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Physics: Post your doubts here!

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A bow of mass 400g shoots an arrow of mass 120g vertically upwards. The potential energy stored in the bow just before release is 80J. The system has an efficiency of 28%. What is the height reached by the arrow when air resistance is neglected?
A 4m
B 19m
C 187m
D 243m
 
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A bow of mass 400g shoots an arrow of mass 120g vertically upwards. The potential energy stored in the bow just before release is 80J. The system has an efficiency of 28%. What is the height reached by the arrow when air resistance is neglected?
A 4m
B 19m
C 187m
D 243m
B?
 
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If the answer is B then this is how.
We know that PE= 80 J just before release, so use efficiency equation and PE=mgh

Efficiency is 28%
We know efficiency = output/input
28/100= output/ 80
Output = 22.4 J
Now use PE= mgh
Mass of arrow= 120 g= 0.12 Kg since arrow is shot alone
22.4= 0.12(10)(h)
h= 18.7 m ~= 19 m
I'm not sure though I might be wrong
 
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Someone please help. MJ 2010 paper 22 question question 7 part b ii 2, question 4 part c, question 3 part c.
 
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As you are a staff member, I would like to ask why this website is down, it has been so long since it was functioning here in the Middle East . I was really dependent on it, but now when CIE are approaching , the website is not working. Pls repair ASAP.
 
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Xylferion

Could you please tell me hoe to take uncertainity in physics p3?

It depends on what type of uncertainty you're working with.

There's the random uncertainty in which you use a range of readings. This is:
(Maximum value - Minimum value) / Number of readings

The random uncertainty is the absolute uncertainty. It's the +- value you see after the main value.
Eg. 28 +- 0.02 seconds, 0.02 seconds is the absolute/random uncertainty

---------------------

Then there's the uncertainty involved with taking readings from instruments, like a voltmeter/scale/stopwatch etc. These are different for each instrument, but basically, the precision of each instrument is the absolute uncertainty.

Voltmeter precision = 0.01 V
If you have a voltmeter with voltage 5.67 V.

Your final value will be 5.67 +- 0.01 V

----------------------------

Next there's percentage uncertainty,

Absolute uncertainty ( this is the precision of the instrument you're using in the exam ) / Your reading * 100

So if your reading is 15 V. The precision of the voltmeter is 0.01

Your percentage uncertainty is [ 0.01/15 ]*100
 
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I need help for question 6(b). If its possible please explain in detail. I understand the basics of phase difference but when it comes to applying it in questions, i go blank
http://www.acethem.com/pastpapers/a...uestion-paper-2011-summer-paper-2-15425.html/

Thank you.

You cannot find the time period from a displacement/distance graph, but you can find the wavelength.

Firstly though, form any relations you can between the values that have changed. They tell you that waves move "forward" by 0.25T.

T = 1/ƒ
0.25T = 1/4ƒ

The above indicates that the frequency will quadruple, If the waves move forward by 0.25T.

v = ƒλ
λ = v/ƒ

If ƒ is quadrupled, you will take a quarter of lambda.

1/4λ = v/4ƒ

From the graph itself, you can see that the wavelength of 1 wave is 80 cm. If the waves move forward by 0.25T, they move forward by 0.25λ according to the above relations.

0.25λ = 80/4 = 20 cm

The waves will move forward by 20 cm. So your peaks move from 20 and 60 ---> 40 and 80, and so do all the other points on the wave.

Screen Shot 2015-04-21 at 12.50.24 PM.png

Hope that made sense! ;)
 
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It depends on what type of uncertainty you're working with.

There's the random uncertainty in which you use a range of readings. This is:
(Maximum value - Minimum value) / Number of readings

The random uncertainty is the absolute uncertainty. It's the +- value you see after the main value.
Eg. 28 +- 0.02 seconds, 0.02 seconds is the absolute/random uncertainty

---------------------

Then there's the uncertainty involved with taking readings from instruments, like a voltmeter/scale/stopwatch etc. These are different for each instrument, but basically, the precision of each instrument is the absolute uncertainty.

Voltmeter precision = 0.01 V
If you have a voltmeter with voltage 5.67 V.

Your final value will be 5.67 +- 0.01 V

----------------------------

Next there's percentage uncertainty,

Absolute uncertainty ( this is the precision of the instrument you're using in the exam ) / Your reading * 100

So if your reading is 15 V. The precision of the voltmeter is 0.01

Your percentage uncertainty is [ 0.01/15 ]*100
Thanks..

Bur recently I did one practical, and in it we had to time the oscillation for 10 swings. The stopwatch provided had precision of o.o1 s. So I took the uncertainity by 0.01/my value*100. But the mark scheme said the uncertainity should be between 0.2s to 0.5s .
The question paper was WINTER 2014 paper 36, Question 2 diii. Do check and please reply.
 
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Thanks..

Bur recently I did one practical, and in it we had to time the oscillation for 10 swings. The stopwatch provided had precision of o.o1 s. So I took the uncertainity by 0.01/my value*100. But the mark scheme said the uncertainity should be between 0.2s to 0.5s .
The question paper was WINTER 2014 paper 36, Question 2 diii. Do check and please reply.

Did you find the time for 1 swing? I'm sure that would work.
 
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No. And that was not required.
Did you see the Question?

Keyword is "estimate" not "calculate", you need to have several readings.

Max - Min / Number of readings = absolute uncertainty

Find the mean value for t as well.

Absolute uncertainty / Mean value * 100 = Percentage uncertainty.
 
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