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AVprozaik [17]
3 years ago
15

The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuc

k in the sandy bottom of the pond. If the stick is 19 cm tall, how long is the shadow?

Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

shadow length 7.67 cm

Explanation:

given data:

refractive index of water is 1.33

by snell's law we have

n_{air} sin30 =n_{water} sin\theta

1*0.5 = 1.33*sin\theta

solving for\theta

sin\theta = \frac{3}{8}

\theta = sin^{-1}\frac{3}{8}

\theta =  22 degree

from shadow- stick traingle

tan(90-\theta) = cot\theta = \frac{h}{s}

s = \frac{h}{cot\theta} = h tan\theta

s = 19tan22 = 7.67 cm

s = shadow length

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1) The equilibrium constant Kc for the reaction N 2(g) + O 2(g) 2NO(g) at 1200 C is 1.00x 10^-5. Calculate the molar concentrati
Elina [12.6K]

Explanation:

1) N₂ + O₂ → 2 NO

Kc = [NO]² / ([N₂] [O₂])

Set up an ICE table:

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_{2}&0.114&-x&0.114-x\\O_{2}&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]

Plug into the equilibrium equation and solve for x.

1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))

1.00×10⁻⁵ = (2x)² / (0.114 − x)²

√(1.00×10⁻⁵) = 2x / (0.114 − x)

0.00316 = 2x / (0.114 − x)

0.00361 − 0.00316x = 2x

0.00361 = 2.00316x

x = 0.00018

The volume is 1.00 L, so the concentrations at equilibrium are:

[N₂] = 0.114 − x = 0.11382

[O₂] = 0.114 − x = 0.11382

[NO] = 2x = 0.00036

2(a) Cl₂ → 2 Cl

Kc = [Cl]² / [Cl₂]

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_{2}&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]

1.2×10⁻⁷ = (2x)² / (2 − x)

1.2×10⁻⁷ (2 − x) = 4x²

2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²

2.4×10⁻⁷ ≈ 4x²

x² ≈ 6×10⁻⁸

x ≈ 0.000245

2x ≈ 0.00049

2(b) F₂ → 2 F

Kc = [F]² / [F₂]

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_{2}&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]

1.2×10⁻⁴ = (2x)² / (2 − x)

1.2×10⁻⁴ (2 − x) = 4x²

2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²

2.4×10⁻⁴ ≈ 4x²

x² ≈ 6×10⁻⁵

x ≈ 0.00775

2x ≈ 0.0155

F₂ dissociates more, so Cl₂ is more stable at 1000 K.

7 0
3 years ago
The creation and use of GM plants can reduce genetic diversity.<br><br>A.True<br>B.False​
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Technically, this is true...

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3 years ago
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(4) A stainless steel storage tank contains 5 kg of carbon dioxide gas and 7 kg of argon gas. How many kmoles are in the tank?
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Answer:

0.2886 Kmoles

Explanation:

Number of moles is calculated by dividing mass of element/molecular mass of element

Mass of Carbon Di oxide = 5000 g or 5 kg

Molecular mass of CO2 = 12 + 16×2 = 44 g

No of moles = 5000/44 = 113.6 moles = 0.1136 Kmoles

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Molecular mass of Argon(Ar) = 40g

No of moles = 7000/40 = 175 moles = 0.175 Kmoles

Total Kmoles = 0.175+0.1136=0.2886 Kmoles

4 0
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A 1000 kg car moves with a constant speed 20.0 m/s
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Answer:

13.33 m/s^2

Explanation:

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