Tell whether the picture below shows a
2 answers:
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For counting x you use simple equation for the distance covered by the object when it moves with constant velocity:
that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.
For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.
For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
Answer:
Multiple answers:
1. Power output P=17.59W
2.Intensity 160m I=17.6W/
3. dB = 77.3
4. f=178.5 Hz
Explanation:
First one comes from the expression
where<em> I </em>is the intensity, <em>P </em>is the power and <em>r </em>is the radio of the spherical wave, or in this case, the distance <em>x</em>. I solved for the Power by multiplying Intensity with the area (4
Second one is done with:
Solving for Intensity 2, the result mentioned.
The third is simply computed with
And finally the last one is done with doppler effect, taking into account the speed of the air as in 10ºC 337m/s.
Where <em>Finitial</em> is the frequency emitted and <em>s</em> is the speed of the sound. The wind blowing in positive is, in principle, going away of the observer.