Answer:
E" = Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})
Explanation:
The electric field due to a charged ring of radius R at a distance x from the center of the ring when the axis of the ring is located on the x - axis is
E = Qx/4πε₀[√(x² + R²)]³
Since the rings are separated by a distance L, the electric field at point x due to the second ring is E' = -Q(L - x)/4πε₀[√((L - x)² + R²)]³. It is negative since it points in the negative x - direction.
So, the resultant electric field at x is E" = E + E' = Qx/4πε₀[√(x² + R²)]³ + {-Q(L - x)/4πε₀[√((L - x)² + R²)]³}
E" = Qx/4πε₀√[(x² + R²)]³ - Q(L - x)/4πε₀√[((L - x)² + R²)]³
E" = Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[((L - x)² + R²)]³})
E" = Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L² - 2Lx + x² + R²)]³})
E" = Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L - 2x)L + (x² + R²)]³})
E" = Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})
So, the electric field at points along the x axis is
E" = Q/4πε₀√[(x² + R²)]³(x - {(L - x)/√[(L - 2x)L/(x² + R²) + 1]³})
Answer:
476.387 Hz
714.583 Hz
Explanation:
L = Length of tube
v = Speed of sound in air = 343 m/s
Frequency for a closed tube is given by
The frequency is 476.387 Hz
If it was one third full 
The frequency is 714.583 Hz
Ans: In team offensive strategy, the baseball coach or manager may make certain calls based on the game situation. One play is called ht hit-and-run. This is case where a base runner starts running as soon as the pitcher throws the pitch. The batter is then supposed to try and make contact with the ball.
