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AURORKA [14]
2 years ago
9

A tuning fork is sounded above a resonating tube (one end closed), which resonates at a length of 0.20 m and again at 0.60 m. If

the tube length were extended further, at what point will the tuning fork again create a resonance condition?

Physics
1 answer:
Nina [5.8K]2 years ago
6 0

To solve this problem we will apply the concepts related to resonance. The velocity with which sound travels in any medium may be determined if the frequency and the wavelength are known. Since the pipe is closed at one end, that produces frequencies ratio 1:3:5, then

n_1 = \frac{V}{4L_1}

Third resonance occurs at 5n_2

n_2 = \frac{V}{4L_2}

At resonance 5n_2=n_1

5\frac{V}{4L_2} = \frac{V}{4L_1}

L_2 = 5L_1

L_2 = 5*0.2

L_2 = 1m

Therefore at 1m will the tuning fork again create a resonance condition

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An object is dropped from rest into a pit, and accelerates due to gravity at roughly 10 m/s2. It hits the ground in 5 seconds. A
vitfil [10]

Answer:

Second pit is 375 m deeper compared to first pit.

Explanation:

We have equation of motion s = ut + 0.5at²

First object hits the ground after 5 seconds,

          Initial velocity, u = 0 m/s

         Acceleration, a = 10 m/s²

         Time, t = 5 s

    Substituting,

                  s = ut + 0.5 at²

                 s = 0 x 5 + 0.5 x 10 x 5²

                    s = 125 m

           Depth of pit 1 = 125 m

Second object hits the ground after 10 seconds,

          Initial velocity, u = 0 m/s

         Acceleration, a = 10 m/s²

         Time, t = 10 s

    Substituting,

                  s = ut + 0.5 at²

                 s = 0 x 10 + 0.5 x 10 x 10²

                    s = 500 m

           Depth of pit 2 = 500 m

Difference in depths = 500 - 125 = 375 m

Second pit is 375 m deeper compared to first pit.

7 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Darya [45]

Answer:

y = -8.37 cm

Explanation:

As we know that the equation of SHM is given as

y = A cos(\omega t)

here we know that

\omega = \frac{2\pi}{T}

here we have

T = 0.66 s

now we have

\omega = \frac{2\pi}{0.66}

\omega = 3\pi

now we have

y = (8.8 cm) cos(3\pi t)

now at t = 2.3 s we have

y = (8.8 cm) cos(3\pi \times 2.3)

y = -8.37 cm

6 0
2 years ago
Read 2 more answers
A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme
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Given that,

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since,

cos ∅ = base/hypotenuse

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East component = 125 km

8 0
2 years ago
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It depends on where you live when you're not visiting Chicago. We need to know the distance of the trip.

Your average speed on the trip is . . .

(total distance in miles) / (3 hours)

miles per hour

5 0
2 years ago
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