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krek1111 [17]
4 years ago
12

An object traveling westward with an initial velocity of 11.2 m/s experiences a constant acceleration of 2.78 m/s^2 ​2 ​​ in the

eastward direction for 2.03 seconds. During this time, the object travels a distance of: ____________
Physics
1 answer:
elixir [45]4 years ago
7 0

Explanation:

Given that,

The initial velocity of the object, u = -11.2 m/s (due west)

Acceleration of the object, a=+2.78\ m/s^2 (due east)

Time, t = 2.03 seconds

We need to find the distance traveled by the object during this time. Let the distance covered is d. Using second equation of motion to find it as :

d=ut+\dfrac{1}{2}at^2\\\\d=-11.2\times 2.03+\dfrac{1}{2}\times (2.78)\times (2.03)^2 \\\\d=-17.007\ m

So, the object cover a distance of 17.007 meters in the westward direction.

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Answer:

83.6°

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So, substituting these values into the equation,

n₁sinθ₁ = n₂sinθ₂

1.5 × sinθ₁ = 1 × sin90

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So, for total internal reflection, an incidence angle of 41.8° is required. So, a full convergence angle of 2 × 41.8° = 83.6° is required for the whole bundle of rays.

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4 years ago
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