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Sunny_sXe [5.5K]
2 years ago
6

Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr

ic field at a point between the two surfaces?
Physics
1 answer:
svlad2 [7]2 years ago
6 0

Answer:

Electric field, E = 45.19 N/C

Explanation:

It is given that,

Surface charge density of first surface, \sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2

Surface charge density of second surface, \sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2

The electric field at a point between the two surfaces is given by :

E=\dfrac{\sigma}{2\epsilon_o}

E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}

E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}

E = 45.19 N/C

So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.

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Two girls,(masses m1 and m2) are on roller skates and stand at rest, close to each other and face-to-face. Girl 1 pushes squarel
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If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
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This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

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a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

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b) the tangential acceleration of the ball is;

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Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

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