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Llana [10]
3 years ago
10

Which of the following values represents an index of refraction of an actual material?

Physics
1 answer:
serg [7]3 years ago
3 0

Sadly, we're forced to answer the question without the benefit of the
list of choices, which, for some reason, you decided not to let us see.

Index of refraction of a substance =

               (speed of light in vacuum) / (speed of light in the substance).

Any number greater than  ' 1 ' can be an index of refraction.  A number
less than ' 1 ' can't be . . . that would be saying that the speed of light
in this substance is greater than the speed of light in vacuum.

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4 0
3 years ago
You drop a ball from a height of 32 meters how much time passes before the ball hits the ground
tankabanditka [31]

Answer:

31 seconds

Explanation:

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5 0
3 years ago
F = 2.0*10^20 N,
natka813 [3]

F=\dfrac{Gm_1m_2}{r^2}

With the given values of F,G,m_1,r, we have

2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}

Try dealing with the powers of 10 first: On the right, we have

\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}

Meanwhile, the other values on the right reduce to

\dfrac{6.67\times5.98}{3.8^2}\approx2.76

Then taking units into account, we end up with the equation

2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2

Now we solve for m_2:

m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}

m_2=7.25\times10^{22}\,\mathrm{kg}

or, if taking significant digits into account,

m_2=7.3\times10^{22}\,\mathrm{kg}

5 0
3 years ago
A car accelerates from rest at a constant rate 5m/s^2. Which of the following statements is true
dusya [7]

Answer:

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8 0
3 years ago
The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in
N76 [4]

Answer:

28.2 m/s

Explanation:

The range of a projectile launched from the ground is given by:

d=\frac{v^2}{g}sin 2\theta

where

v is the initial speed

g = 9.8 m/s^2 is the acceleration of gravity

\theta is the angle at which the projectile is thrown

In this problem we have

d = 81.1 m is the range

\theta=45^{\circ} is the angle

Solving for v, we find the speed of the projectile:

v=\sqrt{\frac{dg}{sin 2 \theta}}=\sqrt{\frac{(81.1 m)(9.8 m/s^2)}{sin (2\cdot 45^{\circ})}}=28.2 m/s

7 0
3 years ago
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