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wolverine [178]
2 years ago
11

A tube with a cap on one end, but open at the other end, produces a standing wave whose fundamental frequency is 130.8 Hz. The s

peed of sound is 343 m/s. a. If the cap is removed , what is the new fundamental frequency?
Physics
1 answer:
Angelina_Jolie [31]2 years ago
8 0

Answer:

A. 261.6 hz.

B. 0.656 m.

Explanation:

A.

When yhe tube is open at one end and closed at the other,

F1 = V/4*L

Where,

F1 = fundamental frequency

V = velocity

L = length of the tube

When the tube is open at both ends,

F'1 = V/2*L

Where

F'1 = the new fundamental frequency

Therefore,

V/2*L x V/4*L

F'1 = 2 * F1

= 2 * 130.8

= 261.6 hz.

B.

F1 = V/4*L

Or

F'1 = V/2*L

Given:

V = 343 m/s

F1 = 130.8

L = 343/(4 * 130.8)

= 0.656 m.

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Explanation:

From Stephan-Boltzmann Law Formula;

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Now, σ = 5.67 x 10^(-8)

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Read 2 more answers
An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init
elena-14-01-66 [18.8K]

Answer:

Approximately 71\%.

Explanation:

The formula for the kinetic energy \rm KE of an object is:

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where

  • m is the mass of that object, and
  • v is the speed of that object.

Important: Joule (\rm J) is the standard unit for energy. The formula for \rm KE requires two inputs: mass and speed. The standard unit of mass is \rm kg while the standard unit for speed is \rm m \cdot s^{-1}. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

m = 63\; \rm g = 0.063\; \rm kg.

Initial \rm KE of this arrow:

\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}.

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%.

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