The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1
Answer:
Velocity = 131 m/s
Speed = 131 m/s
Explanation:
Equation of motion, s = f(t) = 12t² + 35 t + 1
To get velocity of the particle, let us find the first derivative of s
v (t) = ds/dt = 24t + 35
At t = 4
v(4) = 24(4) + 35
v(4) = 131 m/s
Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s
Acceleration = (Vf - Vi)/t
Since Vf= 60m/s
Vi= 15m/s
T= 15s
=> a= (60m/s - 15m/s)/15s
= 3
So the acceleration is 3m/s^2
The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.
<h3>Change in energy level of the electron</h3>
When photons jump from a higher energy level to a lower level, they emit or radiate energy.
The change in energy level of the electrons is calculated as follows;
ΔE = Eb - Ef
ΔE = -2.68 eV - (-5.74 eV)
ΔE = 3.06 eV
Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.
Learn more about energy level here: brainly.com/question/14287666
#SPJ1
