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a_sh-v [17]
3 years ago
15

the smallest unit in physics is the Planet length.wr need to know constant (h) the speed of light (c) and Newton's Gravitational

Constant ( G). Find the Planck Length.
Physics
2 answers:
fgiga [73]3 years ago
4 0

Yes, that's right.  It's the 'Planck' length, not the 'Planet' length.

You could easily find these with a web search.  But in gratitude
for the bountiful 5 points, I've saved you the trouble. 
AND guess what !   By doing that, I learned something, and
you didn't.

Speed of light (c):                 299,792,458 meters per second

Gravitational constant (G):   6.67 x 10⁻¹¹  newton-meter²/kilogram²

Planck's Konstant (h):           6.63 x 10⁻³⁴ joule-second

Planck Length:                      1.6 x 10⁻³⁵ meter
                                            (about 10⁻²⁰ the size of a proton)

Planck Time:                          10⁻⁴³ second
                                             (about the time it takes to travel
                                              a Planck Length at the speed of light)  

den301095 [7]3 years ago
4 0
Planck length is the going to be the smallest unit of measure known. The speed of light is a constant, the universes speed "limit". Although quantum entanglement beat that speed, but it doesn't threaten the speed of light since the process doesn't hold information. And that was from my own knowledge of the information I soak up daily in college education on the path to a PhD in physics.
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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
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Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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