Qalaxia Knowlege Bot
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I found an answer from large.stanford.edu

NASA SP-413


Nov 17, 1975 ... The concept of spacecraft Earth, a sphere of finite resources and ominous ... for ten weeks to construct a convincing picture of how ... Appendix K — The Lunar Gas Gun Mass Driver . ... that space colonization offers large potential benefits and ... kPa or ~100 mm Hg) for good respiration yet low enough to.


For more information, see NASA SP-413

Qalaxia Master Bot
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I found an answer from en.wikipedia.org

Glossary of physics - Wikipedia


This glossary of physics is a list of definitions of terms and concepts relevant to physics, its sub-disciplines, and related fields, including mechanics, materials ...


For more information, see Glossary of physics - Wikipedia

Pravalika
0

Given that

Potential of gun collector = 100  V  

Magnetic fiels B = 2.83 * 10^{-4} T

Radius of the curved path r = 12.0 cm = 12 *10^2 m


Step 1: Determine the velocity of an ejected electron from gun due to voltage

               Kinetic energy of electron K.E = \frac{1}{2} mv^2

                                           v^2 = \frac{2 K.E}{m}

                  According to energy conversion  K.E = energy of each electron = eV  

                                         v^2 = \frac{2 eV}{m}     ....................(1)


Step 2: Velocity of electron in a spherical bulb due to magnetic field.

              Electron moving in a circular path so it acquires centripetal force

                                         F = \frac{mv^2}{r}

                    

              Force on the electron in a magnetic field  

                                           F =evB                            \because \theta = 90\degree


              So, from the two equations we can write, evB = \frac{mv^2}{r}

                                        eB = \frac{mv}{r}

                                       v = \frac{eBr}{m}

                                       v^2 = \frac{e^2B^2r^2}{m^2} ....................(2)


Step 3: Calculating the specific charge ratio

               From equation (1) and (2)

                                     \frac{e^2B^2r^2}{m} = \frac{2 eV}{m}

                                     \frac{e}{m} = \frac{2V}{B^2 r^2}


              Plug in the values in the above equation.

                             [math] \frac{e}{m} = \frac{2 * 100}{[2.83 * 10^{-4}]^2 (12*10^2)^2} [/math]

                             \frac{e}{m} = 1.734 * 10^{11} C/kg


              Hence, the specific charge ratio   \frac{e}{m} = 1.734 * 10^{11} C/kg