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March 29, 2024, 12:04:36 pm

Author Topic: VCE Physics Question Thread!  (Read 603425 times)  Share 

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paper-back

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Re: VCE Physics Question Thread!
« Reply #870 on: February 25, 2015, 08:57:04 pm »
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Normal force always acts perpendicular to the direction of motion. In this case, if its traveling along flat ground, the net force vertically is 0 as the mower isnt moving up or down, but rather left and right. Thus, using f=mg, f=14.5x10 (take gravity as 10m/s^2) gives us 145N (force of gravity acting downwards). Thus the reaction (normal) force is 145N in the opposite direction (acting upwards). In this case, the ground is pushing the mower with a force of 145N. Check the answers because i might of interpreted the question wrong (next time post a photo of the actual whole question with diagram if there is, makes it a bit easier).
Thanks Cosec.
That's what I thought, however I believe the answer for the question adds in the verticle component of the force being exerted on the mower for the verticle force as it's close to 200

Cosec

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Re: VCE Physics Question Thread!
« Reply #871 on: February 25, 2015, 09:21:49 pm »
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Thanks Cosec.
That's what I thought, however I believe the answer for the question adds in the verticle component of the force being exerted on the mower for the verticle force as it's close to 200

Hmm, i thought of that too, but i dont see why it would be added and ive never encountered a question that takes it into consideration. But if thats the answer, i guess thats the way they expect you to work it out in which case you just find the vertical component and add it to the weight force. Someone else with a bit more experience might have to chime in.

Conic

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Re: VCE Physics Question Thread!
« Reply #872 on: February 25, 2015, 09:46:52 pm »
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You do have to consider the component of the pushing force on the lawnmower. If the normal force simply cancelled out the force of gravity, the net force in the vertical direction would be the downward component of the pushing force. This would mean the lawnmower accelerates into the ground, but this is clearly not the case.

What do we know in this situation? We know the gravitational force, and we can work out downward component of the pushing force. We need to relate the normal force to the things we already know. Using Newton's Second Law, we know that the net vertical force is 0, as there is no acceleration in this direction. The normal force is in the opposite direction to the weight force and the pushing force, so



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Re: VCE Physics Question Thread!
« Reply #873 on: February 26, 2015, 05:22:11 pm »
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The average speed for the uphill part of the journey is 4ms^1 while the average speed for travelling the same distance for downhill is 8ms^1. What is the average speed of the rider for whole journey. You may ignore time taken to turn around. Theres an algebraic method, in letting distances equal right, but from there i get confused
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Re: VCE Physics Question Thread!
« Reply #874 on: February 26, 2015, 05:22:31 pm »
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Acceleration is 0
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #875 on: February 26, 2015, 05:42:55 pm »
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The average speed for the uphill part of the journey is 4ms^1 while the average speed for travelling the same distance for downhill is 8ms^1. What is the average speed of the rider for whole journey. You may ignore time taken to turn around. Theres an algebraic method, in letting distances equal right, but from there i get confused

Let the distance be x. Work out the time travelled going up and down hill in terms of x, find the total distance and total time to work out the average speed.
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Re: VCE Physics Question Thread!
« Reply #876 on: February 26, 2015, 08:17:02 pm »
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Let the distance be x. Work out the time travelled going up and down hill in terms of x, find the total distance and total time to work out the average speed.
I tried to find time as 4t=x and 8t=x however i get t=0?
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Re: VCE Physics Question Thread!
« Reply #877 on: February 26, 2015, 10:44:31 pm »
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The two times clearly aren't the same as the speeds aren't the same. Hence you need x = 4t1 = 8t2
Total time = t1 + t2, total distance = 2x
So average speed should be 2x / (t1 + t2)
Simplify with what you have now.
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Re: VCE Physics Question Thread!
« Reply #878 on: March 02, 2015, 08:21:06 pm »
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Can someone help me with question 5. This is from the detailed study materials and structures, Young's Modulus.
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Re: VCE Physics Question Thread!
« Reply #879 on: March 02, 2015, 08:38:48 pm »
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Can someone help me with question 5. This is from the detailed study materials and structures, Young's Modulus.
A strain percentage of 0.075% is equal to a strain of 0.00075. Also, note that while I don't convert the cm measurements to meters, the output is also in cm.










I hope that was correct ahah!
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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #880 on: March 02, 2015, 09:22:32 pm »
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Thank you :) you were correct haha. Also what detailed study do most schools do? My teacher says everyone does materials and structures because it is the easiest.
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Re: VCE Physics Question Thread!
« Reply #881 on: March 02, 2015, 09:26:41 pm »
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Thank you :) you were correct haha. Also what detailed study do most schools do? My teacher says everyone does materials and structures because it is the easiest.
I think the examiners report tells you
But yeah definately structures, it's so straightforward.
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Re: VCE Physics Question Thread!
« Reply #882 on: March 03, 2015, 09:22:23 am »
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I think the examiners report tells you
But yeah definately structures, it's so straightforward.

yeah either structure and sound (what i did) are the easiest afaik
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Re: VCE Physics Question Thread!
« Reply #883 on: March 03, 2015, 05:53:55 pm »
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Could someone please explain to me how the equation for resistance total for parallel circuits proves the validity of Kirchoff's Law?

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Re: VCE Physics Question Thread!
« Reply #884 on: March 03, 2015, 09:15:36 pm »
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When finding the kinetic energy of a projectile in motion do we use the horizontal component, verticle component or the resultant vector component?