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
Explanation:
The light is bent away from the normal
Despite current has a magnitude and a direction, like vectors, it is a scalar because it doesn't obey laws of vector addition. For instance, if we consider a junction of
in a circuit, and two currents entering this junction, we know that the resultant current is just the algebraic sum of the two currents, not the vector sum, so it is not a vector quantity.
Answer:
6 seconds
Explanation:
Just divide 3900 by 650 to get the amount of time.
Ok 3 A9 ok so that is right