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AleksAgata [21]
3 years ago
14

At noon on a clear day, sunlight reaches the earth\'s surface at Madison, Wisconsin, with an average power of approximately 3.00

kJ·s–1·m–2. If the sunlight consists of photons with an average wavelength of 510.0 nm, how many photons strike a 2.70 cm2 area per second?
Physics
1 answer:
alexira [117]3 years ago
4 0
Energy of a wave:
E = nhc/λ
3000 = (n x 6.63 x 10⁻³⁴ x 3 x 10⁸)/(510 x 10⁻⁹)
n = 7.69 x 10 ²¹ photons per second per meter²
2.70 cm² = 2.70/10,000 m²
= 2.7 x 10⁻⁴
Photons per second = 7.69 x 10 ²¹ x 2.7 x 10⁻⁴
= 2.08 x 10¹⁸ photons per second
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Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

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\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

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