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Multiple Choice Help (1 Viewer)
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Fizzy_Cyst
Well-Known Member
Q1:
A geostationary satellite will appear to be motionless from the Earth, much like the stars. Therefore it is unlikely that they could be distinguished.
Q2:
Highest precision = lowest error
Q3:
Induced \, EMF \, \alpha \, - \frac{\Delta \phi }{\Delta t}
The higher the voltage in the primary, the higher the flux running from the primary and assuming 100% efficiency, the higher the flux cutting the secondary.
We know
The larger the gradient of the V vs t graph, the greater the
and remember due to Lenz' Law, the induced EMF opposes the change in flux.
Therefore for all the values of positive gradient, the induced EMF will be negative.
At the apex of the V vs t graph, the induced EMF will be zero (as is zero)
As the gradient becomes negative, the induced EMF will be positive.
A geostationary satellite will appear to be motionless from the Earth, much like the stars. Therefore it is unlikely that they could be distinguished.
Q2:
Highest precision = lowest error
Q3:
Induced \, EMF \, \alpha \, - \frac{\Delta \phi }{\Delta t}
The higher the voltage in the primary, the higher the flux running from the primary and assuming 100% efficiency, the higher the flux cutting the secondary.
We know
The larger the gradient of the V vs t graph, the greater the
Therefore for all the values of positive gradient, the induced EMF will be negative.
At the apex of the V vs t graph, the induced EMF will be zero (as
As the gradient becomes negative, the induced EMF will be positive.
romesh
Master of the Universe
These are interesting questions.
Q1. I don't think this is a very good question. Why? Because normal stars *do* move. This is why you can take long-exposure photos like this http://www.astropix.com/HTML/I_ASTROP/TRIPOD/TRIPOD2.HTM - it is just a result of the Earth rotating. However, the whole point of geostationary satellites is that they *don't* move, so you can point an antenna at the satellite and you never need to move the antenna. So if a geostationary satellite was present in one of those long-exposure photos, it would appear as a dot, not as a trail. So I would say that none of the answers are actually correct (although A is the best answer)
Q3. The other way to see this (perhaps a university-ish way though) is to realise that the induced EMF is related to the derivative of the voltage in the primary coil (this is basically restating that the induced EMF is the change in flux divided by change in time). So, you look at the voltage in the primary coil and see that there are places where the derivative suddenly changes sign, by a large amount. Like the plot steeply approaches zero, and then when it hits zero it starts going upwards fairly steeply. This corresponds to firstly the derivative changing sign, and secondly the derivative having a discontinuity (you can see that there is a discontinuity in the derivative because the voltage curve sharply changes direction like two lines meeting, rather than a continuous curve). The only answer where the induced EMF is discontinuous (has gaps) is C
Q1. I don't think this is a very good question. Why? Because normal stars *do* move. This is why you can take long-exposure photos like this http://www.astropix.com/HTML/I_ASTROP/TRIPOD/TRIPOD2.HTM - it is just a result of the Earth rotating. However, the whole point of geostationary satellites is that they *don't* move, so you can point an antenna at the satellite and you never need to move the antenna. So if a geostationary satellite was present in one of those long-exposure photos, it would appear as a dot, not as a trail. So I would say that none of the answers are actually correct (although A is the best answer)
Q3. The other way to see this (perhaps a university-ish way though) is to realise that the induced EMF is related to the derivative of the voltage in the primary coil (this is basically restating that the induced EMF is the change in flux divided by change in time). So, you look at the voltage in the primary coil and see that there are places where the derivative suddenly changes sign, by a large amount. Like the plot steeply approaches zero, and then when it hits zero it starts going upwards fairly steeply. This corresponds to firstly the derivative changing sign, and secondly the derivative having a discontinuity (you can see that there is a discontinuity in the derivative because the voltage curve sharply changes direction like two lines meeting, rather than a continuous curve). The only answer where the induced EMF is discontinuous (has gaps) is C
Fizzy_Cyst
Well-Known Member
To clarify question 2, precision is to do with the certainty of the measurements, or, the number of 'decimal places' in the measurement.
Think of measuring an A4 sheet of paper, measuring it with a ruler that has mm intervals will be more precise than measuring with a metre ruler, which only has cm intervals.
So, the smaller the measurement intervals, the greater the precision. In your question (D) has the smallest measurement intervals (mg), therefore is most precise.
Think of measuring an A4 sheet of paper, measuring it with a ruler that has mm intervals will be more precise than measuring with a metre ruler, which only has cm intervals.
So, the smaller the measurement intervals, the greater the precision. In your question (D) has the smallest measurement intervals (mg), therefore is most precise.