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2 years ago

11

The distance a toy car travels over time is shown in the graph.

1 answer:

Korvikt [17]2 years ago

6 0

Answer:

a toy car speed is about 2.5 to 3.5 mph

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A light beam is incident on a piece of fluorite (n = 1.434) at the Brewster's angle.

trasher [3.6K]

The Brewster angle is the angle can be found through this equation
θ=arctan(n2/n1).
This came from Snell's Law:
n1sinθ=n2sin(90-θ)
=n2cosθ
n2/n1=tanθ --- θ=arctan(n2/n1)=arctan(1.434)=55.11 degrees

Next, the angle of refraction can be found by using Snell's Law
n1sinθ=n2sinθ'
sin(55.11)=1.434sinθ'
sinθ'=sin(55.11)/1.434=0.572 --- θ'=arcsin(0.572)=34.89 degrees

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3 years ago

The only type of investigation that involves a control group is

OleMash [197]

<span>experimental? I hope that helps :)</span>

4 0

3 years ago

Read 2 more answers

A standard for comparison is called a frame of reference<br>True or False

kramer

Answer:

TRUE.

Explanation:

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2 years ago

Read 2 more answers

If the earth can travel 1800 km in 1 minute what is it’s speed in km/s

Lady bird [3.3K]

Answer:

30km/s

Explanation:

1800/60=30

4 0

3 years ago

Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron $m_{e}=9.11\times10^{-31}\ kg$

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}$

$E=0.582\ Mev$

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}$

$E=0.350\ Mev$

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0

3 years ago