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Usimov [2.4K]
3 years ago
13

The earth travels around the sun once a year in an approximately circular orbit whose radius is 1.50x10^11 m. From this data det

ermine (a) the orbital speed of the earth and (b) the mass of the sun.
Physics
1 answer:
seraphim [82]3 years ago
8 0
(a) Determine the circumference of the Earth through the equation,
            C = 2πr
Substituting the known values, 
           C = 2π(1.50 x 10¹¹ m)
             C = 9.424 x 10¹¹ m

Then, divide the answer by time which is given to a year which is equal to 31536000 s. 
          orbital speed = (9.424 x 10¹¹ m)/31536000 s

               orbital speed = 29883.307 m/s

Hence, the orbital speed of the Earth is ~29883.307 m/s.

(b) The mass of the sun is ~1.9891 x 10³⁰ kg. 
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A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco
Sunny_sXe [5.5K]

Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies 45^o south of east. The y-component will be negative and the x-component will be positive.

x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

x_2=40cos60^o=20km

y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

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r=\sqrt{x^2+y^2}=41.32km

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