Answer:
the answer to the question is calcium
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
Answer:
The incidence angle is 0.145°.
Explanation:
Given that,
Distance due to east = 119 m
Distance due to north= 47.0 km
We need to calculate the incidence angle
Using formula of incidence angle
Where, y = distance due to north
x = distance due to east
Put the value into the formula
Hence, The incidence angle is 0.145°.
Answer:
D. Left
Explanation:
The question is worded weirdly, but based on the fact that heat transfers from hot to cold and the temperature on the right is warmer than the temperature on the left, I would say that the heat is flowing from the box on the right to the box on the left.
I hope this makes sense and was helpful! ☺