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tino4ka555 [31]

3 years ago

12

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 44.0 N at the s

urface of the earth. In this problem, use 9.81 m/s2 for the acceleration due to gravity on earth.

1 answer:

BabaBlast [244]3 years ago

6 0

Answer:

Weight at the surface of Jupiter's moon Io is 8.13 N .

Explanation:

Given :

Acceleration due to gravity at the surface of Jupiter's moon is $g_m=1.81\ m/s^2$ .

Weight of watermelon in earth , $W=44\ N$ .

Acceleration due to gravity at the surface of earth is $g=9.81\ m/s^2$ .

We know , weight is given by :

$W=mg\\m=\dfrac{W}{g}\\\\m=4.49\ kg$

Therefore , mass at the surface of Jupiter's moon Io is :

$W_m=mg_m\\\\W_m=4.49\times 1.81\\\\W_m=8.13 \ N$

Hence , this is the required solution .

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From the question we are told that

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Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

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$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

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=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

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Hence the length of the rod as measured by a stationary observer is

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Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

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=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

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3 years ago

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