Has anyone has the answers for Physical Education # 11 cross word and the word search?
1 answer:
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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>
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that , which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
