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Anton [14]
3 years ago
6

What do you call the material or "stuff" that a wave travels through

Physics
2 answers:
frutty [35]3 years ago
8 0
C I think if not it’s B
statuscvo [17]3 years ago
3 0

Answer:

<em>a</em><em>.</em><em> </em><em>m</em><em>e</em><em>d</em><em>i</em><em>u</em><em>m</em><em>.</em>

Explanation:

<em>t</em><em>h</em><em>e</em><em> </em><em>m</em><em>a</em><em>t</em><em>t</em><em>e</em><em>r</em><em> </em><em>t</em><em>h</em><em>r</em><em>o</em><em>u</em><em>g</em><em>h</em><em> </em><em>w</em><em>h</em><em>i</em><em>c</em><em>h</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>w</em><em>a</em><em>v</em><em>e</em><em>s</em><em> </em><em>t</em><em>r</em><em>a</em><em>v</em><em>e</em><em>l</em><em>s</em><em> </em><em>i</em><em>s</em><em> </em><em>c</em><em>a</em><em>l</em><em>l</em><em>e</em><em>d</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>m</em><em>e</em><em>d</em><em>i</em><em>u</em><em>m</em><em> </em><em>(</em><em> </em><em>p</em><em>l</em><em>u</em><em>r</em><em>a</em><em>l</em><em>,</em><em> </em><em>m</em><em>e</em><em>d</em><em>i</em><em>a</em><em> </em><em>)</em><em>.</em><em> </em><em>b</em><em>u</em><em>t</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>m</em><em>e</em><em>d</em><em>i</em><em>u</em><em>m</em><em> </em><em>o</em><em>f</em><em> </em><em>a</em><em> </em><em>m</em><em>e</em><em>c</em><em>h</em><em>a</em><em>n</em><em>i</em><em>c</em><em>a</em><em>l</em><em> </em><em>w</em><em>a</em><em>v</em><em>e</em><em> </em><em>c</em><em>a</em><em>n</em><em> </em><em>b</em><em>e</em><em> </em><em>a</em><em>n</em><em>y</em><em> </em><em>s</em><em>t</em><em>a</em><em>t</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>m</em><em>a</em><em>t</em><em>t</em><em>e</em><em>r</em><em>,</em><em> </em><em>e</em><em>v</em><em>e</em><em>n</em><em> </em><em>a</em><em> </em><em>s</em><em>o</em><em>l</em><em>i</em><em>d</em><em>.</em>

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A weather balloon is filled to a volume of 6000 L while it is on the ground, at a pressure of 1 atm and a temperature of 273 K.
avanturin [10]

Answer : The final volume of the balloon at this temperature and pressure is, 17582.4 L

Solution :

Using combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 1 atm

P_2 = final pressure of gas = 0.3 atm

V_1 = initial volume of gas = 6000 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 273 K

T_2 = final temperature of gas = 240 K

Now put all the given values in the above equation, we get the final pressure of gas.

\frac{1atm\times 6000L}{273K}=\frac{0.3atm\times V_2}{240K}

V_2=17582.4L

Therefore, the final volume of the balloon at this temperature and pressure is, 17582.4 L

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4 years ago
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Which is an important step in how an electric motor uses magnetic force to produce motion?
Fiesta28 [93]

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Explanation:

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A person walks 9 meters to the right and 8 meters to the left what's the distance and displacement​
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Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
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Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

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