Use this formula to find your answer...
Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms
frequency (f)=1/( Time period).
Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.
Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.
<span><u><em>This begins with the sun’s radiation,
which is absorbed differently on the earth’s surface. The earth's
surface is heated differently because of scenarios like cloud cover,
mountains, valleys, water bodies, vegetation and desert lands.</em></u>
<u><em></em></u>
<u><em>Good luck! </em></u></span><u><em></em></u>
The kinetic energy at the bottom of the swing is also 918 J.
Assume the origin of the coordinate system to be at the lowest point of the pendulum's swing. A pendulum, when raised to the highest point has potential energy since it is raised to a height h above the origin. At the highest point, the pendulum's velocity becomes zero, hence it has no kinetic energy. Its energy at the highest point is wholly potential.
When the pendulum swings down from its highest position, it gains velocity. Hence a part of its potential energy begins to convert itself into kinetic energy. If no dissipative forces such as air resistance exist, then, the law of conservation of energy can be applied to the swing.
Under the action of conservative forces, the total mechanical energy of a system remains constant.This means that the sum of the potential and kinetic energies of a body remains constant.
When the pendulum reaches the lowest point of its swing, it is at the origin of the chosen coordinate system. Its vertical displacement from the origin is zero, hence its potential energy with respect to the origin is zero. Therefore the entire potential energy of 918 J should have been converted into kinetic energy, according to the law of conservation of energy.
Thus, the kinetic energy of the pendulum at the lowest point of its swing is equal to the potential energy it had at its highest point, which is equal to <u>918 J.</u>
.Answer;
Using Fmax=qVB
F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)
ANS=1.29*10^-12 N
2. Using Amax=Fmax/ m
Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)
ANS=1.93*10^15 m/s^2*
3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton
