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March 29, 2024, 01:07:29 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164230 times)  Share 

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Mr. Study

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Re: Specialist 3/4 Question Thread!
« Reply #690 on: September 10, 2012, 04:28:15 pm »
0
Would someone be able to show me how they would do this question?

I've done it but I feel as though my working is unconventional...

. Find in the form .

Thank you. :)
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Truck

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Re: Specialist 3/4 Question Thread!
« Reply #691 on: September 10, 2012, 05:13:51 pm »
0
Would someone be able to show me how they would do this question?

I've done it but I feel as though my working is unconventional...

. Find in the form .

Thank you. :)

sec(pi/5) = 1/cos(pi/5) = 1/cos(2*pi/10)
= 1/1-2sin^2(pi/10), and then you just sub in your value for sin(pi/10) and it's just algebra. Eventually you get sec(pi/5) = root(5) - 1, where a=1 and b=-1.

The only other way I can think to do it is to sketch a triangle and use one of the cos double angle formulas with cos in them, but that would be dumb :P.
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Biceps

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Re: Specialist 3/4 Question Thread!
« Reply #692 on: September 10, 2012, 05:17:20 pm »
0
Would someone be able to show me how they would do this question?

I've done it but I feel as though my working is unconventional...

. Find in the form .

Thank you. :)
use double angle formula to find 1/1-2sin^2(pi/10) sub sin(pi/10) in and there you go. :)
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luke_rulz94

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Re: Specialist 3/4 Question Thread!
« Reply #693 on: September 10, 2012, 07:01:43 pm »
0
Can someone please show me how to do these two questions :


A body of mass 10 kg is attached to a second body of mass 14 kg by a light string passing over a smooth pulley attached to the edge of a horizontal table. The lighter body is on the table and the heavier body hangs vertically. The coefficient of friction between the lighter body and the table is 0.25. The system is held at rest by a horizontal force F applied to the lighter body. What is the least value of F for the body to be in limiting equilibrium?



4   A box of mass 4 kg lies on a level floor. The coefficient of friction between the floor and the box is 0.45. A horizontal force of 15 N is applied to the box. Show that the box does not move.



Truck

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Re: Specialist 3/4 Question Thread!
« Reply #694 on: September 10, 2012, 07:47:45 pm »
+2
Can someone please show me how to do these two questions :


A body of mass 10 kg is attached to a second body of mass 14 kg by a light string passing over a smooth pulley attached to the edge of a horizontal table. The lighter body is on the table and the heavier body hangs vertically. The coefficient of friction between the lighter body and the table is 0.25. The system is held at rest by a horizontal force F applied to the lighter body. What is the least value of F for the body to be in limiting equilibrium?



4   A box of mass 4 kg lies on a level floor. The coefficient of friction between the floor and the box is 0.45. A horizontal force of 15 N is applied to the box. Show that the box does not move.




Clue: For the system to be in limiting equilibrium / your box not moving, your resultant force has to = 0.
That means that if you resolve your forces (for example) vertically and horizontally, Upwards Force = Downwards Force and Right Force = Frictional Force. (alternatively, Upwards Force - Downwards Force = 0, Right Force - Frictional Force = 0).

See if that helps you at all.
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b^3

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Re: Specialist 3/4 Question Thread!
« Reply #695 on: September 10, 2012, 08:00:33 pm »
+10
EDIT: Teewreck's given you the hints, but since I've made the diagram and typed all this out..., I'm going to post it anyway, although try with teewreck's hints before you have a look below.


First of all, diagrams are your best friend in this situation :)

Labelling all the forces, assuming that if block 2 were moving it would be moving downwards. We have a normal and weight force acting on block 1,as well as the force F that is to hold it in equlibirum and a frictional force. This frictional force will be in the opposite direction to which the block will travel. So block 1 will move to the right if its not held, so the fricitonal force is to the left. There is a weigth force acting on block 2, now there is also tension in the rope over the pulley.

So now we want it to be in limiting equilibirum, that is it is on the point of moving. So our net force on each block is 0 N.

The forces acting on block 2.


The forces acting on block 1 in the  vertical direction.


The forces acting on block 1 in the horizontal direction.


For the second question, again draw a diagram out (including your forces). The key for this one is that if the force applied to the block is not greater than , then the block won't move.
« Last Edit: September 10, 2012, 08:20:29 pm by b^3 »
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pi

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Re: Specialist 3/4 Question Thread!
« Reply #696 on: September 10, 2012, 08:28:40 pm »
+4
^Why isn't this guy writing a book? The diagram alone is more than impressive! :O

Jenny_2108

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Re: Specialist 3/4 Question Thread!
« Reply #697 on: September 10, 2012, 09:33:59 pm »
0
EDIT: Teewreck's given you the hints, but since I've made the diagram and typed all this out..., I'm going to post it anyway, although try with teewreck's hints before you have a look below.


First of all, diagrams are your best friend in this situation :)
(Image removed from quote.)
Labelling all the forces, assuming that if block 2 were moving it would be moving downwards. We have a normal and weight force acting on block 1,as well as the force F that is to hold it in equlibirum and a frictional force. This frictional force will be in the opposite direction to which the block will travel. So block 1 will move to the right if its not held, so the fricitonal force is to the left. There is a weigth force acting on block 2, now there is also tension in the rope over the pulley.

So now we want it to be in limiting equilibirum, that is it is on the point of moving. So our net force on each block is 0 N.

The forces acting on block 2.


The forces acting on block 1 in the  vertical direction.


The forces acting on block 1 in the horizontal direction.


For the second question, again draw a diagram out (including your forces). The key for this one is that if the force applied to the block is not greater than , then the block won't move.

You should write Spesh and Methods books for 2013 students, b^3!!!!!!!!!
The diagram is so awesome.

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Re: Specialist 3/4 Question Thread!
« Reply #698 on: September 12, 2012, 07:30:42 pm »
0
I'm doing an extended response question for differential eqns and it asks for a prediction of a flock of emus after 5 yrs. I got 1296.42 but do i round up to 1297 or round down to 1296? The answers rounded down but i disagree as there is 0.42 of an emu left which can't just disappear...

Thanks
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Re: Specialist 3/4 Question Thread!
« Reply #699 on: September 12, 2012, 07:52:50 pm »
0
I'm doing an extended response question for differential eqns and it asks for a prediction of a flock of emus after 5 yrs. I got 1296.42 but do i round up to 1297 or round down to 1296? The answers rounded down but i disagree as there is 0.42 of an emu left which can't just disappear...

Thanks
But then again you can't pull out 0.58 of an emu from no where :P With these kinds of questions you need to round down even if the decimal is larger than 0.5
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Re: Specialist 3/4 Question Thread!
« Reply #700 on: September 12, 2012, 07:53:22 pm »
0
They would be specific about an answer on an exam though, i.e. "rounded down" or "to the nearest ten X." So don't worry too much.
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Biceps

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Re: Specialist 3/4 Question Thread!
« Reply #701 on: September 12, 2012, 09:56:00 pm »
0
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??
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Jenny_2108

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Re: Specialist 3/4 Question Thread!
« Reply #702 on: September 12, 2012, 10:02:48 pm »
0
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??





Then you use formula


Biceps

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Re: Specialist 3/4 Question Thread!
« Reply #703 on: September 12, 2012, 10:06:29 pm »
0
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??





Then you use formula


That would give the answer in terms of t though not x. wouldn't it?
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Jenny_2108

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Re: Specialist 3/4 Question Thread!
« Reply #704 on: September 12, 2012, 10:25:40 pm »
0
find the cartesian equation for: r(t)=2e^0.3t Cos(2t) i + 2e^0.3t Sin(2t) j
How on earth do you do this??





Then you use formula


That would give the answer in terms of t though not x. wouldn't it?

lol  stupid Jenny

I'm thinking we write 2e^0.3t in term of x and y then equate them