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3 years ago

What is the magnitude of the electric field at a point 0.0055 m from a 0.0025

Physics

1 answer:

allochka39001 [22]3 years ago

7 0

Answer:

Explanation:

The equation for the electric field is

$E=\frac{kQ}{r^2}$ so filling in:

$E=\frac{9.00*10^9(.0025)}{(.0055)^2}$ which in the end gives you

E = 7.4 × 10¹¹, choice A

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Answer:

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Now

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If y = 0

$0=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow utsin\theta=\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow t=\dfrac{2musin\theta}{qE}$

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3 years ago

An infrared beam of light as a wavelength of 7.7 mm What is this energy

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Answer:

E=2.58155844×10^-29

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we use this formula

E= hc/ lambda

E= [(6.626×10^-34)( 3×10^8)]/(7.7×10^-3)

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