_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:
![r=\frac{mv}{qB}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bmv%7D%7BqB%7D)
Where:
![\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20r%3D%5Ctext%7B%20radius%7D%20%5C%5C%20v%3D%5Ctext%7B%20velocity%7D%20%5C%5C%20q%3D%5Ctext%7B%20charge%7D%20%5C%5C%20B%3D%5Ctext%7B%20magnetic%20field%7D%20%5Cend%7Bgathered%7D)
We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:
![\frac{r_d}{m_d}=\frac{v}{q_B_}](https://tex.z-dn.net/?f=%5Cfrac%7Br_d%7D%7Bm_d%7D%3D%5Cfrac%7Bv%7D%7Bq_B_%7D)
Now, we multiply both sides by the magnetic field "B":
![\frac{Br_d}{m_d}=\frac{v}{q}](https://tex.z-dn.net/?f=%5Cfrac%7BBr_d%7D%7Bm_d%7D%3D%5Cfrac%7Bv%7D%7Bq%7D)
Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:
![r_p=\frac{m_pv}{qB}](https://tex.z-dn.net/?f=r_p%3D%5Cfrac%7Bm_pv%7D%7BqB%7D)
And substitute the quotient between velocity and charge:
![r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})](https://tex.z-dn.net/?f=r_p%3D%5Cfrac%7Bm_p%7D%7BB%7D%28%5Cfrac%7BBr_d%7D%7Bm_d%7D%29)
Now, we cancel out the magnetic field:
![r_p=\frac{m_pr_d}{m_d}](https://tex.z-dn.net/?f=r_p%3D%5Cfrac%7Bm_pr_d%7D%7Bm_d%7D)
Now, we substitute the values:
![r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}](https://tex.z-dn.net/?f=r_p%3D%5Cfrac%7B%281.67%5Ctimes10%5E%7B-27%7Dkg%29%280.385m%29%7D%7B%283.34%5Ctimes10%5E%7B-27%7Dkg%29%7D)
Solving the operations:
![r_p=0.193m=19.3cm](https://tex.z-dn.net/?f=r_p%3D0.193m%3D19.3cm)
Therefore, the radius is 19.3 cm.
Using the wavelength expressions, such as the distance between two peaks of a wave, that of the frequency as a function of the speed and the wavelength, such as the period inversely proportional to the frequency we have to for the first question
the wavelength is the distance between the two ridges that is
![\lambda = 0.8m](https://tex.z-dn.net/?f=%5Clambda%20%3D%200.8m)
For the second question the frequency is determined as the rate of change of the velocity versus the wavelength, that is
![f = \frac{v}{\lambda}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7Bv%7D%7B%5Clambda%7D)
![f = \frac{2.2}{0.8}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B2.2%7D%7B0.8%7D)
![f= 2.75Hz \approx 2.8Hz](https://tex.z-dn.net/?f=f%3D%202.75Hz%20%5Capprox%202.8Hz)
For the third question the period is inversely proportional to the frequency, therefore
![T = \frac{1}{f}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B1%7D%7Bf%7D)
![T = \frac{1}{2.8}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B1%7D%7B2.8%7D)
![T = 0.36s](https://tex.z-dn.net/?f=T%20%3D%200.36s)
I would say the most beautiful
Answer:
The answer is A
Explanation:
When a rockets thrusters push on the ground the ground pushes back on the rocket with equal force in the opposite direction. Hence the rocket takes off.
Newtons third law of motion states, for every action there is an equal and opposite reaction.