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timofeeve [1]
2 years ago
5

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

. Assume each tick mark represents 1 cm. Calculate the total displacement if a toy car starts at 0, moves 5 cm to the left, then 8 cm to the right, and then 3 cm to the left. The car moves cm, so there is no displacement.
Physics
2 answers:
Natali5045456 [20]2 years ago
6 0

The total displacement of the toy car at the given positions is 0.

The given parameters;

  • <em>First displacement of the car, = 5 cm left</em>
  • <em>Second displacement of the car, = 8 cm right</em>
  • <em>Third displacement of the car, = 3 cm to the left</em>

The total displacement of the car is calculated as follows;

  • <em>Let the </em><em>left </em><em>direction be "</em><em>negative </em><em>direction"</em>
  • <em>Let the </em><em>right </em><em>direction be "</em><em>positive </em><em>direction"</em>

\Delta x = - \ 5\ cm  \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0

Thus, the total displacement of the toy car at the given positions is 0.

Learn more about displacement here: brainly.com/question/18158577

jeka57 [31]2 years ago
5 0

Answer:0

Explanation:

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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
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Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

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If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu
ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , n=1.34 \times 10^{20} .

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