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laila [671]
3 years ago
7

At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de

vice to stop the train so that it will not plow through the station if the engineer misjudges the stopping distance. While waiting, you wonder what would be the fastest train that the spring could stop at its full compression which is L=3 ft . To keep the passengers safe when the train stops, you assume a maximum stopping acceleration of g/2. You also guess that a train weighs half a million lbs. For purpose of getting an estimate, you decide to assume that all frictional force are negligible.
Physics
1 answer:
vazorg [7]3 years ago
5 0

Answer:

   v=2.13 m/s

Explanation:

Given that

L= 3 ft

We know that

1 ft = 0.3048 m

L=0.91 m

a= g/2

F= m a

F= m g/2 = K L

k= (mg)/(2L)

Now from energy conservation

Kinetic energy of train = potential energy of the spring

\dfrac{1}{2}mv^2=\dfrac{1}{2}kL^2

k= (mg)/(2L)

\dfrac{1}{2}mv^2=\dfrac{1}{2}\times \dfrac{mg}{2L}L^2

v^2=\dfrac{g}{2}L

v=\sqrt{\dfrac{g}{2}L}

By putting the values

v=\sqrt{\dfrac{g}{2}L}

v=\sqrt{\dfrac{10\times 0.91}{2}}     ( take g =10 m/s²)

v=2.13 m/s

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Four pairs of objects have the masses as described below, along with the distances between
lord [1]

Answer:

<h2>Mass of 1 Kg and 2 Kg, 1 meters apart.</h2>

Explanation:

The gravitational force is defined as

F=G\frac{m_{1} m_{2} }{r^{2} }

By definition, the gravitational force depends directly on the product of the masses and indirectly on the distance between the masses, which means the further they are, the less gravitational force would be. And, the greater the masses, the greater the gravitational force.

Among the options, the pair that would have the greatest gravitational force is  Mass of 1 Kg and 2 Kg, with 1 meter between them.

Notice that the last choice includes the same masses but with a greater distance between them, that means it would be a weaker graviational force.

Therefore, the right answer is the second choice.

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4 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

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Here, I hope this helps.

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