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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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Molodets [167]

Speed of the tip of the minute hand=V=0.0244 cm/s

Explanation:

The angular velocity of the minute hand is given by

\omega= \frac{2\pi}{T}

T= time period of the minute hand=60 min=3600 s

so ω= 2 π/3600 rad/s

Now linear velocity v= r ω

r= radius of minute hand=14 cm

so v= 14 (2 π/3600)

V=0.0244 cm/s

8 0
3 years ago
Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out
blondinia [14]

Answer:

\theta = 20.98 degree

Explanation:

As we know that the speed of the sound is given as

v = 332 + 0.6 t

now at t = 273 k = 0 degree

v = 332 m/s

so we have

a sin\theta = N\lambda

a sin\theta = N(\frac{v_1}{f})

now when temperature is changed to 313 K we have

t = 313 - 273 = 40 degree

now we have

v = 332 + (0.6)(40)

v_2 = 356 m/s

a sin\theta' = N(\frac{v_2}{f})

now from two equations we have

\frac{sin19.5}{sin\theta} = \frac{332}{356}

so we have

sin\theta = 0.358

\theta = 20.98 degree

7 0
3 years ago
a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou
zmey [24]

Answer:

1.56 J

Explanation:

The potential energy only depends on the vertical height from the ground level.

We consider the ground level to have zero P.E.

So when it is 2 m above the ground level,

P.E. =  mgh

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4 years ago
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statuscvo [17]

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Explanation:

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The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.
tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

W=K_f -K_i

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The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

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5 0
3 years ago
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