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Zorluk: Çok zorScientific Notation and Unit Conversions

During a space weather observation, a satellite detector measures the solar wind proton flux. The detector has an active surface area of 4.0×10−4 m24.0 \times 10^{-4}\text{ m}^2. Over a continuous period of 2.5 hours2.5\text{ hours}, the detector records a total of 1.8×1015 protons1.8 \times 10^{15}\text{ protons}. What is the average proton flux measured by the detector, in units of protons⋅cm−2⋅s−1\text{protons}\cdot\text{cm}^{-2}\cdot\text{s}^{-1}?

  1. A
    5.0×106 protons⋅cm−2⋅s−15.0 \times 10^{6}\text{ protons}\cdot\text{cm}^{-2}\cdot\text{s}^{-1}
  2. 5.0×1010 protons⋅cm−2⋅s−15.0 \times 10^{10}\text{ protons}\cdot\text{cm}^{-2}\cdot\text{s}^{-1}Cevap
  3. C
    5.0×1012 protons⋅cm−2⋅s−15.0 \times 10^{12}\text{ protons}\cdot\text{cm}^{-2}\cdot\text{s}^{-1}
  4. D
    5.0×1014 protons⋅cm−2⋅s−15.0 \times 10^{14}\text{ protons}\cdot\text{cm}^{-2}\cdot\text{s}^{-1}

Cevap

The average proton flux measured by the detector is 5.0×1010 protons⋅cm−2⋅s−15.0 \times 10^{10}\text{ protons}\cdot\text{cm}^{-2}\cdot\text{s}^{-1}.
The correct option is 5.0×1010 protons⋅cm−2⋅s−15.0 \times 10^{10}\text{ protons}\cdot\text{cm}^{-2}\cdot\text{s}^{-1} because it represents the total number of recorded protons (1.8×10151.8 \times 10^{15}) divided by the product of the surface area in square centimeters (4.0 cm24.0\text{ cm}^2) and the time duration in seconds (9.0×103 s9.0 \times 10^3\text{ s}).

Adım Adım Çözüm

1
Convert the continuous time interval from hours to seconds.
2.5 hours×3600 s/hour=9.0×103 s2.5\text{ hours} \times 3600\text{ s/hour} = 9.0 \times 10^3\text{ s}
The target unit requires time in seconds (s−1\text{s}^{-1}).
2
Convert the active detector surface area from square meters (m2\text{m}^2) to square centimeters (cm2\text{cm}^2).
4.0×10−4 m2×104 cm2/m2=4.0 cm24.0 \times 10^{-4}\text{ m}^2 \times 10^4\text{ cm}^2/\text{m}^2 = 4.0\text{ cm}^2
The target unit requires area in square centimeters (cm−2\text{cm}^{-2}).
3
Calculate the average proton flux by dividing the total number of protons by the product of the converted area and time.
Flux=1.8×1015 protons4.0 cm2×9.0×103 s=5.0×1010 protons⋅cm−2⋅s−1\text{Flux} = \frac{1.8 \times 10^{15}\text{ protons}}{4.0\text{ cm}^2 \times 9.0 \times 10^3\text{ s}} = 5.0 \times 10^{10}\text{ protons}\cdot\text{cm}^{-2}\cdot\text{s}^{-1}
Flux is defined as the quantity of particles passing through a unit area per unit time.

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