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3 years ago

7

Describe what happens to the magnitude of the net electrostatic force on the electron as the electron

1 answer:

solniwko [45]3 years ago

7 0

Answer:

hm my mom said the net force changes

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A. Describe one meal that you typically order when eating out.

marta [7]

A. “I usually get 10 piece nugget meal with medium fries at McDonald’s with a McFlurry

B. I could replace the 10 piece nuggets with a grilled chicken wrap and replace the medium fries with a small fries then replace the mcflurry with either water or unsweetened tea.

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3 years ago

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1.) What causes ocean tides

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The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of

Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is

$\theta _d =0.19 ^o$

Explanation:

From the question we are told that

The refractive index of the transparent acrylic plastic for blue light is $n_F = 1.497$

The wavelength of the blue light is $F = 486.1 nm = 486.1 *10^{-9} \ m$

The refractive index of the transparent acrylic plastic for red light is $n_C = 1.488$

The wavelength of the red light is $C = 656.3 nm = 656.3 *10^{-9} \ m$

The incidence angle is $i = 45^o$

Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as

$r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]$

$r_F = 28.18^o$

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

$r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]$

$r_F = 28.37^o$

The angle between the blue beam and the red beam in the acrylic block

$\theta _d = r_C - r_F$

substituting values

$\theta _d = 28.37 - 28.18$

$\theta _d =0.19 ^o$

4 0

3 years ago

What is the acceleration during the pushing-off phase, in m/27 Express your answer in meters per second squared. A bush baby, an

Sindrei [870]

Answer:

$a = 12.78 g's$

Explanation:

Height reached by the object after push off is given as

$H = 2.3 m$

$v_f^2 - v_i^2 = 2 a s$

now we have

$0 - v^2 = 2(-9.81)(2.3)$

$v = 6.72 m/s$

now we know that this push last for total distance of 0.18 m

so during the push we will have

$v_f^2 - v_i^2 = 2 a d$

$6.72^2 - 0 = 2a(0.18)$

$a = 125.35 m/s^2$

now in terms of g = 9.81 m/s/s we have

$a = \frac{125.35}{9.81} gs$

$a = 12.78 g's$

6 0

3 years ago

FASTTT FOR BRAINLIST

telo118 [61]

The answer should be A?

7 0

2 years ago