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rewona [7]
3 years ago
5

On a cloudless day, the sunlight that reaches the surface of the earth has an intensity of about 1.40 × 10 3 1.40×103 W/m². What

is the electromagnetic energy contained in 6.00 m³ of space just above the earth's surface?
Physics
1 answer:
nalin [4]3 years ago
3 0
<h3><u>Answer;</u></h3>

2.8 × 10^-5 Joules

<h3><u>Explanation;</u></h3>

Considering the base area of the 'space' to be 1m²

Therefore; it's 'height' h = 6.0 m  

Time taken (t) for all the light in this space to travel down 6.0 m ..  

t = h/c

  = 6.0 m / 3.0^8m/s

   t = 2 × 10^-8 s  

Energy reaching this 1 m² base = 1.40 × 10^3 J/s .. (1400W)  

Net energy (E) travelling through this space = 1400J/s x 2 × 10^-8 s  

E = <u>2.8 × 10^-5 J</u>

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Answer:

english please

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5 0
3 years ago
Many arlines restrict maximum mass of a case to 23 kg why​
Alexxandr [17]

Answer:

weight

Explanation:

the weight of an object on an airline is one of the most important thing a pilot has to consider when prepping a flight and that is because if there is too much weight then the plane simply can't fly. imagine if everyone wanted to bring a 50 kg box. there are at least  200 people.  that alone is 10,000 lg of weight than you have to factor in all the people, wires on the plane, and certain appliances that some planes have.

7 0
3 years ago
During which type of collision do the two objects stick together?
inysia [295]

Answer:

D. Perfectly inelastic

Explanation:

Kinetic energy is lost so the two bodies stick together.

6 0
3 years ago
A 59 kg man has a total mechanical energy of 150,023. J. If he is swinging downward and is currently 2.6 m above the ground, wha
Alborosie

Answer:

v = 70.95 \ m/s.

Explanation:

Given data:

Mass of the man, m = 59 \ kg

Total mechanical energy, E_{i} = 150,023 \ \rm J

Height, h = 2.6 \ m

Suppose there is no external force acting on the man. In this situation, the total mechanical energy (kinetic + potential) will remain steady.

Let the speed of the man at 2.6 m be <em>v</em>.

Thus,

E_{i} = E_{f}

E_{i} = \frac{1}{2}mv^{2} + mgh

150023 = 0.5 \times 59 \times v^{2} + 59 \times 9.80 \times 2.6

\Rightarrow \ v = 70.95 \ m/s.

6 0
3 years ago
(NEED HELP BADLY)
spin [16.1K]

Answer:

The answer to your question is below

Explanation:

Data 1

mass 1 = 250

mass 2 = 250 kg

gravity constant = 6.67 x 10⁻¹¹ Nm²/kg²

distance = 8 m

Formula

F = G\frac{m1m2}{r^{2} }

Substitution

F = 6.67 x 10^{-11} \frac{250 x 250}{8^{2} }

Result

F = 0.000000065 N

Data 2

mass 1 = 1000 kg

mass 2 = 1000 kg

distance = 5 m

Substitution

F = 6.67 x 10^{-11} \frac{1000 x 1000 }{5^{2} }

Result

F = 0.000002667 N      

8 0
4 years ago
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