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stellarik [79]
3 years ago
14

If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh

at will its speed be, in meters per second, when it reaches the positive plate in this case
Physics
1 answer:
In-s [12.5K]3 years ago
7 0

Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

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A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta
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Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

x = 26.16 m

Part d)

y = 7.49 m

Part e)

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Part f)

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Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

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v_y = 19.9 sin39.9

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Part a)

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x = v_x t

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x = 15.76 m

Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

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Horizontal displacement in 1.71 s

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Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

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Part e)

Horizontal displacement in 5.44 s

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x = (15.3)(5.44)

x = 83.23 m

Part f)

Vertical direction we have

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y = (12.76)(5.44) - 4.9(5.44)^2

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