Answer: The first one
Explanation: My best guess is in the pic attached. Hope this helps.
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Considering the Coulomb's Law, the magnitude of the Coulomb force is 3.1865 N.
<h3>Coulomb's Law</h3>Charged bodies experience a force of attraction or repulsion on approach.
From Coulomb's Law it is possible to predict what the electrostatic force of attraction or repulsion between two particles will be according to their electric charge and the distance between them.
From Coulomb's Law, the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
where:
- F is the electrical force of attraction or repulsion. It is measured in Newtons (N).
- Q and q are the values of the two point charges. They are measured in Coulombs (C).
- d is the value of the distance that separates them. It is measured in meters (m).
- K is a constant of proportionality called the Coulomb's law constant. It depends on the medium in which the charges are located. Specifically for vacuum k is approximately 9×10⁹
.
The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.
<h3>This case</h3>In this case, you know that:
- The two uncharged sphere are separated by the distance of d= 3.50 m
- The number of electrons are 1.30×10¹².
- Electrons is elementary charge and charges on both the sphere is same. The value of electron is 1.602×10⁻¹⁹ C. This is, Q=q=1.30×10¹²×1.602×10⁻¹⁹ C= 2.0826×10⁻⁷ C
Replacing in Coulomb's Law:
Solving:
<u><em>F= 3.1865 N</em></u>
Finally, the magnitude of the Coulomb force is 3.1865 N.
Learn more about Coulomb's Law:
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(1) The linear acceleration of the yoyo is 3.21 m/s².
(2) The angular acceleration of the yoyo is 80.25 rad/s²
(3) The weight of the yoyo is 1.47 N
(4) The tension in the rope is 1.47 N.
(5) The angular speed of the yoyo is 71.385 rad/s.
<h3> Linear acceleration of the yoyo</h3>The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.
∑τ = Iα
rT - Rf = Iα
where;
- I is moment of inertia
- α is angular acceleration
- T is tension in the rope
- r is inner radius
- R is outer radius
- f is frictional force
rT - Rf = Iα ----- (1)
T - f = Ma -------- (2)
a = Rα
where;
- a is the linear acceleration of the yoyo
Torque equation for frictional force;
solve (1) and (2)
since the yoyo is pulled in vertical direction, T = mg
α = a/R
α = 3.21/0.04
α = 80.25 rad/s²
<h3>Weight of the yoyo</h3>W = mg
W = 0.15 x 9.8 = 1.47 N
<h3>Tension in the rope </h3>T = mg = 1.47 N
<h3>Angular speed of the yoyo </h3>v² = u² + 2as
v² = 0 + 2(3.21)(1.27)
v² = 8.1534
v = √8.1534
v = 2.855 m/s
ω = v/R
ω = 2.855/0.04
ω = 71.385 rad/s
Learn more about angular speed here: brainly.com/question/6860269
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