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Lostsunrise [7]
2 years ago
14

A sound wave travels through water at 1,500 m/s. How far would it travel in 5 s?

Physics
2 answers:
hichkok12 [17]2 years ago
8 0
7500m because 1,500 in 1 sec so 1500x5 7500
Andrews [41]2 years ago
7 0

Answer:

7,500 miles

Explanation:

lets see it gos 1,500 per second so if it gos for five seconds it will go 7500, (just 1500 times 5, its that simple).

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Microphones and loudspeakers are used in an auditorium because the sound waves at the stage compared to the sound waves at the b
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Acoustics (sounds)
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6 0
3 years ago
Lauren throws her first pitch of the season for her school's softball team. If the ball travels 80 feet and takes 1.58 seconds t
kow [346]

Answer:

<em><u>5</u></em><em><u>0</u></em><em><u>.</u></em><em><u>6</u></em><em><u>3</u></em><em><u> </u></em><em><u>f</u></em><em><u>t</u></em><em><u>/</u></em><em><u>s</u></em><em><u> </u></em><em><u>(</u></em><em><u>2</u></em><em><u>d</u></em><em><u>p</u></em><em><u>)</u></em>

Explanation:

Speed = Distance/Time

80/1.58 = 50.63291139

= <u>50.63</u><u> </u><u>f</u><u>t</u><u>/</u><u>s</u> (2dp)

6 0
2 years ago
A boat moves up and down as water waves pass under the boat. If the amplitude of the wave gets bigger, then
xeze [42]
The anwser is A or D
7 0
2 years ago
Read 2 more answers
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
A student drops a ball from the top of a tall building; it takes 2.9 s for the ball to reach the ground.
AlexFokin [52]

<h2><u>We have</u>,</h2>

  • Initial velocity (u) = 0 m/s
  • Time taken (t) = 2.9s
  • Acceleration due to gravity (g) = + 10 m/s² [Down]

<h2><u>To calculate</u>,</h2>

  • Final velocity (v)
  • Height (h)

<h2><u>Solution</u><u>,</u></h2>

→ v = u + gt

→ v = 0 + 10(2.9)

→ v = 29 m/s \qquad … ( Ans )

And,

→ h = ut + ½gt²

→ h = 0(2.9) + ½ × 10 × (2.9)²

→ h = 5 × 8.41

→ h = 42.05 m \qquad … ( Ans )

4 0
2 years ago
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