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marishachu [46]
3 years ago
10

A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

tor is oriented 55.2° clockwise from the vertical axis, as shown. if the magnitude of the electric field is 4.82 n/c, how much work is done by the electric field as the particle is made to move a distance of d = 0.556 m straight up?
Physics
1 answer:
miss Akunina [59]3 years ago
3 0
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
W=F d cos \theta
where the force is F=qE, d=0.556 and \theta=55.2^{\circ}. Using the value of q and E given by the problem, we find
W=qEdcos\theta = 6.39\cdot10^{-5}J
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3 years ago
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A spy camera is said to be able to read the numbers on a car's license plate. If the numbers on the plate are 4.30 cm apart, and
Maslowich

Answer:

D = 2.38 m

Explanation:

This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit

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           θ = λ / a

Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant

           θ = 1.22 λ / D

Where D is the circular tightness

       

Let's apply this equation to our case

         D = 1.22 λ /  θ

To calculate the angles let's use trigonometry

         tan  θ = y / x

          θ = tan⁻¹  y / x

          θ = tan⁻¹ (4.30 10⁻² / 140 10³)

          θ = tan⁻¹ (3.07 10⁻⁷)

          θ = 3.07 10⁻⁷ rad

Let's calculate

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4 0
2 years ago
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Answer:

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Explanation:

We are given that

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We know that

v^2-u^2=2as

Using the formula

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v^2=40+1.96

v^2=41.96

v=\sqrt{41.96}

v=6.5 m/s

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

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