This question is incomplete, the complete question is;
A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.
How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
Answer:
it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
Explanation:
Given that;
mass m = 6.15 x 10⁻⁵ kg
q = 8.45μC = 8.45 × 10⁻⁶ C
B = 0.493
we know that
Time period T = 2πr / V
where r = mv/qB
so T = 2πm/qB
we substitute
T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)
T = 0.0003862 / 0.000004165
T = 92.7 sec
Therefore it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
The fibers can be shorten
Answer:
In a normal body cell, the processes that are necessary for continued growth and division is referred to as cell cycle. Cell cycle is the procedure by which cells divide and grow, consisting of specific step by step phases. INTERPHASE is the phase in which the dividing cells spends most of the time resting as it grows to prepare for cell division.
Answer:
x = A sin w t displacement in SHM
v = A w cos w t velocity in SHM
PE = 1/2 k x^2 = 1/2 k A^2 sin^2 w t
KE = 1/2 m v^2 = 1/2 m w^2 A^2 cos^2 w t
If KE = PE then
k sin^2 w t = m w^2 cos^2 w t
sin^2 wt / cos^2 w t = tan^2 w t = m w^2 / k
but k / m = w^2
So tan^2 w t = 1 and tan w t = 1 or w t = pi / 4 or theta = 45 deg
Then x = r sin w t = r sin 45 = .707 r