Respuesta :
Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
[tex]F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N[/tex]
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
[tex]W = F_{net} *h\\\\W = 15.328 *10^{-3} * h[/tex]
W = K.E
[tex]15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m[/tex]
Therefore, the ball traveled 0.827 m
The height at which the ball goes for the given parameters is; 0.827 m
What is the height of the ball?
We are given;
distance between the metal plates; d = 3.1 m
mass of glass; m = 1.1g = 0.0011 kg
charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C
speed of the glass ball; v = 4.8 m/s
voltage of the ceiling; V = +3.0 × 10⁶ V
The repulsive force experienced by the ball is gotten from the formula;
F = qV/d
|F| = (4.7 × 10⁻⁹ × 3 × 10⁶)/3.1
|F| = 4.548 × 10⁻³ N
F = -4.548 × 10⁻³ N (negative because it is repulsive force)
The net horizontal force experienced by the ball is;
F_net = F - mg
F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)
F_net = -15.328 × 10⁻³ N
To get the height of the ball, we will use the formula;
F_net * h = ¹/₂mv²
h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)
We took the absolute value of F_net, hence it is not negative
h = 0.827 m
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