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ElenaW [278]
2 years ago
14

uniform electric field of magnitude 365 N/C pointing in the positive x-direction acts on an electron, which is initially at rest

. The electron has moved 3.00 cm. (a) What is the work done by the field on the electron? 1.753e-18 J (b) What is the change in potential energy associated with the electron? J
Physics
1 answer:
vfiekz [6]2 years ago
3 0

Answer:

a)   W = - 1.752 10⁻¹⁸ J,  b)    U = + 1.752 10⁻¹⁸ J

Explanation:

a) work is defined by

         W = F . x

the bold letters indicate vectors, in this case the force is electric

         F = q E

we substitute

         F = q E x

the charge of the electron is

         q = - e

         F = - e E x

let's calculate

         W = - 1.6 10⁻¹⁹  365  3 10⁻²

         W = - 1.752 10⁻¹⁸ J

b) the change in potential energy is

          U = q ΔV

the potential difference is

          ΔV = - E. Δs

 

we substitute

         U = - q E Δs

the charge of the electron is

           q = - e

          U = e E Δs

we calculate

           U = 1.6 10⁻¹⁹ 365  3 10⁻²

           U = + 1.752 10⁻¹⁸ J

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Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th
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Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

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PLEASE HELP
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8 0
2 years ago
The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
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