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krok68 [10]
3 years ago
7

The path of the earth around the sun is an ellipse with the sun at one focus. The ellipse has a major axis of 186,000,000 miles

and eccentricity of 0.017. Find the distance between the earth and the sun when the earth is (a) closest to the sun and (b) farthest from the sun
Physics
1 answer:
Charra [1.4K]3 years ago
7 0

Answer:

a) r = 182838000\,mi, b) r = 189162000\,mi

Explanation:

a) The distance between the center and one of the foci is:

c = \epsilon \cdot a

c = (0.017)\cdot (186000000\,mi)

c = 3162000\,mi

The shortest distance is:

r = a - c

r = 186000000\,mi -3162000\,mi

r = 182838000\,mi

b) The longest distance is:

r = a + c

r = 186000000\,mi +3162000\,mi

r = 189162000\,mi

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The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

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K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

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= 1/12 *4 * (0.6)²

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on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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