Answer:
Explanation:
Initial speed, v = 10 x 10^3 m/s
Mass of the earth, M = 6 x 10^24 kg
Radius of the earth, R = 6.4 x 10^6 m
Maximum from the surface of earth, h = ?
Let m = Mass of the projectile
Solution:
Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface
![-G M m / ( R + h )=- G M m / R + (1/2) m v^2](https://tex.z-dn.net/?f=-G%20M%20m%20%2F%20%28%20R%20%2B%20h%20%29%3D-%20G%20M%20m%20%2F%20R%20%2B%20%281%2F2%29%20m%20v%5E2)
![- G M / ( R + h ) = - G M / R + (1/2) v^2](https://tex.z-dn.net/?f=-%20G%20M%20%2F%20%28%20R%20%2B%20h%20%29%20%3D%20-%20G%20M%20%2F%20R%20%2B%20%281%2F2%29%20v%5E2)
![-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2](https://tex.z-dn.net/?f=-2%5Ctimes%20G%20M%20%2F%20%28%20R%20%2B%20h%20%29%20%3D%20%28%20-%202%20G%20M%20%2F%20R%20%29%20%2B%20v%5E2)
=![( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2](https://tex.z-dn.net/?f=%28%20%28-%202%5Ctimes%206.67%5Ctimes10%5E%7B-11%7D%5Ctimes6%5Ctimes10%5E%7B24%7D%29%20%2F%286.4%5Ctimes10%5E6%29%7D%20%2B10000%5E2)
=![-2.50625\times10^7 J](https://tex.z-dn.net/?f=-2.50625%5Ctimes10%5E7%20J)
![=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7](https://tex.z-dn.net/?f=%3D-%208%5Ctimes10%5E%7B14%7D%20%2F%20%28%20R%20%2B%20h%20%29%3D-2.50625%5Ctimes%2010%5E7)
![R+h=31.92\times10^{6}](https://tex.z-dn.net/?f=R%2Bh%3D31.92%5Ctimes10%5E%7B6%7D)
![h=31.92\times10^{6}-6.4\times10^6](https://tex.z-dn.net/?f=h%3D31.92%5Ctimes10%5E%7B6%7D-6.4%5Ctimes10%5E6)
The 8 moon phases in order are New moon, Waxing Crescent, First Quarter, Waxing Gibbous, Full moon, Waning Gibbous, Third Quarter, and finally Waxing Crescent.
Magnetic lines of forces make a closed loop because there are no magnetic charges (monopoles). While electric <span>field lines end on electric charges, magnetic field lines have no objects on which they could end. This is why they must form loops. </span>
Answer:
![v_r=5.89\ m.s^{-1}](https://tex.z-dn.net/?f=v_r%3D5.89%5C%20m.s%5E%7B-1%7D)
Explanation:
Given:
- mass of rocket,
![m_r=50\ g](https://tex.z-dn.net/?f=m_r%3D50%5C%20g)
- time of observation,
![t=2\ s](https://tex.z-dn.net/?f=t%3D2%5C%20s)
- mass lost by the rocket by expulsion of air,
![m_a=10\%\ of m_r=5\ g](https://tex.z-dn.net/?f=m_a%3D10%5C%25%5C%20of%20m_r%3D5%5C%20g)
- velocity of air,
![v_a=53\ m.s^{-1}](https://tex.z-dn.net/?f=v_a%3D53%5C%20m.s%5E%7B-1%7D)
<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)
![m_a.v_a=(m_r-m_a)\times v_r](https://tex.z-dn.net/?f=m_a.v_a%3D%28m_r-m_a%29%5Ctimes%20v_r)
![5\times 53=(50-5)\times v_r](https://tex.z-dn.net/?f=5%5Ctimes%2053%3D%2850-5%29%5Ctimes%20v_r)
![v_r=5.89\ m.s^{-1}](https://tex.z-dn.net/?f=v_r%3D5.89%5C%20m.s%5E%7B-1%7D)