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Softa [21]
3 years ago
3

A tuba creates a 4th harmonic of frequency 116.5 Hz. When the first valve is pushed, it opens an extra bit of tubing 0.721 m lon

g. What is the new frequency of the 4th harmonic ? (Hint : Find the original length .)
Physics
1 answer:
lions [1.4K]3 years ago
0 0

Answer:

The new frequency is:

f₄ = 93.54 Hz

Explanation:

The data given to us is:

No. of harmonic frequency = 4

f₄ = 116.5 Hz

We know that the harmonic frequency is given by the formula:

fₙ = (n · v)/4L

where

n = no. of harmonic frequency

v = speed of sound

L = Length of rope

Lets find the initial length of rope for 4rth harmonic frequency (n=4) and v=343 m/s

fₙ = (n · v)/4L

f₄ = 4v/4L

f₄ = v/L

L = v/f₄

L = 343/116.5

L= 2.944m

Find frequency for new length that is 2.944m + 0.721m = 3.667m

f₄ = v/l

f₄ = 343/3.667

f₄ = 93.54 Hz

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4 0
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Find the focal length of a lens of power-2.0D . What type of lens is this ?
olga_2 [115]

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3 0
3 years ago
A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness
Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

\sigma_h = \frac{Pd}{2t}

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

\sigma_l = \frac{Pd}{4t}

The letters have the same meaning as before.

Then he hoop stress would be,

\sigma_h = \frac{Pd}{2t}

\sigma_h = \frac{75 \times 8}{2\times 0.25}

\sigma_h = 1200psi

And the longitudinal stress would be

\sigma_l = \frac{Pd}{4t}

\sigma_l = \frac{75\times 8}{4\times 0.25}

\sigma_l = 600Psi

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

\tau_{max} = \frac{\sigma_h}{2}

\tau_{max} = \frac{1200}{2}

\tau_{max} = 600Psi

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0
3 years ago
The magnitude of the momentum of an object is the product of its mass m and speed v. If m = 3 kg and v = 1.5 m/s, what is the ma
Oliga [24]

Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

p = m v

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Velocity, v = 1.5 m/s

So, momentum of this object is given by :

p=3\ kg\times 1.5\ m/s

p = 4.5 kg-m/s

or

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

5 0
3 years ago
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