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tresset_1 [31]
3 years ago
14

Why do underwater objects appear distorted when you see them from above?

Physics
2 answers:
steposvetlana [31]3 years ago
4 0
REfraction of the rays coming from the object through the water and then being refracted into the air
WITCHER [35]3 years ago
4 0

The correct answer to the question is : Refraction.

EXPLANATION:

Before going to answer this question, first we have to understand refraction.

Refraction is the optical phenomenon in which light rays are bent at the interface of the two media due to their optical density difference.

When light comes from optical rarer medium to optical denser medium, the rays are bent towards the normal at the point of incidence.

Similarly light rays bend away from the normal when they pass from optical denser medium to optical rarer medium.

As per the questions, the object is present inside the water.  The water is optically denser as compared to air. Hence, the light coming from the object is bent at the air-water interface due to refraction of light. That's why the  object is appeared to be bent.

Hence, the correct answer is refraction.

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7 0
2 years ago
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You are standing in front of a flat large mirror. As you step 1.0 m farther from the mirror, the distance between you and your i
Lesechka [4]

Answer:

2 meters

Explanation:

when you step away from a reflection you get farther

5 0
3 years ago
a light beam that hits a mirror at an angle of 36°... what is the Angle of Incidence? Angle of Reflection?​
Kipish [7]

Angle of incidence is 36° and so is the reflection. Both angles are equal.

6 0
3 years ago
What are the two sources of earth's internal heat energy
RUDIKE [14]

Answer:

Hey

Your answer would be Radioactive Decay and Heat of formation.

When earth first formed it was very violant. Some planetary scientists believe that the moon was created from a collision involving earth and Thea (a theoretical dwarf planet). these collisions that have formed earth heated it so much that it is still hot from them.

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6 0
2 years ago
If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder
kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0
3 years ago
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