a) An electron in an atom has a speed of 2.2 × 10^6 m/s and orbits the nucleus at a distance of 5 × 10^-11m. What is its centripetal acceleration?
(already solved)
9.68x 10^22 m/s^2
b) A neutron star of radius 17 km is found to rotate at 5 revolutions per second. What is the centripetal acceleration of a point on its equator?
To find the centripetal acceleration, we can use the formula:
Centripetal acceleration (a) = (velocity (v))^2 / radius (r)
a) Given that the speed of the electron is 2.2 × 10^6 m/s and the distance from the nucleus is 5 × 10^-11 m, we can substitute these values into the formula:
a = (2.2 × 10^6 m/s)^2 / (5 × 10^-11 m) = 9.68 × 10^22 m/s^2
Therefore, the centripetal acceleration of the electron is 9.68 × 10^22 m/s^2.
b) In this case, we are given the number of revolutions per second instead of the linear velocity. To find the linear velocity, we need to convert the number of revolutions per second to radians per second.
1 revolution = 2π radians
So, to convert the 5 revolutions per second to radians per second, we multiply it by 2π:
Linear velocity (v) = 5 revolutions/second * 2π radians/revolution = 10π radians/second
Now, we can use the formula for centripetal acceleration:
a = (v)^2 / r
Given that the radius is 17 km or 17,000 m, we can substitute the values:
a = (10π radians/second)^2 / 17,000 m = (100π^2)/17,000 m ≈ 5.9 m/s^2 (rounded to one decimal place)
Therefore, the centripetal acceleration of a point on the equator of the neutron star is approximately 5.9 m/s^2.