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seraphim [82]
1 year ago
9

The intensity of sunlight at the Earth is about 1367 W/m^2. How many times further away would the Sun have to be from the Earth

before solar radiation was only as intense as a 100W light bulb 10 meters away, assuming that all energy from the bulb goes into radiation and spreads out evenly in all directions?
Physics
1 answer:
grandymaker [24]1 year ago
7 0

The intensity of sunlight on the Earth is about 1367 W/m^2. The earth has to be 131 the distance farther away

This is further explained below.

<h3>What is the distance?</h3>

Generally, bet the power emitted by the sun is P. so. at the distance of earth intensely where $r_E$ is the distance of the earth from the sun. were $I_1=1362 \frac{w}{m^2}$

Therefore

p=4 \pi 1367 r_E^2

let at distance \gamma r_E$ is a positive number) the intensity is the same as the intensity at 10m away from a 100w bulb. So.

In conclusion,

\begin{aligned}&\frac{P}{4 \pi\left(\gamma r_{\varepsilon}\right)^2}=\frac{100}{4 \pi 10^2}\\\\&\text { or, }\left(\gamma r_E\right)^2=P=4 \pi 1367 r_E^2\\\\&\therefore \gamma=\sqrt{4 \pi \times 1367}\\\\&=131.06\\\\ }\end{aligned}

Read more about distance

brainly.com/question/15172156

#SPJ1

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C. green light

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Answer:

276 days

Explanation:

1/4 th of the original means <u><em>2 half lives</em></u>

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An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p
kotykmax [81]

Answer:

a)v=1.77\times 10^6\ m/s

b)v=3.872\times 10^6\ m/s

c)v=5.5\times 10^6\ m/s

d)v=6.7\times 10^6\ m/s

e)v=7.7\times 10^6\ m/s

Explanation:

Given that

d = 2 cm

V = 200 V

u=4\times 10^5\ m/s

We know that

F = E q

F = m a

E = V/d

So

m a =  q .V/d b            

a=\dfrac{q.V}{m.d}                 ---------1

The mass of electron

m=9.1\times 10^{-31}\ kg

The charge on electron

q=1.6\times 10^{-19}\ C

Now by putting the all values in equation 1

a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2

a=1.5\times 10^{15}\ m/s^2

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v^2=u^2+2as

a)

s = 0.1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}

v=\sqrt{3.16\times 10^{12}}\ m/s

v=1.77\times 10^6\ m/s

b)

s = 0.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}

v=\sqrt{1.5\times 10^{13}}\ m/s

v=3.872\times 10^6\ m/s

c)

s = 1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}

v=\sqrt{3.06\times 10^{13}}\ m/s

v=5.5\times 10^6\ m/s

d)

s = 1.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}

v=\sqrt{4.5\times 10^{13}}\ m/s

v=6.7\times 10^6\ m/s

e)

s = 2 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}

v=\sqrt{6.06\times 10^{13}}\ m/s

v=7.7\times 10^6\ m/s

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