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1 year ago

Does the refraction of light make a swimming pool seem deeper or shallower? explain your answer.

Physics

1 answer:

larisa [96]1 year ago

6 0

The refraction of light makes a swimming pool seem <u>shallower</u>.

The swimming pool seems shallower because the rays of light coming from the bottom of the pool do not come with a straight path. The path of light is straight as long as it is in the water.

When lights come out of the water into the air it bents downwards. This bending is called refraction.

Refraction forms a virtual image of the pool and it seems shallower than it actually is to the observer. This only happens when light travels from one transparent medium into another having lower density.

If you need to learn more about why a swimming pool appears <u>shallower</u>, click here

https://brainly.in/question/7136803?referrer=searchResults

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3 years ago

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3 years ago

A 45 kg wagon is being pulled with a rope that makes an angle of 38 degrees with the horizontal. The applied force is 410 N and

mel-nik [20]

Answer:

7.35m/s²

Explanation:

From the question we are not told what to find but we can as well find the acceleration of the wagon.

According to newton second law of motion

$\sum F_x = ma_x\\Fm - Ff = ma_x\\Fm - \mu R = ma_x\\Fm- \mu mg = ma_x\\$

Fm is the moving force = 410N

$\mu$ is the coefficient of friction = 0.18

m is the mass = 45kg

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a is the acceleration of the wagon

Substitute the given data into the equation ang get ax

$Fm- \mu mg = ma_x\\410 - (0.18)(9.8)(45) = 45a_x\\410 - 79.38 = 45a_x\\330.62 = 45a_x\\a_x = 330.62/45\\a_x = 7.35m/s^2\\$

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8 0

3 years ago

A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

nalin [4]

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

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F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

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