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laiz [17]
3 years ago
11

An Earth satellite is in a circular orbit at an altitude of 500 km.

Physics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

Speed Unchanged.

Explanation:

As work is defined as a product of force over a distance. If the distance in altitude is constant = 500km, there's 0 change in distance and force, no work would be done by the gravitational force.

Since potential energy of the satellite is unchanged, unless there's additional internal energy source, the kinetic energy would remain unchanged, so would its speed.

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Read 2 more answers
An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g
HACTEHA [7]

Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

        Po = pf

        Ko = Kf

 Before crash

       Po = m1 Vo1 + 0

       Ko = ½ m1 Vo1²

 

After the crash

       Pf = m1 Vif + Vvf

       Kf = ½ m1 V1f² + ½ m2 V2f²

 

      m1 V1o = m1 V1f + m2 V2f           (1)

      m1 V1o² = m1 V1f² + m2 V2f²      (2)

We see that we have two equations with two unknowns, so the system is solvable,  we substitute in 1 and 2

   

     m1 (V1o -V1f) = m2 V2f      (3)

      m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship      (a + b) (a-b) = a² -b²

     m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide  with 3 and simplify

      (V1o + V1f) = V2f      (4)

Substitute in 3, group and clear

         m1 (V1o - V1f) = m2 (V1o + V1f)

         m1 V1o - m2 V1o = m2 V1f + m1 V1f

         V1f (m1 -m2) = V1o (m1 + m2)

         V1f = V1o  (m1-m2 / m1+m2)

We substitute in (4) and group

         V2f = V1o + (m1-m2 / m1 + m2) V1o

         V2f = V1o [1+ + (m1-m2 / m1 + m2)]

         V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

         V1f = 0.81 (239-513 / 239 + 513)

         V1f = 0.81 (-274/752)

         V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

         V2f = 0.81 [2 239 / (239 + 513)]

        V2f = 0.81 [0.636]

        V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

         Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

         Eo = K = ½ m2 vf2²

         Ef = U = m2 g Y

   

         ½ m2 v2f² = m2 g Y

         Y = V2f² / 2g

         Y = 0.515²/2 9.8

         Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

         tan θ = Y / x

         Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

        Y = 1.2 tan 0.7

        Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

7 0
3 years ago
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