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Elenna [48]
3 years ago
10

A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.

The observed frequency of the hum will be 152 Hz as the bumblebee flies away from you.
Physics
2 answers:
Veronika [31]3 years ago
8 0

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.

lyudmila [28]3 years ago
3 0

Answer:

<em>Less Than</em>

Explanation:

A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s. The observed frequency of the hum will be Less Than 152 Hz as the bumblebee flies away from you.

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Answer:

a) t=6.37s

b) t=3.3333s

Explanation:

The knowable variables are the initial hight and initial velocity

s_{o}=80ft

v_{os}=64ft/s

The equation that describes the motion of the ball is:

s=80+64t-16t^{2}

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

0=80+64t-12t^{2}

a) Solving for t, we are going to have two answers

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a=-16

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c=80

t=-1.045 s or t=6.378s

<em><u>Since time can not be negative the answer is t=6.378s </u></em>

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

v_{y}=v_{o}+gt

at maximum point the velocity is 0

0=64-32.2t

Solving for t

t=1.9875 s

Now, we must know how much distance does it take to reach maximum point

s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft

So, the ball pass the top of the building on its way down at 160 ft

160=80+64t-16t^{2}

Solving for t

t=2s or t=3.333s

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

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Answer:

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