Physics: On a cloudless day, the sunlight that reaches the surface of the earth has an intensity of about W/m². What is the electromagnetic energy contained in 5.20 m³ of space just above the earth s surface?

On a cloudless day, the sunlight that reaches the

Ответ:

kenzietaylre

21.01.2022

Physics

1.907 x 10⁻⁵ J.

Explanation:

Given,

Volume of space, V = 5.20 m³

Assuming the intensity of sunlight(S) be equal to 1.1 x 10³ W/m².

Electromagnetic energy = ?

$E = \mu V$

$E = (\dfrac{S}{c})\times V$

where c is the speed of light.

$E = (\dfrac{1.1\times 10^3}{3\times 10^8})\times 5.20$

$E = 1.907\times 10^{-5}\ J$

Hence, Electromagnetic energy is equal to 1.907 x 10⁻⁵ J.

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Ответ:

Mikey3414

10.06.2020

Physics

a) p = 95.66 cm, b) p = 93.13 cm

Explanation:

For this problem we use the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distances to the object and the image, respectively

the power of the lens is

P = 1 / f

f = 1 / P

f = 1 / 2.25

f = 0.4444 m

the distance to the object is

$\frac{1}{p} = \frac{1}{f} -\frac{1}{q}$

the distance to the image is

q = 85 -2

q = 83 cm

we must have all the magnitudes in the same units

f = 0.4444 m = 44.44 cm

we calculate

$\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}$

1 / p = 0.010454

p = 95.66 cm

b) if they were contact lenses

q = 85 cm

$\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}$

1 / p = 0.107375

p = 93.13 cm

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On a cloudless day, the sunlight that reaches the surface of the earth has an intensity of about W/m². What is the electromagnetic energy contained in 5.20 m³ of space just above the earth's surface?

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