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fredd [130]
3 years ago
12

If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______

Physics
1 answer:
adell [148]3 years ago
3 0

Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

F = \dfrac{GMm}{r^2}............(1)

new gravitational force

F' = \dfrac{GMm}{r'^2}

now from the given condition

F' = \dfrac{GMm}{(3r)^2}

F' = \dfrac{GMm}{9r^2}

F' = \dfrac{1}{9}\dfrac{GMm}{r^2}

now, from equation (1)

F' = \dfrac{1}{9}F

\dfrac{F'}{F} = \dfrac{1}{9}

now, the change in gravitational force factor is equal to \dfrac{1}{9}

Hence, the correct answer is option A

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007 (part 1 of 2) 1.0 points
levacccp [35]

Answer:

a) The angle of refraction is approximately 34.7

b) The angle the light have to be incident to give an angle of refraction of 90° is approximately 53.42°

Explanation:

According to Snell's law, we have;

\dfrac{n_1}{n_2} = \dfrac{sin (\theta_2)}{sin (\theta_1)}

The refractive index of the glass, n₁ = 1.66

The angle of incident of the light as it moves into water, θ₁ = 27.2°

a) The refractive index of water, n₂ = 1.333

Let θ₂ represent the angle of refraction of the light in water

By plugging in the values of the variables in Snell's Law equation gives;

\dfrac{1.66}{1.333} = \dfrac{sin (\theta_2)}{sin (27.2^{\circ})}

sin (\theta_2) = sin (27.2^{\circ}) \times \dfrac{1.66}{1.333} \approx 0.5692292265

θ₂ = arcsin(0.5692292265) ≈ 34.7°

The angle of refraction of the light in water, θ₂ ≈ 34.7°

b) When the angle of refraction, θ₂ = 90°, we have;

\dfrac{1.66}{1.333} = \dfrac{sin (90^{\circ})}{sin (\theta_1)}

sin (\theta_1) = \dfrac{sin (90^{\circ})}{\left( \dfrac{1.66}{1.333}\right)} = sin (90^{\circ}) \times \dfrac{1.333}{1.66} \approx 0.803

θ₁ ≈ arcsin(0.803) ≈ 53.42°

The angle of incident, θ₁, that would give an angle of refraction of 90° is θ₁ ≈ 53.42°

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The magnitude of the second charge given that the first is –6×10¯⁶ C and is located 0.05 m away is +3.0×10¯⁶ C

<h3>Coulomb's law equation </h3>

F = Kq₁q₂ / r²

Where

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  • r is the distance apart

<h3>How to determine the second charge </h3>
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  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 0.05 m
  • Force (F) = 65 N
  • Charge 2 (q₂) =?

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Divide both side by Kq₁

q₂ = Fr² / Kq₁

q₂ = (65 × 0.05²) / (9×10⁹ × 6×10¯⁶)

q₂ = +3.0×10¯⁶ C (since the force is attractive)

Learn more about Coulomb's law:

brainly.com/question/506926

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