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Fiesta28 [93]
2 years ago
12

What are the three longest wavelengths for standing sound waves in a 121-cm-long tube that is (a) open at both ends and (b) open

at one end, closed at the other?
Physics
1 answer:
irina [24]2 years ago
3 0

Answer:A)For open pipe

1stπ=484cm

2nd wavelength=161.3cm

3rd is=96.8cm

B)For closed pipe

1st wavelength=242cm

2nd=121cm

3rd wavelength=80.7cm

Explanation:

An open pipe produces a standing waves whose wavelength is related to it's length(length of pipe) by nL/4=π ,where n represent odd integers 1,2,3...

So the first note called fundamental note or 1st Harmonic is L=π/ 4,121=π/ 4

π= 121×4=484cm

Third harmonic,L=3π /4

So π= 4l/3=4×121/3=161.3cm

The 3rd wavelength is

Fifth harmonic=

π =4l/5=4×121/5=96.8cm

For a pipe open at one end and closed at the other also called closed pipe

Fundamental note is

2L/n= π where n=1,2,3....ie integers

First wavelength

π=2L=2×121=242cm

Second wavelength

π=L=121cm

Third is

π=2L/3=2×121/3=80.7cm

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A neutron star has a mass of between 1.4-2.8 solar masses compressed to the size of:
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The correct answer is D. An average city

Explanation:

A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.

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2 years ago
Definition for compression and rarefraction
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Compression and rarefaction are two phenomenon occurs in longitudunal wave!

when there is denser particle gathering in that wave , there we called it compression and the rarer part of particles is rarefaction !
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3 years ago
Read 2 more answers
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

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2 years ago
Identify and define the four major forms of matter, explain how melting, freezing, boiling,
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There are four forms of matter: Solid, Liquid, gas and Plasma and matter undergoes various state changes termed as melting, freezing, boiling, evaporation, condensation, sublimation and deposition.

<u>Explanation:</u>

Solids

A matter that have a definite shape because of its closely packed molecular structure; are known as Solids. It can be identified as they have a definite shape and cannot flow or float without external forces are applied.

Liquids

These have a an internal molecular structure with comparatively more spaces with one another. Liquids have a property to flow and change shape according to the container it is taken.

Gases

The internal molecular structure of gases has the widest range of space among one another and thus they have a floating property because of least density.

Plasma

A complete ionized gas which has equal amount of positively and negatively charged ions. The best example of plasma is a plasma-ball.

<u>Phase transformation among the four forms of matter</u>

<u>Melting</u>

A matter changing from a solid phase to liquid phase is known as melting. Ex: Ice into water

<u>Freezing</u>

A matter changing from liquid to solid is known as freezing.

<u>Boiling</u>

When the liquid is heated to its boiling point, this gets transformed into the state of gas where liquid's pressure equals to the external pressure.

<u>Evaporation</u>

Once the liquid reached the temperature range above the boiling point ad starts converting into vapours or gaseous state.

<u>Condensation</u>

When the gases changes from the gaseous phase to liquid phase, this is called condensation.

<u>Sublimation</u>

The change of solid into gas is called as sublimation.

<u>Deposition</u>

Deposition refers the thermodynamic process where phase transition takes place as the gas solidifies without passing through the liquid phase. An example: the process of converting water vapour from frozen air directly into ice without initially becoming a liquid.

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3 years ago
A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
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