Ethan drove a 1500 kg car around a circular track with a radius of 75 m. If the magnitude of the centripetal acceleration is 2 m
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Answer:
The distance of the goggle from the edge is 5.30 m
Explanation:
Given:
The depth of pool (d) = 3.2 m
let 'i' be the angle of incidence
thus,
i =
i = 67.75°
Now, Using snell's law, we have,
n₁ × sin(i) = n₂ × 2 × sin(r)
where,
r is the angle of refraction
n₁ is the refractive index of medium 1 = 1 for air
n₂ is the refractive index of medium 1 = 1.33 for water
now,
1 × sin 67.75° = 1.33 × sin(r)
or
r = 44.09°
Now,
the distance of googles = 2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) = 5.30 m
Hence, <u>the distance of the goggle from the edge is 5.30 m</u>
The object’s resultant angle of motion with the +x-axis after the collision is 47°
<span>From object A:
1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.
Now, we know, tan</span>Ф =
⇒tanФ =
⇒tanФ = 1.088
⇒ Ф =
1.088
= 47.4 ≈ 47
<span>From object A:
1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.
Now, we know, tan</span>Ф =
⇒tanФ =
⇒tanФ = 1.088
⇒ Ф =
= 47.4 ≈ 47