Answer:
T= 4.24sec
Explanation:
We are going to use the formula below to calculate.
Where T is period
L is length of rod
g is acceleration due to gravity =
From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this
= 4.4625m
thus
T= 4.24sec
Answer:
E=
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by
Substitute x=a and R=a
Then, we get
Where K=
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=
Answer:
0.000234 seconds
Explanation:
Since the row is 0.15m, its radius of rotation must be 0.15 / 2 = 0.075 m
We can start by calculating the angular speed of the rod:
Since one revolution equals to 2π rad. The speed in revolution per second must be
26800 / 2π = 4265 revolution/s
The number of seconds per revolution, or period, is the inverse:
1/4265 = 0.000234 seconds