Particle A has less mass than particle B. Both are pushed forward across a frictionless surface by equal forces for 1 s. Both st
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Answer:
<em>113.4 J</em>
Explanation:
<u>Elastic Potential Energy</u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.
The spring has a natural length of 0.7 m and a spring constant of k=70 N/m. When the spring is stretched to a length of 2.5 m, the change of length is
Δx = 2.5 m - 0.7 m = 1.8 m
The energy stored in the spring is:
PE = 113.4 J
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: