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blagie [28]
3 years ago
14

An open vertical tube has water in it. a tuning fork vibrates over its mouth. as the water level is lowered in the tube, a reson

ance is heard when the water level is 180 cm below the top of the tube, and again after the water level is 220 cm below the top of the tube a resonance is heard. what is the frequency of the tuning fork? the speed of sound in air is 343 m/s. answer in units
Physics
1 answer:
bagirrra123 [75]3 years ago
6 0
Since any first distanced1=λ(2N1−1)/4, whereN1= 1,2,3,· · ·will do, the first distance only cannot be used to know the wavelength.The space between the resonances is ½ a wavelength, soλ= 2 [d2−d1]= 2 [(35 cm)−(25 cm)]= 20 cm.
The frequency isF = vλ=343 m/s2 [(35 cm)−(25 cm)]= 1715 Hz.
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where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed
lesya [120]

Answer:

E_r(6)=4.35614\ MPa

Explanation:

\epsilon = Strain = 0.49

\sigma _0 = 3.1 MPa

At t = Time = 32 s \sigma = 0.41 MPa

\tau = Time-independent constant

Stress relation with time

\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)

at t = 32 s

0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s

The time independent constant is 16.0787 s

E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}

At t = 6

\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}

From the first equation

\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451

E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa

E_r(6)=4.35614\ MPa

6 0
2 years ago
when a person uses an iron to remove the wrinkles from a shirt, why does heat travel from the iron to the shirt?
artcher [175]
I think the answer to your question is that:
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3 years ago
An object moves from the position +16m to the position +43m in 12s. What us the total displacement
NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

                                     =43m -16m

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Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V =  (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12  ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

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The volume of gas in a flexible container at a depth of 10 m/33 ft will expand to double its original volume if taken to the surface.

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The amount of space that a substance or object occupies.

Volume is denoted with letter V

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Generally, a gas molecule is simply defined as a combination of numbers of atoms that are connected one to another.

here, 10 m/33 feet of sea water is taken to exerts a gauge pressure that is the same pressure as the atmosphere.

This means that we add one atmosphere/bar pressure for every 10 m/33 feet you descend.

The effect of this is that,

at 10 m/33 ft absolute pressure is two ata/bar

Adding another 10 m/33 ft to this depth to make it 20 m/66 ft, puts you under under three ata/bar of pressure and so on.

Descending to a depth it means that you have double the ata/bar pressure acting on you. If you decide to resurface, the pressure on the flexible container will reduce by a factor of two.

Since volume is inversely proportional to the pressure,

If the pressure reduces by a factor of 2, then the volume will increase by a factor of two.

Learn more about volume of gas here:

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