Respuesta :
Answer: 12.67 cm, 8 cm
Explanation:
Given
Normal distance of separation of eyes, d(n) = 6 cm
Distance of separation is your eyes, d(y) = 9.5 cm
Angle created during the jump, θ = 0.75°
To solve this, we use the formula,
θ = d/r, where
θ = angle created during the jump
d = separation between the eyes
r = distance from the object
θ = d/r
0.75 = 9.5 / r
r = 9.5 / 0.75
r = 12.67 cm
θ = d/r
0.75 = 6 / r
r = 6 / 0.75
r = 8 cm
Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye
The object is 12.67 cm far away for unique eyes while 8 cm far away to the normal eye.
Given here,
Normal distance between of eyes= 6 cm
Distance between advanced eyes = 9.5 cm
Angle created, θ = 0.75°
To solve this, we use the formula,
θ = d/r,
Where
θ = angle
d = distance between the eyes
r = distance from the object
For advanced eye,
θ = d/r
0.75 = 9.5 / r
r = 9.5 / 0.75
r = 12.67 cm
For Normal eye,
θ = d/r
0.75 = 6 / r
r = 6 / 0.75
r = 8 cm
Therefore, the object is 12.67 cm far away for unique eyes while 8 cm far away to the normal eye.
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