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Tema [17]
3 years ago
6

Why might compound gear trains be better than two gears alone?

Physics
1 answer:
mina [271]3 years ago
4 0

Answer:

the answer is there is more separation for the vitality to exchange; with just two apparatuses, all the vitality is applied on a shorter separation.  

Explanation:

Work done and vitality moved are estimated in joules (J). The work done on a protest can be figured if the power and separation moved are known. An adjustment in energy happens when a power is connected to a question that is moving or can move.Pulleys are ground-breaking straightforward machines. They can alter the course of a power, which can make it significantly simpler for us to move something. On the off chance that we need to lift a protest that gauges 10 kilograms one meter high, we can lift it straight up or we can utilize a pulley, so we can pull down toward one side to lift the question up.

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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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