don't understand the reasons of photo diode being reverse-biased

I know that b/c of this it can be switched on and off, and current cannot be reversed.

[silverpixeli]

Just giving this thread a bump so it comes up in my 'new replies to your posts' section. Rod, I don't check AN a lot, but if I'm passing by and see something I can help with I'll give you my input.

[Rod]

Just giving this thread a bump so it comes up in my 'new replies to your posts' section. Rod, I don't check AN a lot, but if I'm passing by and see something I can help with I'll give you my input.

Thanks so much Silver,

Really appreciate all this help .

Good night

[Rod]

An enormous truck is about to hit three stationary carriages. The truck, and each of the three carriages are the same mass. Will the force applied by the truck onto the carriages be higher? Or will the force applied by the stationary carriages be higher?

By calculation, the experience the same impulse. Maybe because of the newton's third law? Not 100% sure.

Thanks

[Rod]

Can someone help please? Two questions before and nobody answered . Is everyone busy at uni?

I've just got my head around amplitude modulation and nek minit I see the study design and they want us to know about 'light intensity modulation'. Is that the same thing as amplitude modulation? If not what's the difference? Can someone also post some links thanks