Answer:
a)
b)
Explanation:
a)
The width of the central bright in this diffraction pattern is given by:
when m is a natural number.
here:
So we have:
b)
Now, if we do m=2 we can find the distance to the second minima.
Now we need to subtract these distance, to get the width of the first bright fringe :
I hope it heps you!
a) -0.259 rad/s/y
b) 1732.8 years
c) 0.0069698 s
Explanation:
a)
The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.
Mathematically, it is given by
where
is the change in angular velocity
is the time elapsed
The angular velocity can be written as
where T is the period of rotation of the object.
Therefore, the change in angular velocity can be written as
In this problem:
T = 0.0140 s is the initial period of the pulsar
The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is
Therefore, the change in angular velocity after 1 year is
So, the angular acceleration of the pulsar is
b)
To solve this part, we can use the following equation of motion:
where
is the final angular velocity
is the initial angular velocity
is the angular acceleration
t is the time
For the pulsar in this problem:
is the initial angular velocity
, since we want to find the time t after which the pulsar stops rotating
is the angular acceleration
Therefore solving for t, we find the time after which the pulsar stops rotating:
c)
As we said in the previous part of the problem, the rate of change of the period of the pulsar is
which means that the period of the pulsar increases by
For every year:
From part A), we also know that the current period of the pulsar is
T = 0.0140 s
The current period is related to the initial period of the supernova by
where is the original period and
is the time that has passed; solving for T0,