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aniked [119]
3 years ago
13

A projectile is fired straight upward from the Earth's surface at the South Pole with an initial speed equal to one third the es

cape speed. (The radius of the Earth is 6.38 106 m.)
Physics
1 answer:
7nadin3 [17]3 years ago
3 0
What is the question? if you are looking for the initial velocity of the projectile, then V escape of Earth = 11.2 km / s. divided by three
Vi = 3.73 km/s
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Explanation:

Formula to calculate the electric potential is as follows.

            V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}

Putting the given values into the above formula as follows.

       V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}

               = \frac{9 \times 10^{9}}{0.25}[\frac{-3.3}{\sqrt{2}} + 4.2] \times 10^{-6}

               = 6.72 \times 10^{4} V

Hence, electric potential at point A is 6.72 \times 10^{4} V.

Now, the electric potential at point B is as follows.

         V_{B} = \frac{kq_{1}}{L} + \frac{kq_{2}}{\sqrt{2}L}

                  = \frac{9 \times 10^{9}}{0.25} [-3.3 + \frac{4.2}{\sqrt{2}}] \times 10^{-6}

                  = -1.19 \times 10^{4} V

Hence, electric potential at point B is -1.19 \times 10^{4} V.

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