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Inessa [10]
3 years ago
7

Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th

e planets combined, this is essentially the position of the center of mass of the solar system.)
Physics
1 answer:
8090 [49]3 years ago
3 0

Answer:

r_{cm} = 0.074 m from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

M_s = 1.98 \times 10^{30} kg

M_j = 1.89 \times 10^{27} kg

distance between Sun and Jupiter is given as

r = 7.78 \times 10^{11} m

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}

r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}

r_{cm} = 0.074 m from the position of the center of the Sun

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Answer: V_{f}=2.96m/s    

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As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

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In addition, we know N=w, then \sum F_{y}=0

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\sum F_{x}=m.a

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Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

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On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

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We already know the value of  d and calculated a, we have to find t:

t=\sqrt{\frac{2d}{a}}   (11)

t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

If V_{o}=0

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Finally we get the value of the Final Velocity of the block:

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Answer:

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Explanation:

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Explanation:

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e
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