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truthfinder
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I feel stupid for asking this, but the book I'm working through doesn't explain it at all, and I want to make sure I am correct.
The book is Fundamentals of Physics, by Halliday. This is problem #16 after the first chapter.
My solutions for A and B (please forgive me if my tex isn't very good):
A:
Convert 7 days to ms
[tex]7 days \times \frac{24 hr}{1 day} \times \frac{60 min}{1 hr} \times \frac{60 s}{1 min} \times \frac{1000 ms}{1 s} = 604800000 ms[/tex]
[tex]\frac{604800000 ms}{1.55780644887275 ms} = 3.88\times 10^8 rotations[/tex]
B:
[tex]1.55780644887275 ms \times 1000000 = 1.56E6 ms[/tex]
[tex]1.56\times 10^6 ms \times \frac{1 s}{1000 ms} = 1.56\times 10^3 s[/tex]
C:
This is the part I am not sure about.
My logic is that if the last digit is ±3, that is 14 decimal places. So I took:
[tex]3\times 10^{-14} \times 1000000[/tex]
That gives 3e-8 ms
I wanted to simplify this a bit, so I converted it to seconds:
[tex]\frac{3\times 10^{-8}}{1000} = 3\times 10^{-11}[/tex]
Simplifying it to simpler units:
[tex]3\times 10^{-11} \times 1\times 10^12 = 30 ps[/tex]
Therefore my answer is ±30 ps.
But seriously, the book never mentions uncertainty, so I did the first thing that was logical to me. It seems to have the tendency to ask about things it doesn't discuss, though so far it has only been various common formulas.
The book is Fundamentals of Physics, by Halliday. This is problem #16 after the first chapter.
Time standards are now based on atomic clocks.
A promising second standard is based on pulsars, which are rotating neutron stars.
Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon.
Pulsar PSR 1937+21 is an example. It rotates once every 1.55780644887275 ±3 ms, where the trailing ±3 indicates the uncertainty in the last decimal place. It does not mean +-3 ms.
A. How many rotations does PSR 1937+21 make in 7.00 days?
B. How much time does the pulsar take to rotate exactly 1000000 times.
C. What is the associated uncertainty?
My solutions for A and B (please forgive me if my tex isn't very good):
A:
Convert 7 days to ms
[tex]7 days \times \frac{24 hr}{1 day} \times \frac{60 min}{1 hr} \times \frac{60 s}{1 min} \times \frac{1000 ms}{1 s} = 604800000 ms[/tex]
[tex]\frac{604800000 ms}{1.55780644887275 ms} = 3.88\times 10^8 rotations[/tex]
B:
[tex]1.55780644887275 ms \times 1000000 = 1.56E6 ms[/tex]
[tex]1.56\times 10^6 ms \times \frac{1 s}{1000 ms} = 1.56\times 10^3 s[/tex]
C:
This is the part I am not sure about.
My logic is that if the last digit is ±3, that is 14 decimal places. So I took:
[tex]3\times 10^{-14} \times 1000000[/tex]
That gives 3e-8 ms
I wanted to simplify this a bit, so I converted it to seconds:
[tex]\frac{3\times 10^{-8}}{1000} = 3\times 10^{-11}[/tex]
Simplifying it to simpler units:
[tex]3\times 10^{-11} \times 1\times 10^12 = 30 ps[/tex]
Therefore my answer is ±30 ps.
But seriously, the book never mentions uncertainty, so I did the first thing that was logical to me. It seems to have the tendency to ask about things it doesn't discuss, though so far it has only been various common formulas.
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