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Neko [114]
3 years ago
11

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

1.55 × 105 m. Suppose further that a second group of engineers programmed the orbiter to go to 1.55 × 105 ft. What was the difference in kilometers between the two altitudes? How low did the probe go?
Physics
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

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2.0 Hz

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(b) A Blu-ray laser has a power of 5 milliwatts (1 watt = 1 J s−1). How many photons of light are produced by the laser in 1 hou
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Answer:

The number of photons of light produced by the laser in 1 hour is

1 Photon / hour

Explanation:

Number of photons of light produced is given by

Number of photons = \frac{Power}{Energy}

From the question,

Power = 5 mW (milliwatts) = 5 × 10⁻³ W

Since 1 Watt = 1 Js⁻¹

Then, 5 × 10⁻³ W = 5 × 10⁻³ Js⁻¹

For the Energy,

As given from the question

Energy=(Power)x(time)

Time = 1 hour = (1 × 60 × 60) s = 3600 s

∴ Energy = 5 × 10⁻³ Js⁻¹ x 1 x 3600s

Energy =18.0 J

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Number of photons = \frac{Power}{Energy}

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Number of photons = \frac{0.005}{18}

Number of photons = 2.78 × 10⁻⁴ Photons / sec

This is the number of photons produced in 1 second.

For the number of photons produced in 1 hour, we will multiply the result by 3600

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Number of photons = 2.78 × 10⁻⁴ Photons / sec

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