MC Answers (1 Viewer)

[spice girl]

Phys teachers at ruse are NOTORIOUS at getting things wrong. Therefore 5 is D.

Besides, orbital radius is distance from the star: the radius of the orbit of the planet!?

So if orbital radius is doubled, orbital period must increase, so either C or D. and u calculate it to be D

 

[superhubert]

i definetly agree with spice girl. unfortunatly i'm also beginning to agree with Lazarus...

 

[-=«MÄLÅÇhïtÊ»=-]

i agree wiv alexz

edit:
laz, if ur inducing a current, say even in a dc motor, the back emf is greatest with greatest torque. Greatest torque in this case is horizontal, so the greatest induced current is at max.

 

[McLake]

Originally posted by ske
eeep!..

why iznt 2 b?..like i was tossin up between a n b but doesnt time run faster on the SPACESHIP..?...coz from his perspective time is dilated for him?...coz i alwaiz thot it was the external observer that is dialated..but then again relativity works both ways ><..eh..som1 plz explain..

n for 4..iznt it c..coz at q gforce = 1g?..well eh..otha than that mah answers seem to match..

I agree with you ...

Somone expalin why we are wrong ...

 

[kaseita]

I didn't use lenz's law to figure q10 out
I used positive charges.
Consider a positive charge on the left side (say)
Moving from P to Q, its effectively moving up, so the 'current' is moving up, the magnetic field is moving right, so the positive charges are forced clockwise.
The induced emf is formed first, but because its in a circuit, then the current forms.
At Q to R however, the positive charge is effectively moving right. The current is therefore parallel with the magnetic field, and there is no force on the positive charge. Thus, no induced emf

 

[kaseita]

From a stationary observor, they will see that time slows down on the spaceship, and that it contracts in length

For the person in the ship, everything appears normal. That's because they're accelerating at the same speed as the ship, so they will see that the time appears normal, and that the length is normal.

Therefore the effect is on the observor, who sees them move near the speed of light, not on the person who's in the ship. They will be affected, but they just won't notice.

Thus 2 is a)

 

[McLake]

Originally posted by kaseita
From a stationary observor, they will see that time slows down on the spaceship, and that it contracts in length

For the person in the ship, everything appears normal. That's because they're accelerating at the same speed as the ship, so they will see that the time appears normal, and that the length is normal.

Therefore the effect is on the observor, who sees them move near the speed of light, not on the person who's in the ship. They will be affected, but they just won't notice.

Thus 2 is a)

DAMN, that makes sense (although the equation T = t/(...) worries me ...)

 

[Lazarus]

Originally posted by -=MLhtʻ=-
laz, if ur inducing a current, say even in a dc motor, the back emf is greatest with greatest torque. Greatest torque in this case is horizontal, so the greatest induced current is at max.

Is what I said wrong, then? If so, why? Or would it be used in a different scenario? Or what?

 

[McLake]

I read this formula somewhere:

EMF = BAcosx/t

so, make of that what you will ....

 

[BlackJack]

Originally posted by Lazarus
emf is a potential difference but it's not a voltage (it's not conservative)
you can have an emf in an open circuit.
Grr. I have a physics exam in 2 weeks as well, I should be able to do this stuff.

In a closed circuit like this it is in effect voltage I think. Potential diffeerence is created by the charges moving to one side of the bar when it moves perpendicular to a magnetic field. This wouuld in theory be the same, especially in Q, the side of the coil is moving with the magnetic field... which means there is no force on the electrons, and no emf generated...

For q2, the experiment where they got an atomic clock on a plane and flew it around the earth at a high speed.. it came back slower.

If 4 is the roller coaster question, think about when you were riding a rollercoaster, or even a car. when you travel through the bottom of a dip, or indeed on any part where the incline is increasing, you'll feel the extra force.

 

[McLake]

Originally posted by BlackJack

If 4 is the roller coaster question, think about when you were riding a rollercoaster, or even a car. when you travel through the bottom of a dip, or indeed on any part where the incline is increasing, you'll feel the extra force.

So why isn't halway up the next slope ?

 

[1234567]

i agree with bug except q4, i got c......

and yes, there is only one D in my answer.

 

[BlackJack]

When you'r halfway up the next slope, the gradient is not increasing. If it's not increasing, then there's no acceleration

[Hooray for maths explanations.]

(Look at the track and think of it as a gaph. Derive it twice and you get a graph of acceleration over the distance. That would be the force experienced vertically as you go just from the rollercoaster alone. )

 

[kaseita]

lol BJ, that was what I was trying to do...for the roller coaster one

I so knew they should've made that australia's wonderland excursion at the beginning of the year! Then I'd actually remember

 

[BlackJack]

When you'r halfway up the next slope, the gradient is not increasing. If it's not increasing, then there's no acceleration
Think of the direction of your momentum as you ride the coaster, when it's pointing into the ground, you know the ground will resist you.

[Hooray for maths explanations.]

(Look at the track and think of it as a gaph. Derive it twice and you get a graph of acceleration over the distance. That would be the force experienced vertically as you go just from the rollercoaster alone. )

 

[McLake]

Your squashing my spirit ....

 

[kaseita]

Laz, your not wrong. you've just taken the wrong period of time.

You've taken the time period from when its parallel to the field, to when its 90 degrees to the field.
However when your using that method (of changing flux), you have to take an instant in time.

So at P, when the armature is parallel to the field, in the next instant, when it begins to move upwards, its going to experience an increase in flux, so an induced emf will form that will oppose the increase in flux.

Consider just before Q, Q and just after Q (Q being 90 degrees)
Just before Q, the flux is still increasing. At Q the flux is at a maximum. Just after Q, the flux is beginning to decrease.
Somewhere inbetween, the induced emf has gone from being positive, to negative (in a directional sense). Therefore at Q, though its maximum flux, it must have a zero induced emf, as at some point inbetween the change, its gone from positive change in flux, to zero, to negative change in flux.

 

[wogboy]

For Q10, the reason the answer is B is because:

1) At Q & S, the emf is zero. Why? Because the catapult forces acting on the coil are in the direction of the magnetic field. This rules out
A & D. So the answer is either B or C.

2) At P,R,& T the emf is non-zero (maximum). Why? Because the catapult forces on the coil are PERPENDICULAR to the magentic field. This rules out C, since in C P,Q,R,S,T are all zero in emf.

This leaves the correct answer to be B. In a motor such as this, the current and voltage are always in phase.

 

[Dario]

It only goes around 360 degrees, so it is either the Sine or Cosine curve. Also, it would have to be the same amount in P,R,T and same in Q,S. I thought it has to be B because when flux threading is maximum as in position Q,S emf is 0.

 

[Dario]

Originally posted by ske
eeep!..

why iznt 2 b?..like i was tossin up between a n b but doesnt time run faster on the SPACESHIP..?...coz from his perspective time is dilated for him?...coz i alwaiz thot it was the external observer that is dialated..but then again relativity works both ways ><..eh..som1 plz explain..

n for 4..iznt it c..coz at q gforce = 1g?..well eh..otha than that mah answers seem to match..

As Jacaranda says, time dilation can expressed simply as "moving clocks run slow."