-3.5912m/s^2Use vf^2 = vi^2 + 2ax in your calculations, knowing vf, vi, and x.
Complete Question
A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)
Answer:
The velocity is
Explanation:
From the question we are told that
The magnitude of the electric field is ![E = 144 \ kV /m = 144*10^{3} \ V/m](https://tex.z-dn.net/?f=E%20%20%3D%20%20144%20%5C%20kV%20%2Fm%20%20%20%3D%20%20144%2A10%5E%7B3%7D%20%5C%20%20V%2Fm)
The magnetic field is ![B = 0.38 \ T](https://tex.z-dn.net/?f=B%20%20%3D%20%200.38%20%5C%20T)
The force due to the electric field is mathematically represented as
![F_e = E * q](https://tex.z-dn.net/?f=F_e%20%3D%20%20E%20%20%2A%20q)
and
The force due to the magnetic field is mathematically represented as
![F_b = q * v * B * sin(\theta )](https://tex.z-dn.net/?f=F_b%20%20%3D%20%20q%20%2A%20v%20%20%2A%20%20B%20%2A%20sin%28%5Ctheta%20%29)
Now given that it is perpendicular , ![\theta = 90](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%2090)
=> ![F_b = q * v * B * sin(90)](https://tex.z-dn.net/?f=F_b%20%20%3D%20%20q%20%2A%20v%20%20%2A%20%20B%20%2A%20sin%2890%29)
=> ![F_b = q * v * B](https://tex.z-dn.net/?f=F_b%20%20%3D%20%20q%20%2A%20v%20%20%2A%20%20B)
Now given that it is not deflected it means that
![F_ e = F_b](https://tex.z-dn.net/?f=F_%20e%20%20%3D%20%20F_b)
=> ![q * E = q * v * B](https://tex.z-dn.net/?f=q%20%2A%20%20E%20%3D%20q%20%2A%20%20v%20%20%2A%20B)
=> ![v = \frac{E}{B }](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7BE%7D%7BB%20%7D)
substituting values
![v = \frac{ 144 *10^{3}}{0.38 }](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7B%20144%20%2A10%5E%7B3%7D%7D%7B0.38%20%7D)
![v = 3.79 *10^{5} \ m/s](https://tex.z-dn.net/?f=v%20%3D%20%203.79%20%2A10%5E%7B5%7D%20%5C%20m%2Fs)
<span>
False. A Red Giant evolves from a main sequence star after it commences
Helium fusion and undergoes a great expansion; in the case of our Sun
it would expand to encompass Earth's orbit. Thus, it would have
approximately the same mass (a small amount being expelled in the helium
flash), and vastly greater volume, equating to vastly lower density.
You're welcome :)
</span>
Hello!
The kind of gas cloud that is most likely to give birth to stars is a cold, dense gas cloud.
Stars are formed in giant molecular clouds, called "star nurseries". These are regions with molecular Hydrogen, Helium, and little amounts of other elements. These are cold and dense regions, and the gas cloud collapses, forming a protostar. The protostar gradually heats, powering nuclear reactions in its center, which increases the temperature even more and giving birth to a star.
Have a nice day!
The hotter molecules become, the faster they move around. The colder they are, the more slow and lethargic they are