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ikadub [295]
3 years ago
13

A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th

e star, determine (a) the angular speed of the planet, (b) the tangential speed of the planet, and (c) the magnitude of the planet's centripetal acceleration.
Physics
1 answer:
PIT_PIT [208]3 years ago
8 0

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

   \omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}

   \omega = 1.864\times 10^{-7}\ rad/s

b) tangential speed

   v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}

       v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

      a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}

          a = 3.62 x 10⁻³ m/s²

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Answer:

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Explanation:

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The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

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Answer:

K=0.023J

Explanation:

From the question we are told that:

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Answer:

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a. 1st law

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