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Oduvanchick [21]
3 years ago
9

A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi

nish line?
Physics
1 answer:
Ahat [919]3 years ago
6 0
<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.8 s    

        Displacement, s = 1/4 mi =    400 meters

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.8 + 0.5 x a x 6.8²

                      a = 17.30 m/s²

Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.8 s

      Acceleration, a = 17.30 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.30 x 6.8

                      v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

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3 years ago
The binding energies of K-shell and L-shell electrons in a certain metal are EK and EL, respectively, If a Kαx ray from this met
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Answer:

The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

Explanation:

The relationship between energy and wavelength is expressed below:

E = hc/λ

λ = hc/EK - EL

Considering the condition of Bragg's law:

2dsinθ = mλ

For the first order Bragg's law of reflection:

2dsinθ = (1)λ

2dsinθ = hc/EK - EL

d = hc/2sinθ(EK - EL)

Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.

Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

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Assume we’re able to travel to your planet and decide to take some fireworks with us to celebrate our journey.
julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be \frac{1876.8}{g}

Explanation:

Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.

writing equation of motion in vertical direction:

v_{y}=u_{y}+(-g) t

u_{y}= u\sin \phi

and v_{y}=0

therefore \frac{u\sin \phi }{g} =t

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s_{x}=u_{x}t

u_{x} = u\cos \phi

therefore the equation becomes s_{x}=\frac{u^{2}   \sin \phi  \cos \phi}{g}

therefore horizontal distance traveled =\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}

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Answer: Im not entirly sure but I think It's D all the above. I think all but B because I never really heard of that but if you look in our history I think that happen im not sure I would wait untill you know that somone knows for sure.

Explanation:

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