Ask question

Ask question

All categories

3 years ago

5

A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson

ance. The speed of sound in air is 345 m/s. The common resonant frequency of the string and the pipe is closest to
Answer please with procedure

a. 630 Hz.
b. 450 Hz.
c. 700 Hz.
d. 540 Hz.
e. 360 Hz.

1 answer:

Andrews [41]3 years ago

7 0

Answer:

option D

Explanation:

given,

length of the pipe, L = 0.96 m

Speed of sound,v = 345 m/s

Resonating frequency when both the end is open

$f = \dfrac{nv}{2L}$

n is the Harmonic number

2nd overtone = 3rd harmonic

so, here n = 3

now,

$f = \dfrac{3\times 345}{2\times 0.96}$

f = 540 Hz

The common resonant frequency of the string and the pipe is closest to 540 Hz.

the correct answer is option D

You might be interested in

A5 kg picture is framed and hung on a wall by two strings at 45 degrees. What is the tension in each string?

Sholpan [36]

Answer:

Explanation:

On the east coast east to the brown area in the east west west of west east west corner east of the west west of the brown mountains west east west west

7 0

3 years ago

A 0.42 kg shuffleboard disk is initially at rest when a player uses a cue to increase its speed to 4.2 m/s at constant accelerat

MariettaO [177]

Answer

given,

mass of the shuffleboard disk = 0.42 kg

speed of the cue is increased to = 4.2 m/s

acceleration takes over 2 m then acceleration is zero.

the disk additionally slide to 12 m

final speed of disk = 0 m/s

a) increase in thermal energy

$\Delta E_t = \dfrac{1}{2}m(v_1^2-v_2^2)$

$\Delta E_t = \dfrac{1}{2}\times 0.42 \times (4.2^2-v_2^2)$

$\Delta E_t = 3.704\ J$

b) $\Delta E_t = F_f.d$

F_f is the frictional force

$3.704= 12.F$

$F_f = 0.308\ N$

increase in thermal energy for entire movement of 14 m

$\Delta E_t = 0.308\times 14$

$\Delta E_t =4.312 \ J$

c) Work done on the disk by the cue

$W = \Delta KE + \Delta E_{t}$

$W = \dfrac{1}{2}\times 0.42 \times (4.2^2-0^2)+ F_f \times d$

$W = 3.704+ 0.308 \times 2$

W = 3.704 + 0.616

W = 4.32 J

3 0

4 years ago

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force F exerted by the

Sever21 [200]

Answer:

a

$a = 12.32 \ m/s^2$

b

$F = 1017.52 \ N$

Explanation:

From the question we are told that

The length of the Chinook salmon is $l = 1.50 \ m$

The mass of the Chinook salmon is $m = 46.0 \ kg$

The upward velocity in water is $u = 3.00 \ m/s$

The upward velocity in air is $v = 5.80 \ m/s$

Generally from kinematic equations

$v^2 = u^2 + 2as$

=> $5.80^2 = 3.00^2 + 2 * a * [ [\frac{2}{3} * l ]$

=> $5.80^2 = 3.00^2 + 2 * a * [ [\frac{2}{3} * 1.5 ]$

=> $5.80^2 = 3.00^2 + 2 * a * [1 ]$

=> $a = 12.32 \ m/s^2$

Generally magnitude of the force F during this interval in N is mathematically represented as

$F = m (a + g )$

=> $F = 46.0 (12.32 + 9.8 )$

=> $F = 1017.52 \ N$

5 0

4 years ago

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

Leya [2.2K]

<u>Solution and Explanation:</u>

The following calculation is made in order to find out the area and the final velocity vector.

Using the given information and the data in the question,

m1u1 + m2u2 = (m1 + m2) multiply with v

40000 multiply with ( -20i ) + 60000 multiply with ( 10j ) = 100000 multiply with v

Therefore, v = -8i + 6j

That is |v| = 10 knots towards the 36.86 degree north of the west

6 0

3 years ago

Explain how the gravitational potential energy of an object can be changed.

inna [77]

Answer:

Change in Potential energy, pe = Final potential energy - Initial potential energy

Explanation:

∆p.e = Final p.e - Initial p.e

From the equation ∆p.e = (mgh) final - (mgh) initial

4 0

3 years ago