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1 year ago

Use the passage to answer the question.

Physics

1 answer:

Studentka2010 [4]1 year ago

3 0

This is not something that waves do because they need a medium to travel through, while particles do not.

<h3>How light travels in space?</h3>

A light travels without any medium while on the other hand, a medium is required for sound waves to move from oe place to another. Sound is a mechanical wave that cannot travel through a vacuum.

So we can conclude that electromagnetic waves like light do not require medium for its propagation.

Learn more about light here: brainly.com/question/19697218

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Which two options are examples of waves reflecting?

Umnica [9.8K]

Answer:

transverse

Explanation:

none

5 0

2 years ago

I need someone to give me 2 answers,

yanalaym [24]

Uh.. what's the question..?

8 0

3 years ago

A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t

Liono4ka [1.6K]

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

$d = d_1 + d_2$

given that

$d_1 = 200 cos20 i + 200 sin20 j$

$d_1 = 187.94 i + 68.4 j$

$d_2 = -230 sin30 i + 230 cos30 j$

$d_2 = -115 i + 199.2 j$

now the total distance is

$d = (187.94 - 115)i + (199.2 + 68.4)j$

$d = 72.94 i + 267.6 j$

now the magnitude of the distance is given as

$d = \sqrt{72.94^2 + 267.6^2}$

$d = 277.4$

also the direction is given as

$\theta = tan^{-1}\frac{267.6}{72.94}$

$\theta = 74.7 degree$

<em>so it is 277.4 km at 74.7 degree North of East</em>

8 0

2 years ago

12000 inches to yards

Nutka1998 [239]

ANSWER

$\begin{equation*} 333.33\text{ yds} \end{equation*}$

EXPLANATION

We want to convert 12000 inches to yards.

To do this, divide the value in inches by 36:

$\begin{gathered} 1\text{ in }=\frac{1}{36}\text{ yd} \\ \\ 12000\text{ in }=\frac{12000}{36}\text{ yds }=333.33\text{ yds} \end{gathered}$

That is the answer.

4 0

9 months ago

A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).

yawa3891 [41]

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

$f = n\frac{v}{2L}$