At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?
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Answer:
F = 7.68 10¹¹ N, θ = 45º
Explanation:
In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges
The net force is
F_ {net} = F₂₁ + F₂₃ + F₂₄
bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.
let's use trigonometry
cos 45 = F₂₄ₓ / F₂₄
sin 45 = F_{24y) / F₂₄
F₂₄ₓ = F₂₄ cos 45
F_{24y} = F₂₄ sin 45
let's do the sum on each axis
X axis
Fₓ = -F₂₁ + F₂₄ₓ
Fₓ = -F₂₁₁ + F₂₄ cos 45
Y axis
F_y = - F₂₃ + F_{24y}
F_y = -F₂₃ + F₂₄ sin 45
They indicate that the magnitude of all charges is the same, therefore
F₂₁ = F₂₃
Let's use Coulomb's law
F₂₁ = k q₁ q₂ / r₁₂²
the distance between the two charges is
r = a
F₂₁ = k q² / a²
we calculate F₂₄
F₂₄ = k q₂ q₄ / r₂₄²
the distance is
r² = a² + a²
r² = 2 a²
we substitute
F₂₄ = k q² / 2 a²
we substitute in the components of the forces
Fx =
Fx = ( -1 + ½ cos 45)
F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)
We calculate
F₀ = 9 10⁹ 4.25² / 0.440²
F₀ = 8.40 10¹¹ N
Fₓ = 8.40 10¹¹ (½ 0.707 - 1)
Fₓ = -5.43 10¹¹ N
remember cos 45 = sin 45
F_y = - 5.43 10¹¹ N
We can give the resultant force in two ways
a) F = Fₓ î + F_y ^j
F = -5.43 10¹¹ (i + j) N
b) In the form of module and angle.
For the module we use the Pythagorean theorem
F =
F = 5.43 10¹¹ √2
F = 7.68 10¹¹ N
in angle is
θ = 45º
Answer:
1) The mass of the student is approximately 73.39 kg
2) The net force on the student is approximately 947.523 N
3) The value the scale will read is approximately 96.59 kg
Explanation:
The given parameters are;
The weight of the student = 720 N
The speed at which the elevator is decreasing = 3.1 m/s²
1) The weight of the student = The mass of the student × The acceleration due to gravity
The acceleration due to gravity is a constant = 9.81 m/s²
Substituting the known values gives;
720 N = The mass of the student × 9.81 m/s²
∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg
2) The forces acting on the student are;
i) The force of gravity which is the weight of the student acting downwards
ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student
The net force, = The weight of the student + The inertia force of the slowing elevator
∴ The net force, = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N
3) The scale will read the mass of the student as follows;
Mass reading of student on the scale = Force on scale/9.81
∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg
The value the scale will read = 96.59 kg.