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german
3 years ago
8

At noon on a clear day, sunlight reaches the earth's surface at Madison, Wisconsin, with an average intensity of approximately 5

.00 5.00 kJ·s−1·m−2. If the sunlight consists of photons with an average wavelength of 510.0 nm, how many photons strike a 7.00 7.00 cm2 area per second?
Physics
1 answer:
Bas_tet [7]3 years ago
6 0

Answer: 7.175*10^18

Explanation:

Energy of a wave:

E = nhc/λ, where

E = energy of the wave

n = no of photons

h = Planck's Constant

c = speed of light

λ = wavelength of the wave

4000 = (n * 6.63*10^-34 * 3*10^8) / (510*10^-9)

4000 = 1.989*10^-25 * n / 510*10^-9

4000 = 3.9*10^-19 * n

n = 1.025*10^22 photons per second per meter²

7.0 cm² = 7.0 / 10,000 m²

= 7 x 10⁻⁴

Photons per second = 1.025*10^22 * 7 x 10⁻⁴

= 7.175*10^18 photons per second

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Two equipotential surfaces surround a +1.70 x 10-8-C point charge. How far is the 120-V surface from the 54.0-V surface?
UNO [17]

Answer:

1.55 m

Explanation:

The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:

V =\frac{k*q}{r}

As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.

Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:

r1 =\frac{k*q}{V1} = \frac{9e9 N*m2/C2*1.7e-8C}{120 V} = 1.28 m

r2 =\frac{k*q}{V2} = \frac{9e9 N*m2/C2*1.7e-8C}{54V} = 2.83 m

The distance between both points, is just the difference between the radius of both spheres, as follows:

r₂ - r₁ = 1.55 m

5 0
3 years ago
A 3-kg mass is in free fall. What is the velocity of the mass after 7 seconds??
dmitriy555 [2]
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6 0
3 years ago
What does an object in equilibrium has a net force of .
Vlad1618 [11]
From our studies of the work of Professor Newton,
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In either case, since the object has no acceleration, we glance
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4 years ago
If a flea can jump straight up to a height of 0.550 m , what is its initial speed as it leaves the ground?
olga55 [171]
Let u = upward initial speed of the flea.

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3 years ago
A student drops a 0.4kg ball from a height a of 49m above the ground. Neglect drag. Answer each of the following questions about
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<em>t</em> ² = (98 m) / <em>g</em>

<em>t</em> = √((98 m) / <em>g</em>) = √(10) s ≈ 3.2 s

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so that at the time found previously, the ball will have attained a velocity of

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4 0
3 years ago
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