Answer:
3.4 x 10⁴ m/s
Explanation:
Analyze the circular path of the electron
B = magnetic field = 80 x 10⁻⁶ T
m = mass of an electron = 9.1 x 10⁻³¹ kg
v = speed in the radial direction
r = radius of the circular trajectory = 2 mm = 0.002 m
q = charge of an electron = 1.6 x 10⁻¹⁹ C
For the electron’s circular movement
qBr = mv
(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v
v = 2.8 x 10⁴ m/s
Now, consider the electron's movement in a straight line:
v' = speed in linear motion
x = distance traveled horizontally = 9 mm = 0.009 m
t = duration = 
= 
= 4.5 x 10⁻⁷ sec
Using the formula
x = v' t
0.009 = v' (4.5 x 10⁻⁷)
v' = 20000 m/s
v' = 2 x 10⁴ m/s
The resultant speed is given by
V = sqrt(v² + v'²)
V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)
v = 3.4 x 10⁴ m/s
Based on my findings, within a period of 2 hours, there are certain atoms remaining.
N = N0 * 2^(-t/6.020) = N = N0 * 2^-0.33223 = 0.7943 N0
Thus, the quantity of atoms that undergo disintegration is N0 - N = N0 * (1 - 0.79430) = 0.2057 N0
This must equate to 15 mCi = 15 * 3.7 * 10^7 = 5.55 * 10^8 atoms
N0 = 5.55 * 10^8 / 0.2057 = 2.698 * 10^9 atoms
Consequently, 2.698 * 10^9 atoms represents the value of N0.
Response:
210.3 degrees
Justification:
The total force acting on charge A is 59.5 N
Apply the x and y components of the net force to determine the direction
atan (y/x)
The derived frequency equals 2.63 Hz. Explanation: For an object weighing 8.0 kg with a spring stretching 3.6 cm, calculations involving the spring constant and oscillation frequency lead to this specific oscillation rate.
