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3 years ago

A positron undergoes a displacement ∆r= 2i -3j+6k, ending with the position vector r= 3j-4k, in meters. what was the positron's

initial position

Physics

1 answer:

Margaret [11]3 years ago

3 0

$\Delta{r}=2i-3j+6k \\r_f = 3j-4k \\\Delta{r}=r_f-r_0\Longrightarrow2i-3j+6k=3j-4k-r_0 \\r_0=2i-3j+6k+3j-4k=\boxed{2i+6j+10k}$

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navik [9.2K]

The solution would be like this for this specific problem:

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ω=(kqQ/R3)1/2

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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3 years ago

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A small lead ball, attached to a 1.75-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

soldier1979 [14.2K]

Answer:

h = 57.6 m

Explanation:

First, we find the linear speed of the ball while in circular motion:

v = rω

where,

v = linear speed of ball = ?

r = radius of circle = length of rope = 1.75 m

ω = angular speed = (3 rev/s)(2π rad/1 rev) = 18.84 rad/s

Therefore,

v = (1.75 m)(18.84 rad/s)

v = 32.98 m/s

Now, we apply the 3rd equation of motion on the ball, when it breaks:

2gh = Vf² - Vi²

where,

g = - 9.8 m/s² (negative sign due to upward motion)

h = height covered = ?

Vf = Final Velocity = 0 m/s (since, the ball finally stops at highest point for a moment)

Vi = Initial Velocity = 32.98 m/s

Therefore,

2(- 9.8 m/s²)h = (0 m/s)² - (32.98 m/s)²

h = ( - 1088.12 m²/s²)/( - 19.6 m/s²)

h = 55.5 m

since, the ball was initially at a height of 2.1 m from ground. So, the total height from ground, will now become:

h = 55.5 m + 2.1 m

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