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ozzi
3 years ago
9

A glass tube (open at both ends) of length L is positioned near an audio speaker of frequency f = 770 Hz. For what values of L w

ill the tube resonate with the speaker? (Assume that the speed of sound in air is 343 m/s.) m (lowest possible value) m (second lowest possible value) m (third lowest possible value)
Physics
1 answer:
Schach [20]3 years ago
4 0

Answer:

- The lowest possible value of the length L1 = 0.45/2

L1 = 0.225m

- Second lowest possible value

L2 = 0.45m

- Third lowest possible value

L3 = 3(0.45)/2

L3 = 0.675m

Explanation:

Before calculating the length of the resonance tube, we need to know the wavelength produced by the wave.

Using the expression

v = Foλ

V is the velocity of wave

Fo is the resonance frequency

λ is the wavelength

From the formula

λ = v/Fo

λ = 343/770

λ = 0.45m

For an open pipe, the first resonant length L1 = λ/2

Second resonant length L2= λ

Third resonant length L3 = 3λ/2

The lowest possible value of the length L1 = 0.45/2

L1 = 0.225m

Second lowest possible value

L2 = 0.45m

Third lowest possible value

L3 = 3(0.45)/2

L3 = 0.675m

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3 years ago
The only force acting on a 3.1 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a
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33.5J

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The magnitude of vector V'= Vxi + Vyj + Vzk

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From Kinectic energy and work theorem.

Net work = Kinectic energy of the canister

ΔK= W

(Kf - Ki)= W

Where Kf= final Kinectic energy

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3 years ago
. A light bulb glows because of it’s resistance, and the brightness of the bulbincreases with the electrical power delivered to
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Complete Question

The complete question is shown in the first uploaded image

Answer:

a

When the both bulb are in the circuit  bulb B glows equally brighter to bulb A

This because the power delivered to the both bulb are equal

b

The bulb A on the right will glow brighter than the bulb A on the left due to the fact that the power supplied to bulb A on the right is higher than that gotten by bulb A on the left.

Explanation:

From the question we are been told that the two bulbs are identical

So their resistance denoted by R is the same

Considering the left circuit  where the two bulbs are connected in series which mean that the same current is passing through them

               R_A  =R_B =R

                i_A = i_B  =i

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