Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
Speed = (wavelength) x (frequency
Speed = (.020 m) x (5 / sec)
Speed = 0.1 m/s
Depends on who and where I’m just answering
Acceleration (magnitude anyway) = (change in speed) / (time for the change) .
Change in speed = (10 - 30) = -20 m/s
Time for the change = 4.0sec
Magnitude of acceleration = -20/4 = <em>-5 m/s² </em>
Answer:
i think its because u gave the almost every answer the same exact thing. all the questions have different ways of moving which means different forces for each one i hope this helps :)
Explanation: