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Sauron [17]
3 years ago
6

Which type of wave is not a light wave

Physics
1 answer:
AlekseyPX3 years ago
4 0

<em><u>A</u></em><em><u>. </u></em><em><u>R</u></em><em><u>E</u></em><em><u>D</u></em><em><u> </u></em><em><u>W</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em><em><u>S</u></em><em><u> </u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>N</u></em><em><u>O</u></em><em><u>T</u></em><em><u> </u></em><em><u>A</u></em><em><u> </u></em><em><u>L</u></em><em><u>I</u></em><em><u>G</u></em><em><u>H</u></em><em><u>T</u></em><em><u> </u></em><em><u>W</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em><em><u> </u></em><em><u>B</u></em><em><u>E</u></em><em><u>C</u></em><em><u>A</u></em><em><u>U</u></em><em><u>S</u></em><em><u>E</u></em><em><u> </u></em><em><u>THE</u></em>RE<em><u> </u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>N</u></em><em><u>O</u></em><em><u>T</u></em><em><u> </u></em><em><u>RED</u></em><em><u> </u></em><em><u>W</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em><em><u>.</u></em>

<em><u>A</u></em><em><u>L</u></em><em><u>S</u></em><em><u>O</u></em><em><u> </u></em><em><u>I</u></em><em><u>F</u></em><em><u> </u></em><em><u>Y</u></em><em><u>O</u></em><em><u>U</u></em><em><u> </u></em><em><u>D</u></em><em><u>O</u></em><em><u>N</u></em><em><u>T</u></em><em><u> </u></em><em><u>B</u></em><em><u>E</u></em><em><u>L</u></em><em><u>I</u></em><em><u>E</u></em><em><u>V</u></em><em><u>E</u></em><em><u> </u></em><em><u>S</u></em><em><u>E</u></em><em><u>A</u></em><em><u>R</u></em><em><u>C</u></em><em><u>H</u></em><em><u> </u></em><em><u>I</u></em><em><u>T</u></em><em><u> </u></em><em><u>F</u></em><em><u>R</u></em><em><u>O</u></em><em><u>M</u></em><em><u> </u></em><em><u>G</u></em><em><u>O</u></em><em><u>O</u></em><em><u>G</u></em><em><u>L</u></em><em><u>E</u></em>

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v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

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