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Alla [95]
3 years ago
12

A bugle can be thought of as an open-end pipe. If a bugle were straightened out, it would be 2.65 m long. a. If the speed of sou

nd is 343 m/s, find the lowest frequency (i.e., the largest wavelength) that can resonate in a bugle. (64.72 Hz) b. Find the next two higher resonant frequencies in the bugle. What harmonics are they
Physics
1 answer:
vitfil [10]3 years ago
5 0

Answer

given,

Length of pipe, L = 2.65 m

speed of sound, v = 343 m/s

Pipe is open end Pipe

a) Lowest frequency

  condition, for open end pipe

     \lambda_1 = 2 L

     \lambda_1 = 2\times 2.65  

     \lambda_1= 5.30 m

   For lowest frequency

    f_1 = n\dfrac{v}{\lambda_1}

     n = 1

    f_1 = \dfrac{343}{5.30}

           f₁ = 64.72 Hz

b) For second frequency in bugle

        n = 2

   f_2 =2\times \dfrac{343}{5.30}

          f₂ = 129.43 Hz

   for n = 3

   f_3 =3\times \dfrac{343}{5.30}

          f₃ = 194.16 Hz

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Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

arc length is given as

angle \times Radius = Arc

now we know that

Arc = 2.1 m

Angle = 21 degree

Angle = 21\times \frac{\pi}{180}

so now we have

21\times \frac{\pi}{180} = \frac{2.1}{R}

R = \frac{2.1 \times 180}{21 \times \pi}

R = 5.73 m

6 0
3 years ago
2 cannons fire projectiles upwards with the same velocity. The
dybincka [34]

Answer:

Explanation:

Given

Two projectile is fired vertically upward

One has 4 times the mass of other

When Projectile is fired their trajectory is independent of mass of object. Also if they launched with same speed then both achieved same maximum height in same time and will hit the ground at the same moment.

3 0
3 years ago
The frequency of a certain sound is 440 Mz. What is the wavelength of this sound when the temperature of the air is (a) 20°C; (b
Serggg [28]

Answer:

Explanation:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

λ

:

λ

=

v

f

Let's plug in our given values and see what we get!

λ

=

340

m

s

440

s

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0.773

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3 0
2 years ago
In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
A person drives 20 miles north and 45 miles east. What is the angle of the resultant straight to their destination?
Alex787 [66]
Let us take east and north as the positive x and y-axes should the motion be plotted in a cartesian plane. Thus, the x value is 45 miles and the y value is 20. The tangent of an angle is equal to the ratio of y to x.

   tanθ = y / x

Substituting,

   tanθ = 20/45 = 0.44

The value of θ is 23.96°.
7 0
3 years ago
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