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sladkih [1.3K]
3 years ago
7

Assume that light of wavelength 6000A is coming from a star. What is the limit of resolution of a telescope whose objective has

a diameter of 100 inch?
Engineering
1 answer:
Paul [167]3 years ago
4 0

Answer:

θ=0.0288 radian

Explanation:

resolution limit is the minimum angular separation of  two sources that can be viewed  distinctly    by telescope

\theta =\frac{1.22\times \lambda}{D}

\lambda=6000\times 10^{-8} cm=6 \times 10^{-5} cm

d=100 inch=100\times 2.54=254cm

\theta = \frac{1.22 \times 6 \times 10^{-5}}{254}

θ=0.0288 radian

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Anna007 [38]

Answer:

COP of the heat pump is 3.013

OP of the cycle is  1.124

Explanation:

W = Q₂ - Q₁

Given

a)

Q₂ = Q₁ + W

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     = 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124

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4 years ago
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Answer:

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a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What
Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, dq/dt = 100 W

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Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

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A = 0.24 m²

Temperature of the surrounding, T_0 = 20^0 C = 273 + 20 = 293 K

Temperature of the inside of the box, T_{in} = ?

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The formula for the rate of heat conduction is given by:

dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C

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4 years ago
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