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trapecia [35]
4 years ago
15

A diver is swimming underneath an oil slick with a thickness of 200 nm and an index of refraction of 1.50. A white light shines

straight down towards the diver from above the oil slick. The index of refraction of water is 1.33. What is the longest wavelength of the light in water, ????????????????????????, that is transmitted most easily to the diver?
Physics
1 answer:
Tcecarenko [31]4 years ago
4 0

Answer:

Explanation:

thickness of oil t = 200 nm

index of refraction μ = 1.5

For transmitted light :---

path difference = 2μ t

For constructive interference

path difference = n λ , λ is wavelength  of light

2μ t = n λ

λ = 2μ t /  n

For longest λ , n = 1

λ = 2μ t

= 2 x 1.5 x 200 nm

= 600 nm

Wavelength in water

= 600 / refractive index of water

= 600 / 1.33

= 451.1 nm Ans

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Answer:

a) System aceleration:

  • a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}

b) Direction of movement:

The block m_1 down the plane when the acceleration is negative. This occur when:

m_2-m_1sin(\theta)

The block m_2 up the plane when the acceleration is positive. This occur when:

m_2-m_1sin(\theta)>0

Explanation:

For the block m_1 the move direction is parallel (||) to the plane

\sum F_{||}=m_1a=T-sin(\theta)mg  (1)

For the block m_2  the move direction is vertical (y)

\sum F_y=m_2a=m_2g-T  (2)

Both blocks are connected with the same cable, therefore, the tension on the cable and the acceleration is the same for both.

Solving the equation 2 for T:

T=m_2g-m_2a   (3)

replacing (3) in the equation (1)

m_2g-m_2a- m_1gsin(\theta)=m_1a  (4)

solving (4) for a:

a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}

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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi
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Answer:

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Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

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4 years ago
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Answer:

14000 mm

Explanation:

multiply the length value by 1000

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