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02.11.2020 11:52
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Two parallel, metal plates with separation distance d = 1.00 cm carry charges of

Two parallel, metal plates with separation distance d = 1.00 cm carry charges of equal magnitude but opposite sign. The plates are oriented horizontally. Assume the electric field between the plates is uniform, and it has a magnitude of 1,920 N/C. A charged particle with mass 2.00 â 10^â16 kg and charge 1.02 â 10%â6 C is projected from the center of the bottom negative plate with an initial speed of 1.07 â 10%5 m/s at an angle of 37.0° above the horizontal. Assume the plates are square with side length = 20.0 cm. Required:
a. Describe the trajectory of the particle.
b. Which plate does it strike?
c. Where does it strike, relative to its starting point?
d. What is the minimum initial speed (in m/s) the particle must have in order to change the answer to part (b)?
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vickReis06
vickReis06
4,6(93 marks)

We have that  the trajectory of the particle,The plate the particle strikes,Where it strikes, relative to its starting point and the minimum initial speed (in m/s) the particle must have in order to change the answer to part (b) are as follows

Parabola Negative platea=9.792x10^{12}m/s^2V_m=570452.106m/s

From the question we have that

Distance d=1.00cm=0.01m

Magnitude  M=1,920 N/C

Mass m=2.00x10^{-16} kg

Charge q=1.02x10^{-6} C

initial speed V_0= 1.07x10^5 m/s

Angle \theta= 37

Length l= 20.0 cm.

a)

The Particle follow a Parabola trajectory through its travel as a Parabola Creates a Curved Plane and has its initial half side same as the final side.with the particle is a \theta= 37 angle create on its elevation and depression

Parabola

b)

It strikes the Negative plate at a specific distance to be discovered in the C part of the Question

c)

We are tasked to determine where the particle strikes the plate.

Generally the equation for strike distance S  is mathematically given as

s=\frac{v_0^2 sin(2\theta)}{a}

Where

a=\frac{qM}{m}

a=\frac{1.02x10^{-6}x1920}{2.00x10^{-16} }

a=9.792x10^{12}m/s^2

Therefore

s=\frac{v_0^2 sin(2\theta)}{a}

\frac{(1.07x10^5)^2 xsin(2*37)}{9.792x10^{12}}

s=0.0011239m

d)

Generally the equation for  the minimum Velocity  is mathematically given as

V_m=\sqrt{\frac{2ah}{sin(\theta)}}

V_m=\sqrt{\frac{2x9.792*10^{12}*0.01}{sin(37)}}

V_m=570452.106m/s

In conclusion

The trajectory of the particle,The plate the particle strikes,Where it strikes, relative to its starting point and the minimum initial speed (in m/s) the particle must have in order to change the answer to part (b) are as follows

Parabola Negative platea=9.792x10^{12}m/s^2V_m=570452.106m/s

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svarner2001
svarner2001
4,8(48 marks)

g

Explanation:

tina7659
tina7659
5,0(15 marks)
They are magnesium, aluminum, iron, zinc, and tin. you welcome

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