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3 years ago

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Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.

55 × 10 9 J to arrive on an area of 4.25 m 2 ?

1 answer:

alina1380 [7]3 years ago

4 0

Answer:

The time taken by $3.55\times 10^9\ J$ to arrive on an area of $4.25\ m^2$ is $6.14\times 10^5\ seconds$ .

Explanation:

Given the intensity of the energy is $1.36\ kW/m^2$

And the arriving energy is $3.55\times 10^9\ J$

Also, the area in which energy is being arriving is $4.25\ m^2$

Now, we will use relation between energy $(E)$ , intensity of energy $(p)$ , area $(A)$ and time $(T)$ .

Where energy is in Joule, intensity is in $kW/m^2$ , area is in $m^2$ and time is in seconds.

The equation is

$E=pAT\\\\T=\frac{E}{pA}\\\\T=\frac{3.55\times 10^9}{1.36\times 10^3\times 4.25}\\\\T=0.614\times 10^6\ s\\T=6.14\times 10^5\ seconds$

So, the time taken by $3.55\times 10^9\ J$ to arrive on an area of $4.25\ m^2$ is $6.14\times 10^5\ seconds$ .

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