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Sav [38]
3 years ago
6

Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan

et increases linearly with the height of the atmosphere as measured from the top of a visible boundary layer, defined as 0 kilometers in altitude. The instruments on board can withstand a temperature of 601 K. At what altitude will the probe's instruments fail? A. 50 kilometers B. 80 kilometers C. 83 kilometers D. 100 kilometers E. 111 kilometers
Physics
1 answer:
Elan Coil [88]3 years ago
8 0

Answer:

Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous planet increases linearly with the height of the atmosphere as measured from the top of a visible boundary layer, defined as 0 kilometers in altitude. The instruments on board can withstand a temperature of 601 K. At what altitude will the probe's instruments fail? A. 50 kilometers B. 80 kilometers C. 83 kilometers D. 100 kilometers E. 111 kilometers

Explanation:

A. 50 kilometers

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Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

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\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

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