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Juliette [100K]
3 years ago
9

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha

t is the asteroid's orbital radius?
Physics
1 answer:
anzhelika [568]3 years ago
6 0

To solve the problem, use Kepler's 3rd law : 

T² = 4π²r³ / GM 

Solved for r : 

r = [GMT² / 4π²]⅓ 

but first covert 6.00 years to seconds : 

6.00years = 6.00years(365days/year)(24.0hours/day)(6... 
= 1.89 x 10^8s 

The radius of the orbit then is : 

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓ 
= 6.23 x 10^11m

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Explanation:

4 0
3 years ago
A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s for 6.2 s. It then accelerates at a rate of-1.2
yanalaym [24]

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.27 s

c) Total distance traveled = 402.43 m

Explanation:

a) Maximum speed is obtained after the end of acceleration

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        v = 13.5 + 1.9 x 6.2 = 25.28 m/s

    Maximum speed = 25.28 m/s

b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.

         v = u + at  

         0 = 25.28 - 1.2 t

         t = 21.07 s

    Total time = 6.2 + 21.07 = 27.27 s

c) Distance traveled for the first 6.2 s

          s = ut + 0.5 at²

          s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m

   Distance traveled for the second 21.07 s

          s = ut + 0.5 at²

          s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m

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3 years ago
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Explanation:

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In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is {\displaystyle {\begin{smallmatrix}{\frac {1}{2}}mv^{2}\end{smallmatrix}}}{\begin{smallmatrix}{\frac {1}{2}}mv^{2}\end{smallmatrix}}. In relativistic mechanics, this is a good approximation only when v is much less than the speed of light.

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