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 complete question in the attached figure
Let
c ------------------- > is the speed of light
v ------------------- > is the speed in medium
n ------------------- > is the refractive index of medium
we know that
c/v = n
n = (3 x 10^8)/(2.04 x 10^8) = 1.47
1.47 is the refractive index of glycerol.
therefore
the answer is (4) glycerol
Answer:
those who are attracted to magnets. ex: iron, cobalt and nickel
Answer:
Explanation:
Let assume that direction is positive when football travels to the player. The situation can be described properly by applying the definition of Momentum and Impulse Theorem. That is to say:
The average force needed to stop is obtained after some algebraic manipulations:
The impulse delivered to the ball is:
Answer:
B hopefully this helps you with work