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Dmitry_Shevchenko [17]
3 years ago
9

The speed of a light wave in a certain transparent material is 0.589 times its speed in vacuum, which is 3.00 x108 m/s. When yel

low light with a frequency of 5.25 x104 Hz passes through this material, what is its wavelength in nanometers? Number 囗 n m
Physics
1 answer:
irakobra [83]3 years ago
8 0

Answer:

The wavelength of the yellow light in given transparent material is <em><u>337 nm</u></em>

Explanation:

Speed of the light in vacuum , c=3.00\times 10^{8}\frac{m}{s}

Therefore speed of light in the given material is v=0.589\times c

=>v=0.589\times 3.00\times 10^{8}\frac{m}{s}=1.767\times 10^{8}\frac{m}{s}

frequency of yellow light , f=5.25\times 10^{14}Hz

Wavelength in the given transparent material , \lambda =\frac{v}{f}=\frac{1.767\times 10^{8}}{5.25\times 10^{14}}m

=> \lambda =3.366\times 10^{-7}m=336.6nm

=>\lambda \simeq 337nm

Thus the wavelength of the yellow light in given transparent material is <em><u>337 nm</u></em>

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4 years ago
A student team is to design a human powered submarine for a design competition. The overall length of the prototype submarine is
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Answer:

a) The speed is 61.42 m/s

b) The drag force is 10.32 N

Explanation:

a) The Reynold´s number for the model and prototype is:

Re_{m} =\frac{p_{m}V_{m}L_{m}   }{u_{m} }

Re_{p} =\frac{p_{p}V_{p}L_{p}   }{u_{p} }

Equaling both Reynold's number:

\frac{p_{p}V_{p}L_{p}   }{u_{p} }=\frac{p_{m}V_{m}L_{m}   }{u_{m} }

Clearing Vm:

V_{m} =\frac{p_{p}V_{p}L_{p} u_{m}   }{u_{p} p_{m} L_{m} }=\frac{999.1*0.56*8*1.849x10^{-5} }{1.138x10^{-3}*1.184*1 } =61.42m/s

b) The drag force is:

\frac{F_{Dm} }{p_{m}V_{m}^{2}L_{m}^{2}     } =\frac{F_{Dp} }{p_{p}V_{p}^{2}L_{p}^{2}     } \\F_{Dp} =\frac{F_{Dp}p_{p}V_{p}^{2}L_{p}^{2} }{p_{m}V_{m}^{2}L_{m}^{2}     } \\F_{Dp}=\frac{2.3*999.1*0.56^{2} *8^{2} }{1.184*61.42^{2}*1^{2}  } =10.32N

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3 years ago
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2 years ago
Which method involves using tools to control invasive species?
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Answer;

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3 years ago
Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volu
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Answer:

1) ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

2) W₁ = 0 J

3) P = 41.66 x 10³ Pa = 41.66 KPa

4) v = 1618.72 m/s

5) ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

6) W₂ = - 7.33 KJ

Explanation:

1)

The heat transfer for a constant volume process is given by the formula:

ΔQ₁ = ΔU = n Cv ΔT

where,

ΔQ₁ = Heat transfer during constant volume process

ΔU = Change in internal energy of gas

n = No. of moles = 4.4 mol

Cv = Molar Specific Heat at Constant Volume = 12.47 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 500 k - 300 k = 200 k

Therefore,

ΔQ₁ = (4.4 mol)(12.47 J/mol.k)(200 k)

<u>ΔQ₁ = 10.97 x 10³ J = 10.97 KJ</u>

<u></u>

2)

Since, work done by gas is given as:

W₁ = PΔV

where,

ΔV = 0, due to constant volume

Therefore,

<u>W₁ = 0 J</u>

<u></u>

4)

The average kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

but, K.E is also given by:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

mv² = 3KT

v = √(3KT/m)

where,

v = average speed of gas molecue = ?

K = Boltzman Constant = 1.38 x 10⁻²³ J/k

T = Absolute Temperature = 500 K

m = mass of a molecule = 7.9 x 10⁻²⁷ kg

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/k)(500 k)/(7.9 x 10⁻²⁷ kg)]

<u>v = 1618.72 m/s</u>

<u></u>

3)

From kinetic molecular theory, we know that or an ideal gas:

P = (1/3)ρv²

where,

P = pressure of gas = ?

m = Mass of Gas = (Atomic Mass)(No. of Atoms)

m = (Atomic Mass)(Avogadro's Number)(No. of Moles)

m = (7.9 x 10⁻²⁷ kg/atom)(6.022 x 10²³ atoms/mol)(4.4 mol)

m = 0.021 kg

ρ = density = mass/volume = 0.021 kg/0.44 m³ = 0.0477 kg/m³

Therefore,

P = (1/3)(0.0477 kg/m³)(1618.72 m/s)²

<u>P = 41.66 x 10³ Pa = 41.66 KPa</u>

<u></u>

5)

The heat transfer for a constant pressure process is given by the formula:

ΔQ₂ =  n Cp ΔT

where,

ΔQ₂ = Heat transfer during constant pressure process

n = No. of moles = 4.4 mol

Cp = Molar Specific Heat at Constant Pressure = 20.79 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 300 k - 500 k = -200 k

Therefore,

ΔQ₂ = (4.4 mol)(20.79 J/mol.k)(-200 k)

<u>ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ</u>

<u>Negative sign shows heat flows from system to surrounding.</u>

<u></u>

6)

From Charles' Law, we know that:

V₁/T₁ = V₂/T₂

V₂ = (V₁)(T₂)/(T₁)

where,

V₁ = 0.44 m³

V₂ = ?

T₁ = 500 K

T₂ = 300 k

Therefore,

V₂ = (0.44 m³)(300 k)/(500 k)

V₂ = 0.264 m³

Therefore,

ΔV = V₂ - V₁ = 0.264 m³ - 0.44 m³ = - 0.176 m³

Hence, the work done , will be:

W₂ = PΔV = (41.66 KPa)(- 0.176 m³)

<u>W₂ = - 7.33 KJ</u>

<u>Negative sign shows that the work is done by the gas</u>

4 0
3 years ago
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