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Nataly [62]
3 years ago
5

What are examples of a medium

Physics
2 answers:
Serggg [28]3 years ago
4 0

Examples of medium are solid ,liquid ,and gas basically matter. :D

Mila [183]3 years ago
4 0

Answer:

solid

liquids

gas

Explanation:

water and air (e2020)

Forever,

Cammie

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The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y
xeze [42]

Answer:

The amount of caffeine left after one half life of 5 hours is 15 oz.

Explanation:

Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.

The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.

So that:

After one half life of 5 hours, the value of caffeine that would be left is;

                                    \frac{30}{2} = 15 oz

The amount of caffeine left after one half life of 5 hours is 15 oz.

8 0
4 years ago
What charge do electrons have?<br>Negative<br>Positive<br>Neutral​
hoa [83]

Answer electrons are negative. Protons are positive, and neutrons are neutral

8 0
3 years ago
El espectro visible en el aire está comprendido entre la longitud de onda 450 nm del color azul, Determina la velocidad de propa
TiliK225 [7]

Answer:

v = 2,99913 10⁸ m / s

Explanation:

The velocity of propagation of a wave is

         v = λ f

in the case of an electromagnetic wave in a vacuum the speed that speed of light

         v = c

When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium

         λ = λ₀ / n

where n is the index of refraction of the material medium.

Therefore the expression is

           v = \frac{\lambda_o}{n} f

Let's look for the frequency of blue light in a vacuum

           f =\frac{c }{\lambda_o}  

           f = \frac{3 \ 10^8}{450 \ 10^{-9}}

           f = 6.667 10¹⁴ Hz

the refractive index of air is tabulated

           n = 1,00029

let's calculate

           v = \frac{450 \ 10^{-9} }{1.00029}  \ 6.667 \ 10^{14}450 10-9 / 1,00029 6,667 1014

            v = 2,99913 10⁸ m / s

we can see that the decrease in speed is very small

8 0
3 years ago
A surfactant with a Hydrophile-
zlopas [31]

Answer:

Explanation:

True

8 0
3 years ago
On the surface of planet y, which has a mass of 4.83x1024 kg, a 30.0 kg object weighs 50.0 n. what is the radius of the planet?
Aleksandr [31]
We can solve the problem in two steps:

1) From the weight W=50.0 N of the object, we can find the value of the gravitational acceleration g of the planet. In fact, the weight is equal to
W=mg
where m=30 kg is the mass of the object. From this, we find g:
g= \frac{W}{m}= \frac{50.0 N}{30 kg}=1.67 m/s^2

2) The gravitational acceleration of a planet with mass M and radius r is given by
g= \frac{GM}{r^2}
where G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational  constant. In our problem, the mass of the planet is 
M=4.83 \cdot 10^{24} kg, and we found g in step 1), g=1.67 m/s^2, so we have everything to solve and find the value of the radius r:
r= \sqrt{ \frac{GM}{g} }= \sqrt{ \frac{(6.67\cdot 10^{-11})(4.83 \cdot 10^{24})}{1.67} }=1.39\cdot 10^7 m
5 0
3 years ago
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