A positive charge and a negative charge held a certain distance apart are released. as they move, the force on each particle:___
1 answer:
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In triangle ABC
Sin5 = BC/AC
BC = (AC) Sin5
BC = (50) Sin5 = 4.36 cm
r = distance between the two balls = 2 BC = 2 x 4.36 = 8.72 cm = 0.0872 m
q = charge on each ball
m = mass of each ball = 5 g = 0.005 kg
electric force between the two balls is given as
F =
using equilibrium of force in vertical direction
T Cos5 = mg eq-2
Using equilibrium of force in horizontal direction
T Sin5 = F eq-3
dividing eq-3 by eq-2
T Sin5 /(T Cos5) = F/mg
F = mg tan5
using eq-1
= mg tan5
inserting the values
= 0.005 x 9.8
q = 2.03 x 10⁻⁷ C
sign of charge is same on both the balls. either it is negative or positive
Answer:
The angular acceleration of the top is 0.18 rad/s²
Explanation:
Given;
angular velocity of the top, ω = 16 rad/s
time it takes to come to a complete stop, t = 88.9s
The angular acceleration of the top, is calculated as;
α is the angular acceleration
ω is angular acceleration
t is the time
Therefore, the angular acceleration of the top is 0.18 rad/s²