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3 years ago

Do you think that the formation of a rainbow is more a demonstration of the wave-like or particle-like behavior of light

Physics

1 answer:

kogti [31]3 years ago

6 0

Answer:

The formation of a rainbow is more a demonstration of the wave-like behavior of light.

Explanation:

A rainbow is caused by the interaction of sunlight with atmospheric conditions. A rainbow formation results from the refraction and reflection of light due to changes in the light's wavelength direction. Naturally, light enters a water droplet, slowing down and bending as it goes from air to denser water in a refraction as if "bent." Then light reflects off the inside of the droplet, separating into its component wavelengths or colors. With light exiting from the droplet, a rainbow is formed.

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What is the equilibrium concentration of ibr?

asambeis [7]

IBR is the thermal decomposition of iodine(I) bromide to produce iodine and bromine. This reaction takes place at a temperature of over 40,5°C and is written as:

<span>2IBr ⇄ I2 + Br2
</span>
Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.<span> You can calculate the equilibrium concentration if you know the equilibrium constant Kc (Kc=I^2*Br^2/IBR^2) and the initial concentration for the reaction. The initial concentration is obtained from ICE Table.</span>

3 0

3 years ago

Can someone please help me?

Alina [70]

Answer:

D would be it

Explanation:

cause none of the rest Makes sense to the book of the picture and I'm 100% sure =)

6 0

4 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

To determine the types of equipment your boat must carry, you need to make a measurement of the boat. What is this measurement?

jenyasd209 [6]

Explanation:

In total, the length is measured from the tip of the bow in a linear fashion to the stern of the formation of delight including any back-deck extensions. The measurement involves bow sprits; rudders; detachable engines and engine sections; handles; and various fittings and connections.

Importance in calculating a boat's length:

it affects the transportation costs (the longer the length, the higher the cost). The pontoon's length counts as you find out how much rope you need to wrestle.

The cost of vessel settlement on marinas depends in part on the pontoon length. As more area is consumed by a more drawn pontoon, the docking charges are higher.

Transportation guidelines will probably not allow pontoons past a specific length on specific occasions of the day.

6 0

3 years ago

Mass Center Determine the coordinates (x, y) of the center of mass of the area in blue in the figure below. Answers: x=(3)/(8)a

Naya [18.7K]

Explanation:

The x and y coordinates of the center of mass are:

xcm = ∫ x dm / m = ∫ x ρ dA / ∫ ρ dA

ycm = ∫ y dm / m = ∫ y ρ dA / ∫ ρ dA

Assuming uniform density, the center of mass is also the center of area.

xcm = ∫ x dA / ∫ dA = ∫ x y dx / A

ycm = ∫ y dA / ∫ dA = ∫ ½ y² dx / A

First, let's find the area:

A = ∫ y dx

A = ∫₀ᵃ (-h/a² x² + h) dx

A = -⅓ h/a² x³ + hx |₀ᵃ

A = -⅓ h/a² (a)³ + h(a)

A = ⅔ ha

Now, let's find the x coordinate of the center of mass:

xcm = ∫ x y dx / A

xcm = ∫₀ᵃ x (-h/a² x² + h) dx / (⅔ ha)

xcm = ∫₀ᵃ (-h/a² x³ + hx) dx / (⅔ ha)

xcm = (-¼ h/a² x⁴ + ½ hx²) |₀ᵃ / (⅔ ha)

xcm = (-¼ h/a² (a)⁴ + ½ h(a)²) / (⅔ ha)

xcm = (¼ ha²) / (⅔ ha)

xcm = ⅜ a

Next, we find the y coordinate of the center of mass:

ycm = ∫ y² dx / A

ycm = ∫₀ᵃ ½ (-h/a² x² + h)² dx / (⅔ ha)

ycm = ∫₀ᵃ ½ (h²/a⁴ x⁴ − 2h²/a² x² + h²) dx / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ x⁵ − ⅔ h²/a² x³ + h² x) |₀ᵃ / (⅔ ha)

ycm = ½ (⅕ h²/a⁴ (a)⁵ − ⅔ h²/a² (a)³ + h² (a)) / (⅔ ha)

ycm = ½ (⁸/₁₅ h²a) / (⅔ ha)

ycm = ⅖ h

6 0

3 years ago