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3 years ago

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The mean radius of earth is 6,371.0 kilometers and the mean radius of Earth's Moon is 1,737.5 kilometers.What is the approximate

difference in the mean circumference,in kilometers, of earth and Earth's moon? Round your answer to the nearest tenth of a kilometer.

1 answer:

Dmitrij [34]3 years ago

4 0

D≈2π(6371-1737.5)

d≈9267π

d≈29113.1km (to nearest tenth of a kilometer)

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3 years ago

Read 2 more answers

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Katen [24]

Answer:

Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

$T_1cos\theta_1-T_2cos\theta_2=0$ (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

$T_1sin\theta_1+T_2sin\theta_2-W=0$ (2)

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$T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2$

Then, you sum the equations:

$T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2$ (3)

Next, you use the following identity:

$sin^2\theta+cos^2\theta=1$

and you obtain in the equation (3):

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With this values you can calculate the value of the another angle, by using the equation (1):

$\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°$

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$l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\$

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By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

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3 years ago