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ohaa [14]
2 years ago
7

An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period

in years and average orbital distance in AU 233 years 1.46 AU 5.45 years 3.09 years
Physics
1 answer:
KatRina [158]2 years ago
7 0

Answer:

The value is  x =  45.99 \  Au

Explanation:

From the question we are told that

   The period of the  asteroid is   T =  176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s

Generally the average distance of the asteroid from the sun is mathematically represented as

            R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }

Here M is the mass of the sun with a value  

        M  =  1.99*10^{30} \  kg

         G  is the gravitational constant with value  G  =  6.67 *10^{-11}  \  m^3 \cdot kg^{-1} \cdot  s^{-2}

           R = \sqrt[3]{ \frac{6.67 *10^{-11}  * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }

=>       R = 6.88 *10^{12} \  m

Generally

         1.496* 10^{11}  \  m  \to  1 Au (Astronomical \  unit )

So

          R = 6.88 *10^{12} \  m \ \ \ \ \to \ \   x \  Au

=>      x =  \frac{6.88 *10^{12}}{1.496 *10^{11}}

=>       x =  45.99 \  Au

       

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5.
iris [78.8K]

Answer:

t = 1.659s

Explanation:

We can use the kinematics equations to solve this questions:

v = u + at

v^{2} = u^{2} +2as

where v = Final Velocity, u = initial velocity, a = acceleration, t = time, s = displacement

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u = \frac{70km}{h} =\frac{(70*1000)m}{(1*3600)s} = 19.444m/s (Convert km/h to m/s first)

a = 2m/s^{2}

s = 35m

Now we can substitute these values into the 2nd kinematics equation to find v, final velocity.

v^{2} =(19.444)^{2} +2(2)(35)\\v=\sqrt{(19.444)^{2} +2(2)(35)} \\v= 22.761m/s (5.sf)\\

b) Now we have the final velocity, we can substitute the values into the first kinematics equation to find t , the time taken.

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2 years ago
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