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Anon25 [30]
3 years ago
13

What is the speed of sound at sea level?

Physics
1 answer:
svetlana [45]3 years ago
7 0
The speed of sound at sea level is 340.29 m/s (meters per seconds).
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LOTS OF BRAINLIST WILL BE GIVING TO THOSE WHO HELP
Alex_Xolod [135]
You know you can skip those and just submit them, they don’t even check them
4 0
2 years ago
Read 2 more answers
A stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s. What is the frequency perceived by
inn [45]

For a stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s, the frequency perceived  is mathematically given as'

F=81.721Hz

<h3>What is the frequency perceived by a firefighter racing toward the station at 11km/h?</h3>

Generally, the equation for the doppler effect  is mathematically given as

F'=\frac{vs+v}{vs}*f

Therefore

F=81(343+3.05556)/343

F=81.721Hz

In conclusion, the frequency is

F=81.721Hz

Read more about frequency

brainly.com/question/24623209  

4 0
1 year ago
Your friend just challenged you to a race through an obstacle course. You know in order to beat him, you must run 30 meters with
seropon [69]

Answer:

Velocity = 0.5 m/s South (A)

Explanation:

You need to determine the average rate of velocity.  

The equation you will use is velocity = displacement/time

The displacement is 30m South.

The time is 60 seconds.

Plug into the equation  Velocity = 30m South/60 s

Velocity = 0.5 m/s South

3 0
3 years ago
A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow. What is the skier’s change in velocity
nikdorinn [45]

Answer:

\Delta v=5.77m/s

Explanation:

Newton's 2nd Law relates the net force <em>F</em> on an object of mass <em>m </em>with the acceleration <em>a</em> it experiments by <em>F=ma.</em> In our case the net force is the friction force, since it's the only one the skier is experimenting horizontally and the vertical ones cancel out since he's not moving in that direction. Our acceleration then will be:

a=\frac{F}{m}

Also, acceleration is defined by the change of velocity \Delta v in a given time t, so we have:

a=\frac{\Delta v}{t}

Since we want the change in velocity, <em>mixing both equations</em> we conclude that:

\Delta v=at=\frac{Ft}{m}

Which for our values means:

\Delta v=\frac{Ft}{m}=\frac{(25N)(15s)}{(65Kg)}=5.77m/s

4 0
3 years ago
A running mountain lion can make a leap 10.0 mlong, reaching a maximum height of 3.0 m. What is the speed of the mountain lion j
nika2105 [10]

To solve this problem we will use the kinematic equations of descriptive motion of a projectile for which both the height reached and the distance traveled are defined. From this type of movement the lion reaches a height (H) of 3m and travels a horizontal distance (R) of 10 m. Mathematically the equations that describe this movement are given as,

H = \frac{v_0^2sin^2\theta}{2g}

R = \frac{v_0^2 sin 2\theta}{g}

Dividing the two equation we have that

\frac{H}{R}=\frac{\frac{v_0^2sin^2\theta}{2g}}{\frac{v_0^2 sin 2\theta}{g}}

\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{sin2\theta}

\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{2sin\theta cos\theta}

\frac{H}{R}= \frac{1}{4} \frac{sin\theta}{cos\theta}

\frac{H}{R}= \frac{1}{4} tan\theta

Substituting values of H and R, we get

\frac{3}{10} = \frac{1}{4} tan\theta

\theta = tan^{-1} \frac{12}{10}

\theta = 50.2\°

Substituting the value of \theta in equation we get,

H = \frac{v_0^2sin^2\theta}{2g}

v_0^2 = \frac{H 2g}{sin^2\theta}

v_0^2 = \frac{3*2*9.8}{sin^2(50.2)}

v_0^2 = 99.62

v_0 = \sqrt{99.62}

v_0 = 9.98m/s

Therefore the speed of the mountain lion just as it leaves the ground is 9.98m/s at an angle of 50.2°

5 0
2 years ago
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