How to Calculate Time and Uncertainty for Rotating Pulsars

  • Thread starter Thread starter scmejla
  • Start date Start date
  • Tags Tags
    Units
AI Thread Summary
The discussion focuses on calculating the time and uncertainty for a pulsar that rotates every 1.400 806 448 872 75 ms, with a specified uncertainty. The first part, calculating the number of rotations in 30 days, was correctly solved as approximately 1.85 billion. For the second part, the correct time for 2 million rotations is approximately 2801.6129 seconds, emphasizing the need for precision in calculations. The uncertainty associated with this time can be determined by calculating the time using the maximum and minimum rotation times and averaging the results. The final answer highlights the importance of proper unit conversion and precision in calculations.
scmejla
Messages
4
Reaction score
0
Hello, perhaps you could help with a problem.

Suppose a pulsar rotates once every 1.400 806 448 872 75 + or - 5 ms, where the trailing + or - 5 indicates the uncertainty in the last decimal place (it does not mean + or - 5ms).

a. How many times does the pulsar rotate in 30.0 days?
I solved this one, 30days(60s/min)(60min/hr)(24hr/day) = 1.85(10^9)

b. How much time does the pulsar take to rotate 2.0(10^6) times?
(Give your answer to at least 4 decimal places.)

c. What is the associated uncertainty of this time?

It's these last two that are giving me issues, for b. I figured you just multiply the amount of rotations by the time it takes to rotate, I came up with 2801.61 s but according to webassign is wrong. I even stuffed it with 2 zeros to go out 4 decimal places. I'm just not sure about c, though I need the answer to b. to start.
 
Physics news on Phys.org
scmejla said:
Hello, perhaps you could help with a problem.

Suppose a pulsar rotates once every 1.400 806 448 872 75 + or - 5 ms, where the trailing + or - 5 indicates the uncertainty in the last decimal place (it does not mean + or - 5ms).

a. How many times does the pulsar rotate in 30.0 days?
I solved this one, 30days(60s/min)(60min/hr)(24hr/day) = 1.85(10^9)
Your notation is misleading! (30)(60)(60)(24)= 2592000. Then you multiplied by 1000 ms/s and divided by 1.400 806 448 872 75 ms.

b. How much time does the pulsar take to rotate 2.0(10^6) times?
(Give your answer to at least 4 decimal places.)

c. What is the associated uncertainty of this time?

It's these last two that are giving me issues, for b. I figured you just multiply the amount of rotations by the time it takes to rotate, I came up with 2801.61 s but according to webassign is wrong. I even stuffed it with 2 zeros to go out 4 decimal places.
What do you mean "stuffed it with 2 zeros? 2(10^6)= 2000000 and since you are multiplying by approximately 1.4, you should get about 2800000.
I get 2801612.8977455.

I'm just not sure about c, though I need the answer to b. to start.
1.400 806 448 872 75 + or - 5 ms, could be as large as 1.400 806 448 872 75 +5 = 1.400 806 448 872 80 ms or a small as 1.400 806 448 872 75 -5 = 1.400 806 448 872 70 ms. Do your calculation using each of these, then find the average of the two extremes. The uncertainty will be that average + or - something.
 
Thank you for your help, you've led me to the correct answer. It is 2801.6128977455, you must have left out the (ms) or 10^-3 when multiplying. My calculator was only set up to display 6 digits. Thanx again.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top