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Aneli [31]
3 years ago
12

A binary star system consists of two stars of masses m1 and m2. The stars, which gravitationally attract each other, revolve aro

und the center of mass of the system. The star with mass m1 has a centripetal acceleration of magnitude a1. Note that you do not need to understand universal gravitation to solve this problem.Find a2, the magnitude of the centripetal acceleration of the star with mass m2.
Physics
1 answer:
zubka84 [21]3 years ago
8 0

Answer:

 a₂ = m₁ / m₂ a₁

Explanation:

For this exercise we note that the attraction between the two stars is an action and reaction force, therefore it has the same magnitude, but it is applied to each of the bodies

Let's apply Newton's second law on the star 1

     F₁ = m₁ a₁

Newton's second law in star 2

       F₂ = m₂ a₂

       | F₁ | = | F₂ |

      m₁  a₁ = m₂  a₂

      a₂ = m₁ / m₂ a₁

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Answer:

The time it will take for the object to hit the ground will be 4.

Explanation:

You have:

h(t)=−16t²+v0*t+h0

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h(t)=−16t²+54*t+40

To know how long it will take for the object to touch the ground, the height h(t) must be zero. So:

0=−16t²+54*t+40

Being a quadratic function or parabola: f (x) = a*x² + b*x + c, the roots or zeros of the quadratic function are those values ​​of x for which the expression is 0. Graphically, the roots correspond to the points where the parabola intersects the x axis. To calculate the roots the expression is used:

\frac{-b+-\sqrt{b^{2}-4*a*c } }{2*a}

In this case you have that:

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Replacing in the expression of the calculation of roots you get:

\frac{-54+\sqrt{54^{2}-4*(-16)*40 } }{2*(-16)}  Expresion (A)

and

\frac{-54-\sqrt{54^{2}-4*(-16)*40 } }{2*(-16)} Expresion (B)

Solving the Expresion (A):

\frac{-54+\sqrt{5476 } }{2*(-16)}= \frac{-54+74}{2*(-16)}=\frac{20}{2*(-16)}=\frac{20}{-32}= -\frac{5}{8}

Solving the Expresion (B):

\frac{-54-\sqrt{5476 } }{2*(-16)}= \frac{-54-74}{2*(-16)}=\frac{-128}{2*(-16)}=\frac{-128}{-32}= 4

These results indicate the time it will take for the object to hit the ground can be -5/8 and 4. Since the time cannot be negative, then <u><em>the time it will take for the object to hit the ground will be 4.</em></u>

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