Answer is 3.1*10^7 seconds in a year
The Earths sidereal rotation period is 23 hours and 56 seconds. = =
How does a planet's distance from the sun affect its period of revolution?
Mars - 24.6 hrs
Depending on the number of leap years in the period (14 or 15), there will be 1893369600 or 1893456000 seconds - plus any leap seconds which may be added.
The polar rotational period of Jupiter is 9 hours, 55 minutes, 40.6 seconds. The equatorial rotational period is 9 hours, 50 minutes, 30 seconds.
The earth rotates 360 degrees every 23 hours, 56 minutes and 4.100352 seconds.
Period = 1/Frequency = 0.00175 seconds (approx)Period = 1/Frequency = 0.00175 seconds (approx)Period = 1/Frequency = 0.00175 seconds (approx)Period = 1/Frequency = 0.00175 seconds (approx)
27
27
The SI unit for period is seconds and the symbol is t (because the period is a time measurement, it is expressed in the SI unit seconds)
500 metres is 0.31 miles: whether that distance is covered in a nanosecond or over a period of years.
Period = 1/78.6 seconds = 0.01272 seconds
The Earths sidereal rotation period is 23 hours and 56 seconds. = =
Frequency = (1)/(period) .If the period is still 4 seconds, then the frequency = (1)/(4 seconds) = 0.25 per second = 0.25 Hz.
The frequency is the reciprocal of the period. In other words, divide 1 by the period. If the period is in seconds, the frequency is in hertz.
None. 2.5 seconds is a period of time. It is not a speed since there is no measure of distance involved. By contrast, yards per minte is a measure of speed. The two measures are incompatible.
(8 seconds) / (5 periods) = 1.6 seconds per period