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jok3333 [9.3K]
2 years ago
11

A point charge is placed 3m from a 4uc charge what is the strength of the electric field on the point charge at this distance ro

und to the nearest thousand
Physics
1 answer:
Vlada [557]2 years ago
7 0

The strength of the electric field on the point charge at this distance will be 4000 V/m.

<h3>What is the strength of the electric field?</h3>

The strength of the electric field is the ratio of electric force per unit charge.

The given data in the problem is;

Qis the unit charge = 4.0 × 10⁻⁶ C

E is the strength of the electric field

R is the distance from point charge = 3 m

The strength of the electric field is;

\rm E = \frac{KQ}{R^2} \\\\ \rm E = \frac{9 \times 10^9 \times 4 \times 10^{-6} \ C}{3^2} \\\\ E= 4000 V/m

Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.

To learn more about the strength of the electric field refer to the link;

brainly.com/question/15170044

#SPJ1

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Answer:

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Explanation:

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3 0
3 years ago
When you blink your eye, the upper lid goes from rest with your eye open to completely covering your eye in a time of 0.024 s.
dusya [7]

a) Distance moved by the top lid during a blink: 1 cm (estimate)

b) The acceleration is 34.7 m/s^2

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Explanation:

a)

For the purpose of this problem, we can estimate the size of the eye from the top lid to the bottom lid to be 1 cm, so this is the distance moved by the top lid during a blink.

b)

Assuming the motion of the eyelid to be at constant acceleration, we can find the acceleration by using the following suvat equation:

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t is the time

a is the acceleration

For the eyelid, we have:

u = 0 (it starts from rest)

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c)

The final speed of the upper eyelid can be found by using another suvat equation:

v=u+at

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

For the eyelid in this problem, we have

u = 0

a=34.7 m/s^2

t = 0.024 s

Substituting, we find the final speed:

v=0+(34.7)(0.024)=0.83 m/s

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Answer:

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Explanation:

given,

speed of the ball thrown by the pitcher = 100 mph

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R = \dfrac{u^2sin2\theta}{g}

R = \dfrac{44.704^2sin2\times 45}{9.81}

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Suppose you wanted to be able to see astronauts on the moon. what is the smallest diameter of the objective lens required to res
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inserting the values

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Explanation:

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