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3 years ago

What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.

1 answer:

7 0

There is an indirect relationship between length and frequency. The longer the length the pipe has, the higher frequency it is. The shorter the length the pipe has, the lower frequency it is.

Explanation:

The four properties of the string that affect its frequency are length, diameter, tension, and density. These properties are described below: When the length of a string is changed, it will vibrate with a different frequency. Shorter strings have higher frequency and therefore higher pitch.

The longer the tube is the lower the pitch of the note that it can emit. When a tube is heated it expands and so is longer! As the gas in the tube gets warmer the molecules move faster, that means they can carry the vibrations of the sound wave more rapidly and so the pitch goes up.

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A ball is suspended by a lightweight string, as shown in the figure above.

Irina18 [472]

Answer

A. 1 and 2

Explanation

At point 1 we have the highest potential energy and the kinetic energy is zero.

At 2 the potential energy is minimum and the kinetic energy is maximum.

The law of conservation of energy says that energy cannot be created nor destroyed. So, the change in P.E = Change in K.E.

P.E = height × gravity × mass. The height referred here is the perpendicular height. Gravity and mass are constant in this case.

From the diagram it can be seen clearly that the vertical height from 2 to 1 is much greater than from 4 to 3.

This shows that the change in P.E is greater between 1 and 2 and so is kinetic energy.

6 0

2 years ago

Read 2 more answers

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

2 years ago

A blender takes in 180 J of electrical energy. Of the 180 J, 90 J are transformed into kinetic energy, 60 J are transformed into

a_sh-v [17]

Fa sho it's 30j edge said so

6 0

2 years ago

Read 2 more answers

An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of

Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

$3\times \frac{\lambda}{2} = 0.39$

$\lambda = 0.26 m$

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

$d = \frac{\lambda}{4}$

$d = \frac{0.26}{4} = 0.065 m = 6.5 cm$

So the distance will be 6.5 cm

3 0

2 years ago

The weight of a rock is 8.5 pounds. Does this statement include

Kaylis [27]

Answer:

quantitative

Explanation:

4 0

2 years ago