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Vera_Pavlovna [14]
3 years ago
8

Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

ing these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. 10 times earth's distance from the sun in watts/m^2
Physics
1 answer:
nalin [4]3 years ago
5 0

Answer:

13 W/m^2

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

I \propto \frac{1}{r^2}

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2} (1)

where in this problem, we have:

I_1 = 1300 W/m^2 apparent brightness at a distance r_1, where

r_1 = 150 million km

We want to estimate the apparent brightness at r_2, where r_2 is ten times r_1, so

r_2 = 10 r_1

Re-arranging eq.(1), we find I_2:

I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2

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The actual mass of a 1 cm² piece of copy paper is determined as 0.0074 g.

<h3>Area of the sheet of paper in square centimeters</h3>

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Mass = 0.0074 g

The complete question is below:

A sheet of 8.5 inch by 11inch paper has a mass of 4.50 grams one square centimeter , 1cm2 of that sheet of paper will have a mass closets to ____grams ( (hint use 1inch = 2.54 centimeters and area = length times)

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The speed of the object is the rate where it changes position relative to the frame of reference. The speed equates to a stipulated speed and motion direction of the item, and the further calculation can be defined as follows:

Given:

\bold{\theta = 25^{\circ}}\\\\\bold{Velocity= 50\ \frac{m}{s}}\\\\

To find:

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Solution:

When the angle is north-east, the angle is the east measured

Using formula:

\bold{\text{eastward component = (velocity)}\times \cos \theta}}

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So, the final answer is "45.3155 m/s".

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Answer:

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