The Milky Way galaxy is believed to be composed of over 200billion stars and to have a diameter of over 100000 light years. What
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Answer:
13.6 cm
Explanation:
From Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.
Given n₂ = 4/3:
1 = 4/3 sin θ
sin θ = 3/4
If x is the radius of the circle, then sin θ is:
sin θ = x / √(x² + 12²)
sin θ = x / √(x² + 144)
Substituting:
3/4 = x / √(x² + 144)
9/16 = x² / (x² + 144)
9/16 x² + 81 = x²
81 = 7/16 x²
x ≈ 13.6
The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.