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vodka [1.7K]
3 years ago
6

Planet Tatoone is about 1.7 AU from its Sun. Approximately how long will it take for light to travel from the Sun to Tatoone in

minutes? Use 3 × 108 m/s for the speed of light.
Physics
1 answer:
Radda [10]3 years ago
8 0

Answer:

The value is   t =  14.129 \  minutes    

Explanation:

From the question we are told that

  The distance of planet Tatoone is  d =  1.7 \ AU  =  1.7 *1.496* 10^{11}=2.543*10^{11} \ m

   The  speed of light is  c =  3.0*10^{8} \  m/  s

Generally the time taken is mathematically represented as

     t =  \frac{d}{c}

=> t =  \frac{2.543*10^{11}}{3.0*10^{8} }

=>    t =  847.7 \  s

Now converting to minutes

       t =  \frac{847.7}{60}

   =>     t =  14.129 \  minutes    

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a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,

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Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

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It is delayed with 5/36 sec with respect to displacement.

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c) The formula for calculating the period of an ideal pendulum anywhere is

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6 0
1 year ago
A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
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Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

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Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

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Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

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E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0
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