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ivanzaharov [21]
3 years ago
11

Electromagnetic waves are often distinguished from mechanical waves. The distinction is based on the fact that electromagnetic w

aves ______.
can travel through materials and mechanical waves cannot

come in a range of frequencies and mechanical waves exist with only certain frequencies

can travel through a region void of matter and mechanical waves cannot

electromagnetic waves cannot transport energy and mechanical waves can transport energy
Physics
1 answer:
noname [10]3 years ago
7 0

Answer:

the correct one is the third :

can travel through a region void of matter and mechanical waves cannot

Explanation:

Mechanical waves travel supported by a material medium, for example sound waves travel in air, the wave on a string travels on the string.

Electromagnetic waves are produced by the variation of electric and magnetic fields, after producing this variation by the lenz law the variation of one field induces the other, so these waves do not need a material medium to travel.

Of the final statements the correct one is the third :

can travel through a region void of matter and mechanical waves cannot

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Two planets, Dean and Sam, orbit the Sun. They each have with circular orbits, but orbit at different distances from the Sun. De
lyudmila [28]

Answer:

The correct answer is Dean has a period greater than San

Explanation:

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3 0
3 years ago
Topic Gravitational force amd firld strength.. help me please
I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

x \approx 6.74\,\mathrm m

5 0
3 years ago
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