Answer:Broadly speaking, all energy in the universe can be categorized as either potential energy or kinetic energy. Potential energy is the energy associated with position, like a ball held up in the air. When you let go of that ball and let it fall, the potential energy converts into kinetic energy, or the energy associated with motion.
EXAMPLES: There are five types of kinetic energy: radiant, thermal, sound, electrical and mechanical. Let's explore several kinetic energy examples to better illustrate these various forms.
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.
The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.
When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.
a)
Magentic force, F = q*v*B
q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T
Centripetal force, F = m*Ac = m * v^2 / R
where,
Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit
Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)
=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]
=> R = 0.114 m
b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.
R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]
=> R = 10.4 m
Elements that give up electrons easily are called <u>metals.</u>
hope this helps!
Answer:
(a) the observed frequency is 200 Hz
(b) the observed frequency is 188 Hz.
Explanation:
speed of the truck, Vs = 27 m/s
frequency of the truck as it approaches, Fs = 185 Hz
(a) Apply Doppler effect to determine the frequency you will hear.
As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.
![F_s = F_o [\frac{V}{V_S + V} ]](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7BV%7D%7BV_S%20%2B%20V%7D%20%5D)
Where;
Fo is the observed frequency which is the frequency you will hear.
V is speed of sound in air
![F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7BV%7D%7BV_S%20%2B%20V%7D%20%5D%5C%5C%5C%5C185%20%3D%20F_o%20%5B%5Cfrac%7B340%7D%7B27%20%2B%20340%7D%20%5D%5C%5C%5C%5C185%20%3D%20F_o%20%280.926%29%5C%5C%5C%5CF_o%20%3D%20%5Cfrac%7B185%7D%7B0.926%7D%5C%5C%5C%5CF_o%20%3D%20199.78%20%5C%20Hz)
(b) Apply the following formula for a moving observer and a moving source;
](https://tex.z-dn.net/?f=F_o%20%3D%20F_s%5B%5Cfrac%7BV-V_o%7D%7BV%7D%20%5D%28%5Cfrac%7BV%7D%7BV-V_S%7D%20%29)
The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.
\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz](https://tex.z-dn.net/?f=F_o%20%3D%20F_s%5B%5Cfrac%7BV-V_o%7D%7BV%7D%20%5D%28%5Cfrac%7BV%7D%7BV-V_S%7D%20%29%5C%5C%5C%5CF_o%20%3D%20185%5B%5Cfrac%7B340-22%7D%7B340%7D%20%5D%28%5Cfrac%7B340%7D%7B340-27%7D%20%29%5C%5C%5C%5CF_o%20%3D%20185%280.9353%29%281.0863%29%5C%5C%5C%5CF_o%20%3D%20188%20%5C%20Hz)
The answer is A. The weak nuclear force is usually an attractive force, and the other two answer choices are, indeed, true of the weak nuclear force.
