Answer:
8.79*10^6 rad/s
Explanation:
To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:
(1)
r: radius of the trajectory
m: mass of the electron = 9.1*10^-31 kg
v: speed of the electron = 1.0*10^6 m/s
q: charge of the electron = 1.6*10^-19 C
B: magnitude of the magnetic field = 5.0*10^-5 T
You use the fact that the angular frequency in a circular motion is given by:
Then, you solve the equation (1) in order to obtain v/r:
Finally, you replace the values of the parameters:
hence, the angular frequency is 8.79*10^6 rad/s
The frequency is:
False. The nuclear energy is found within the nucleus. Electrons are located outside the nucleus.
I’m pretty sure it does most of the time ig
Answer:
The second ball lands 1.5 s after the first ball.
Explanation:
Given;
initial velocity of the ball, u = 12 m/s
height of fall, h = 35 m
initial velocity of the second, v = 12 m/s
Time taken for the first ball to land;
determine the maximum height reached by the second ball;
v² = u² -2gh
at maximum height, the final velocity, v = 0
0 = 12² - (2 x 9.8)h
19.6h = 144
h = 144 / 19.6
h = 7.35 m
time to reach this height;
Total height above the ground to be traveled by the second ball is given as;
= 7.35 m + 35m
= 42.35 m
Time taken for the second ball to fall from this height;
total time spent in air by the second ball;
T = t₁ + t₂
T = 1.23 s + 2.94 s
T = 4.17 s
Time taken for the second ball to land after the first ball is given by;
t = 4.17 s - 2.67 s
T = 1.5 s
Therefore, the second ball lands 1.5 s after the first ball.
