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Bond [772]
2 years ago
9

Consider a cylindrical pipe with both ends open and fundamental resonance frequency 150 Hz. If this particular pipe has a length

of L, what the wavelengths of three lowest modes produced by this pipe? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a. L,2L, 3L b. 2L, L, 2L/3 с. 2L/3, L, 2L d. L, L/2, L/3 e. 4L, 24, 4L/3 f. 4L, 4L/3, 4L/5 g. Neither of above.
Physics
1 answer:
Naya [18.7K]2 years ago
8 0

Answer:

answers, the correct one is b

Explanation:

For this problem we use the fact that since the ends are open at these points we have a maximum,

fundamental L =λ/2 λ = 2L

second harmonic L = λ λ = 2L / 2

third harmonic L = 3 λ/2 λ = 2L / 3

n harmonic lam = 2L / n

 with n an integer

therefore the length of the first three harmonics are: 2L, L, 2L / 3

when examining the different answers, the correct one is b

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What is difference between non uniform and uniform circular motion?
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Answer:

The object in a uniform motion covers same distances in an equal time period. Objects in a non-uniform motion cover dissimilar distances in an equal time period.

Explanation:

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3 years ago
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0
3 years ago
How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•
Vika [28.1K]

Answer:

Heat capacity, Q = 781.74 Joules

Explanation:

Given the following data;

Mass = 12g

Initial temperature = 28.3°C

Final temperature = 43.87°C

Specific heat capacity of water = 4.184J/g°C

To find the quantity of heat needed?

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 43.87 - 28.3

dt = 15.57°C

Substituting into the equation, we have;

Q = 12*4.184*15.57

Q = 781.74 Joules

7 0
3 years ago
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