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3 years ago

Identify the vibrating media in three different types of musical instruments.

Physics

1 answer:

mart [117]3 years ago

3 0

Explanation:

Hole. Hole. Different notes can be played on the flute by blocking holes. ...

Drum skin. Drum skin. Hitting the bongo drum makes its tight elastic skin vibrate.

String. String. ...

Sound. hole. ...

Bow. Bow.

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What does a magnetic field look like <br>

Lelechka [254]

Answer:Magnetic fields are invisible, at least usually. But scientists from NASA's Space Sciences Laboratory have made them visible as "animated photographs," using sound-controlled CGI and 3D compositing.

Explanation:

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4 years ago

Air is compressed adiabatically in a piston-cylinder assembly from 1 bar, 300 K to 10 bar, 600 K. The air can be modeled as an i

julia-pushkina [17]

Answer:

1) the entropy generated is Δs= 0.0363 kJ/kg K

2) the minimum theoretical work is w piston = 201.219 kJ/kg

Explanation:

1) From the second law of thermodynamics applied to an ideal gas

ΔS = Cp* ln ( T₂/T₁) - R ln (P₂/P₁)

and also

k= Cp/Cv , Cp-Cv=R → Cp*( 1-1/k) = R → Cp= R/(1-1/k)= k*R/(k-1)

ΔS = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

where R= ideal gas constant , k= adiabatic coefficient of air = 1.4

replacing values (k=1.4)

ΔS = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

ΔS = 1.4/(1.4-1) *8.314 J/mol K * ln( 600K/300K) - 8.314 J/mol K * ln (10 bar/ 1bar)

ΔS = 1.026 J/ mol K

per mass

Δs = ΔS / M

where M= molecular weight of air

Δs = 1.026 J/ mol K / 28.84 gr/mol = 0.0363 J/gr K = 0.0363 kJ/kg K

2) The minimum theoretical work input is carried out under a reversible process. from the second law of thermodynamics

ΔS =∫dQ/T =0 since Q=0→dQ=0

then

0 = k*R/(k-1)* ln ( T₂/T₁) - R ln (P₂/P₁)

T₂/T₁ = (P₂/P₁)^[(k-1)/k]

T₂ = T₁ * (P₂/P₁)^[(k-1)/k]

replacing values

T₂ = 300K * ( 10 bar/1 bar)^[0.4/1.4] = 579.2 K

then from the first law of thermodynamics

ΔU= Q - Wgas = Q + Wpiston ,

where ΔU= variation of internal energy , Wgas = work done by the gas to the piston , Wpiston = work done by the piston to the gas

since Q=0

Wpiston = ΔU

for an ideal gas

ΔU= n*Cv*(T final - T initial)

and also

k= Cp/Cv , Cp-Cv=R → Cv*( k-1) = R → Cv= R/(k-1)

then

ΔU= n*R/(k-1)*(T₂ - T₁)

W piston = ΔU = n*R/(k-1)*(T₂ - T₁)

the work per kilogram of air will be

w piston = W piston / m = n/m*R/(k-1)*(T₂ - T₁) = (1/M*) R/(k-1)*(T₂ - T₁) ,

replacing values

w piston = (1/M*) R/(k-1)*(T₂ - T₁) = 1/ (28.84 gr/mol)* 8.314 J/mol K /0.4 * ( 579.2 K - 300 K) = 201.219 J/gr = 201.219 kJ/kg

6 0

3 years ago

Draw an accurate, labelled diagram of a thermometer

olga nikolaevna [1]

You wear very specific on what type of thermometer he wanted and what specific information you wanted so I just made one like this, hope it helps anyway! -Sam

7 0

3 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

olchik [2.2K]

Answer:

0.833 N

Explanation:

Formula for Kinetic Energy $E_k = \frac{mv^2}{2}$

Formula for Potential Energy $E_p = mgy$

First we need to find the vertical distance between the maximum-angle position and the pendulum lowest point:

Using the swinging point as the reference, the vertical distance from the maximum-angle (34 degree) position to the swinging point is:

$L * cos(34^o) = 1.2cos(34^o) = 1.2*0.83 = 0.995 \approx 1 m$

At the lowest position, pendulum is at string length to the swinging point, which is 1.2 m. Therefore, the vertical distance between the maximum-angle position and the pendulum lowest point would be

y = 1.2 - 1 = 0.2 m.

As the pendulum is traveling from the maximum-angle position to the lowest point position, its potential energy would be converted to the kinetic energy.

By law of energy conservation:

$E_k = E_p$